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In the figure above, points P and Q lie on the circle with center O. What is the value of s?

A. B. C. D.

E.

First draw a perpendicular from the x-axis to the point P. Lets call the point on the x-axis N. Now we have a right

angle triangle . We are given the co-ordinates of P as . i.e. and . Also we

know angle . If you notice this is a triangle with sides and

angle .

Similarly draw another perpendicular from the x-axis to point Q. Lets call the point on the x-axis R. Now we have a

right angle triangle . We know angle .

Now angle NOP + angle POQ (= 90 -- given) + angle QOR = 180

If you notice is also a triangle with sides and angle

The side opposite to this angle is OR = 1.

Since OP and OQ are perpendicular to each other and since OP=OQ=2(radius) , the coordinates of Q are same as P but in REVERSE order, keep the sign +/- based on Quandrant it lies. in this case Q(1,\sqrt{3)}.

Yes you are right.

Point Q in I quadrant ;

Point P in II quadrant ;

Point T in III quadrant (OT perpendicular to OP);

Point R in IV quadrant (OR perpendicular to OQ).

The method I suggested above is very intuitive and useful. I have been thinking of how to present it so that it doesn't look ominous. Let me try to present it in another way:

Imagine a clock face. Think of the minute hand on 10. The length of the minute hand is 2 cm. Its distance from the vertical and horizontal axis is shown in the diagram below. Let's say the minute hand moved to 1. Can you say something about the length of the dotted black and blue lines?

Attachment:

Ques1.jpg [ 7.87 KiB | Viewed 3056 times ]

Isn't it apparent that when the minute hand moved by 90 degrees, the green line became the black line and the red line became the blue line. So can I say that the black line cm and blue line = 1 cm

So if I imagine xy axis lying at the center of the clock, isn't it exactly like your question? The x and y co-ordinates just got inter changed. Of course, according to the quadrants, the signs of x and y coordinates will vary. If you did get it, tell me what are the x and y coordinates when the minute hand goes to 4 and when it goes to 7?

There's a simple reason and misconception as to why people find this question confusing:--> the 90 degree angle between P and Q seems like it is cut perfectly right down the centre >> to make two 45 degree angles.--> that's why your brain instinctively thinks the answer is r = sq(3), since the distance seems the same from the left and to the right of the origin!

However, that's the trick to this question and that's why it is CRUCIAL that you train yourself to be highly critical and weary of accepting the simple answer!

Upon closer inspection of the question:

P(-sq(3),1)Which means the triangle beneath point P has:height = 1base = sq(3)meaning >> radius = 2 (30-60-90 triangle)

This follows the ideal case of the 30-60-90 triangle, which means that opposite to the height of 1, there is an angle of 30 degrees from the x-axis to point P. This means that from the y-axis to point P, there is an angle of 60 degrees (to make a 90 degrees total from x-axis to y-axis). Therefore on the right side, since the figure shows there is a 90 degree angle between point P and Q, 60 degrees have been taken up from the left of the y-axis, which leaves 30 degrees on the right of the y-axis. Furthermore, this leaves 60 degrees from Point Q to the positive x-axis. Therefore, making the exact same 30-60-90 triangle once again, since the radius is the same (the two points lie on the semi-circle). The 60 degree angle is now where the 30 degree angle was on the other triangle, which makes the height now sq(3) and base now 1 >> therefore, base r = 1!

**Have a look at the attachment and then it should make perfect sense!

Attachments:

Line OP => y=mx => 1 =

=> slope of OP = Since OP is perpendicular to OQ => m1*m2 = -1 => slope of OQ = equation of line OQ => y =

since s and t lie on this line t=

also radius of circle = = 2 =

=> => s=1

Using Similar triangles.In triangle APO , angle APO + AOP = 90 ---------------------1In triangle BOQ , angle BOQ + BQO= 90 ---------------------2

Also SINCE angle POQ = 90 as given.

angle AOP + angle BOQ = 90 ------------------3

Using equations 1 and 3 we get Angle APO = Angle BQO -----------4

using equation 2 ,3, 4we get Angle APO = Angle BQO and AOP = OQB

Since hypotenuse is common => both the triangle are congruent

=> base of triangle APO = height of triangle AQO and vise versa.

Thus s = y co-ordinate of P = 1

On the coordinate plane (6, 2) and (0, 6) are the endpoints of the diagonal of a square. What is the distance between point (0, 0) and the closest vertex of the square?

(A) (B) 1(C) (D) (E)

Distance between two points

Given coordinates of two points, distance D between two points is given by:

(where is the difference between the x-coordinates and is the difference between the y-coordinates of the points)

As you can see, the distance formula on the plane is derived from the Pythagorean theorem.

Above formula can be written in the following way for given two points and :

Vertical and horizontal linesIf the line segment is exactly vertical or horizontal, the formula above will still work fine, but there is an easier way. For a horizontal line, its length is the difference between the x-coordinates. For a vertical line its length is the difference between the y-coordinates.

Distance between the point A (x,y) and the originAs the one point is origin with coordinate O (0,0) the formula can be simplified to:

Example #1Q: Find the distance between the point A (3,-1) and B (-1,2)Solution: Substituting values in the equation we'll get

Midpoint of a Line Segment

A line segment on the coordinate plane is defined by two endpoints whose coordinates are known. The midpoint of this line is exactly halfway between these endpoints and it's location can be found using the Midpoint Theorem, which states:• The x-coordinate of the midpoint is the average of the x-coordinates of the two endpoints.• Likewise, the y-coordinate is the average of the y-coordinates of the endpoints.

Coordinates of the midpoint of the line segment AB, ( and )

are and

Given endpoints of diagonal of a square: B(0,6) and D(6,2). Let other vertices be A (closest to the origin, left bottom vertex) and C (farthest to the origin).

Length of the diagonal would be:

Coordinates of the midpoint M of the diagonal would be: .

Slope of the line segment AM*Slope of the line segment BD=-1 (as they are perpendicular to each other) --

> --> -->

Distance between the unknown vertices to the midpoint is half the diagonal:

--> --> --> or --

> or

Hence the coordinates of the point A(1,1) and point C (5,7). Closest to the origin is A. Distance Answer: C.

You can actually answer the question without calculating any distances at all (except at the very end), provided you have a good understanding of the meaning of perpendicular slopes. The background: perpendicular slopes are negative reciprocals. If one line has a slope of a/b, a perpendicular line has a slope of -b/a. Now, if a line has a slope of a/b, that is just the ratio of the 'vertical change' to the 'horizontal change' of the line as you move to the right; if a line has a slope of a/b, that means if you move b units to the right, the line will rise by a units. On a perpendicular line, to travel the same distance, you'd *reverse* the horizontal and vertical changes (because the slopes are reciprocals) and also reverse the direction of one of the two changes (because of the negative sign). That is, on a perpendicular line, we'd travel the same distance if we moved right by a units and moved down by b units, or if we moved left by a units and up by b units.

So, in this question, we have a square. The diagonals of a square are perpendicular, and meet at their midpoints. The midpoint of the diagonals in this question is (3,4). Moving from (0,6) to (3,4), we go right by 3 units and down by 2 units. On the perpendicular diagonal, starting from the midpoint, to travel the same distance we need to either go right by 2 units and up by 3 units (to get to (5,7)), or left by 2 units and down by 3 units (to get to (1,1)). So the closest vertex to the origin is (1,1) and the distance is sqrt(2).

You don't need to use the distance formula here; instead we can use slopes. I posted this solution a while back to another forum:

We have two endpoints of a diagonal of a square. We can use the following:

-the midpoint of one diagonal is the midpoint of the other diagonal; -the diagonals are perpendicular.

If (0,6) and (6,2) are endpoints, (3,4) is the midpoint.

From (0,6) to (3,4), we go right 3 and down 2; that is, we increase x by 3 and decrease y by 2: the slope is -2/3. Consider the perpendicular diagonal- its slope is the negative reciprocal, i.e. 3/2. From (3,4), on a perpendicular line, to find a point the same distance from (3,4) as (0,6) is, we can decrease x by 2 and decrease y by 3, or we can increase x by 2 and increase y by 3. The endpoints of the other diagonal are (1,1) and (5,7).

The distance from the origin to (1,1) is sqrt(2).

The mid point between (6,2) and (0,6) is (3,4) You know that the vertexes of the diagonal mentioned has distance of 3 in the x coordinate and distance of 2 in the y coordinate. Since both diagonals must be perpendicular, using the perpendicular property (inverse of slope), it must be that those distances are reversed.

This means that from point (3,4), the new vertexes will have distance of 3 in the y coordinate and distance of 2 in the x coordinate.

So two other vertexes will be at: (3-2, 4-3) = (1,1) and (3+2, 4+3) = (5,7)

The closer to the (0,0) is (1,1) Distance between (0,0) and (1,1) = sqrt(2)

I use a method that is out of scope of the GMAT... I prefer to use the properties of perpendicular vectors...

First of all, I search the mid point I that is the center of gravity of the square.

I((6+0)/2 , (2+6)/2) <=> I(3,4)

Lets call: o A : the vertice at (6,2) o C : the vertice at (0,6) o B : the vertice "above" the line AC o D : the vertice "below" the line AC

Then, I calculate the vectors that we can have: o Vector(IA) = (6-3,2-4) = (3,-2) o Vector(IC) = (0-3,6-4) = (-3,2)

Then, as Vector(IA) is perpendicular to Vector(IB) and Vector(ID) is perpendicular to Vector(IC), we have the rules of perpendicular vectors: o Vector(IB) = (-y(IA) , x(IA)) = (2,3) o Vector(ID) = (-y(IC) , x(IC)) = (-2,-3)

Then, we rebuild the coordonate of B and D: o B(3+2 , 4+3) <=> B(5 , 7) o D(3-2 , 4-3) <=> D(1 , 1)

So, we have o A(6,2) o B(5,7) o C(0,6) o D(1,1)

Thus, OD is the shortest distance and OD = sqrt(2)

To me, we should: > Draw a quick XY Plane to have a better idea of what is going on > Know the middle point of A(6, 2) and B(0, 6) > Determine the nearest vertex to 0(0,0) by looking in which cadran this middle point is > Create the equation of circle on which all vertice lie on, centered so at the middle point of AB > Create the equation of the line AB > Create the equation of the line perpendicular to AB and passing by the middle point > Calculate the coordonate of the nearest vertex to 0 by using the equation of the circle and the perpendicular line to AB

Another way can be to use an approach with vectors

Beauty of Graphs - Example 2+ Stunning Symmetry of Squares

If AC is the given diagonal with co-ordinates (0, 6) and (6, 2), we see it is a sloping line. The center point of the line is (3, 4) obtained by averaging x and y co-ordinates as given: (0 + 6)/2 = 3 and (6 + 2)/2 = 4. Think of a horizontal line PQ whose y co-ordinates were shifted by 2 units up and 2 units down to obtain AC

Attachment:

Graph1.jpg [ 9.53 KiB | Viewed 1422 times ]

Diagonal BD will be perpendicular to AC and will pass through (3, 4). It will have a corresponding line RS whose x co-ordinates will be shifted by 2 units right and 2 units left to obtain the co-ordinates of BD. (This is so because y co-ordinates of PQ were shifted by 2 units to obtain AC and PQ is perpendicular to RS)

Attachment:

Graph2.jpg [ 12.98 KiB | Viewed 1422 times ]

The closest co-ordinate to (0, 0) is (1, 1) and its distance from center is

A certain square is to be drawn on a coordinate plane. One of the vertices must be on the origin, and the square is to have an area of 100. If all coordinates of the vertices must be integers, how many different ways can this square be drawn?

(A) 4(B) 6(C) 8(D) 10(E) 12

Best way is to find for one quadrant and multiply by 4.

6,8 satisfy the point for the vertex of the square.

=> 8,6 will also satisfy => 2 squares per quadrant ---> if you are confused why this is true then draw the x-y axis and try to visualize what happens when x is replaced with y.=> 4*2 = 8 squares

Now 10,0 also satisfy the point or the vertex.but when we will replace x with y the same square is generated=> 10,0 and 0,10 are part of same squares.=> 1 per quadrant=> 4*1 = 4 squares

total = 4+8 = 12 hence E

Let the origin be O and one of the vertices be A. Now, we are told that length of OA must be 10 (area to be 100). So if

the coordinates of A is (x, y) then we would have (distance from the origin to the point A(x, y) can be

found by the formula )

Now, has several integer solutions for and , so several positions of vertex A, note that when vertex A has integer coordinates other vertices also have integer coordinates. For example imagine the case when square rests on X-axis to the right of Y-axis, then the vertices are: A(10,0), (10,10), (0,10) and (0,0).

Also you can notice that 100=6^2+8^2 and 100=0^2+10^2, so can tale 7 values: -10, -8, -6, 0, 6, 8, 10. For and , can take only 1 value 0, but for other values of , can take two values positive or negative. For example: when then or . This gives us 1+1+5*2=12 coordinates of point A:

and , imagine this one to be the square which rests on X-axis and to get the other options rotate OA anticlockwise to get all possible cases;

and ; and ; and ;

and ; and ;

and ; and ; and ;

and ; and ; and .

As for your question I doubt that this is a realistic GMAT question. Though if you find that # of squares should be multiple of 4 you'll be left with A, C and E choices right away. Next, you can also rule out A as at least 2 squares per quadrant can be easily found and then make an educated guess for E thus "solving" in less than 2 minutes. Refer for complete solution to the posts above.

What is the least possible distance between a point on the circle x^2 + y^2 = 1 and a point on the line y = 3/4*x - 3?

A) 1.4B) sqrt (2)C) 1.7D) sqrt (3)E) 2.0

First note that min distance from the circle to the line would be: length of perpendicular from the origin to the line (as the circle is centered at the origin) - the radius of a circle (which is 1).

Now we can do this by finding the equation of a line perpendicular to given line (we know it should cross origin and cross given line, so we can write the formula of it), then find the cross point of these lines and then the distance between the origin and this point. But it's very lengthy way.

There is another, shorter one. Though I've never seen any GMAT question requiring the formula used in it.

We know the formula to calculate the distance between two points and

: BUT there is a formula to calculate the distance between the point (in our case origin) and the line:

DISTANCE BETWEEN THE LINE AND POINT:

Line: , point

DISTANCE BETWEEN THE LINE AND ORIGIN:

As origin is -->

So in our case it would be:

So the shortest distance would be: Answer: A.

P.S. Also note that when we have , we have circle (as we have and ), it's centred at the origin (as coefficients of and are ) and the radius of that circle .

Use the formula

D = | Am+Bn+C|/ SQRT(A^2 + B^2) where Ax+By+C = 0put (m,n) =0,0 = center of circlewe get D = 12/5 thus required distance is D-1 = 12/5 -1 = 7/5 = 1.4we don't require points

I have learnt this in my school time that distance from one point to a line is what stated above.

Actually its same like finding the line which originates from the distant point and then intersect at the required line, and then we find the point of intersection and then using the formula we calculate distance between the two points.

But even if u dont rem the formula what u can do is....

suppose line is ax+by-c=0 now when y=0 , x=c/a and when x=0 , y = c/b

area of triangle formed by these 2 points and center (0,0) is 1/2 * c/a * c/b = c^2/2ab

now this is equal to 1/2 * D1 * D2 , where D1 is distance between points on x and y coordinates of the line which is sqrt [ (c/a)^2 + (b/a) ^2 ] and D2 is the required perpendicular distance on the line.

Equate them and u will get the ans

Minimum distance from the circle to the line would be:Length of perpendicular from the origin to the line (as the circle is centered at the origin) - The radius of a circle (which is 1).

(0,0) is not the point on the circle, it's the center of the circle with radius 1 (circle is centered at the origin).

2.4 is the distance from the line to the origin (the center of the circle), so we should subtract the length of the radius to get the distance from the line to the circle.

ANOTHER SOLUTION:

In an x-y Cartesian coordinate system, the circle with center (a, b) and radius r is the set of all points (x, y) such that:

If the circle is centered at the origin (0, 0), then the equation simplifies to:

So, the circle represented by the equation is centered at the origin and has the radius of .

Then note that min distance from the circle to the line would be: length of perpendicular from the origin to the line (as the circle is centered at the origin) - the radius of a circle (which is 1).

So we should find the length of perpendicular, or the height of the right triangle formed by the X and Y axis and the

line .

The legs would be the value of x for y=0 (x intercept) --> y=0, x=4 --> . and the value of y for x=0 (y

intercept) --> x=0, y=-3 --> .

So we have the right triangle with legs 4 and 3 and hypotenuse 5. What is the height of this triangle (perpendicular from right angle to the hypotenuse)? As perpendicular to the hypotenuse will always divide the triangle into two

triangles with the same properties as the original triangle: --> --

> .

A rather nice question and I think Bunuel has already explained sufficient alternative solutions. I just want to add a method of finding the altitude of a right triangle here that I think is particularly neat (discussed in Veritas Geometry book)

Attachment:

Ques2.jpg [ 8.1 KiB | Viewed 3532 times ]

I want to find x here since (x - 1) will be the minimum distance from the circle to the line (as explained above)

Area of the given right triangle = (1/2)*3*4 (3 is base and 4 is altitude)= (1/2)*5*x (5 is base and x is altitude)So x = 2.4Finding the area of the original triangle in two different ways and equating it will help you find the altitude.

Triangle PQR is right angled at Q. QT is perpendicular to PR, where T is the point on PR. If PQ = 3cm and QR = 6cm. Find QT

1. 3/52. 6\sqrt{5}3. 6/\sqrt{5}4. 45. None

Look at the diagram:

Attachment:

untitled.PNG [ 5.73 KiB | Viewed 1538 times ]

You can solve this question using the fact that: perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle, which means that all 3 triangles PQT, QRT and PRQ are similar. Now, as in similar triangles, corresponding sides are all in the same proportion then QT/QR=PQ/PR, as

then --> .

Or you can equate the area: --> the same here: --

> .

Answer: C.

Hi Bunuel, I have a small doubt regarding similarity of triangles. Could you please explain the concept in little more detail or point me to a place where i can learn the concept.

eg lets say same diagram except now QT = 4 , PT = 3 .....we have to find TR = ?

Could you please explain this ?

If QT=4 and PT=3 --> corresponding legs in QTR and in PTQ are in the same ratio: TR/QT=QT/PT --> TR=16/3.

You can use area equation approach if you are not comfortable with similarity.

In the xy-plane, point (r, s) lies on a circle with center at the origin. What is the value of

?(1) The circle has radius 2.(2) The point (\sqrt{ }, -\sqrt{ }) lies on the circle.

THEORY:In an x-y Cartesian coordinate system, the circle with center (a, b) and radius r is the set of all points (x, y) such that:

This equation of the circle follows from the Pythagorean theorem applied to any point on the circle: as shown in the diagram above, the radius is the hypotenuse of a right-angled triangle whose other sides are of length x-a and y-b.

If the circle is centered at the origin (0, 0), then the equation simplifies to:

BACK TO THE ORIGINAL QUESTION:

In the xy-plane, point (r, s) lies on a circle with center at the origin. What is the value of ?

Now, as then the question asks about the value of radius^2.

(1) The circle has radius 2 --> radius^2=4. Sufficient.

(2) The point lies on the circle --> substitute x and y coordinates of a point in --

> . Sufficient.Answer: D.

A certain circular area has its center at point P and has radius 4, and points X and Y lie in the same plane as the circular area. Does point Y lie outside the circular area?

(1) The distance between point P and point X is 4.5.(2) The distance between point X and point Y is 9.

(1)+(2) From (1) X is outside the circle (as radius is 4). Closest point of this circle to the point X equals to 4.5-4=0.5 and the furthest equals to 0.5+8(diameter)=8.5. As XY=9, then Y must be outside the circle. Sufficient.

Answer: C.

Circle C and line K lie in the XY plane. If circle C is centered at the orgin and has a radius 1, does line K intersect circle C?(1) The X-Intercept of line k is greater than 1(2) The slope of line k is -1/10

The best way to solve this question would be to visualize/draw it.

No matter what the slope is, it’s possible for line not to cross the circle as the x intercept can be + infinite.

(1) The X-Intercept of line k is greater than 1 --> Just says that X-intercept is to the right of the circle. Not sufficient(2) The slope of line k is -1/10 --> Just says that slope is negative -1/10 --> line is just going down. Not sufficient.

(1)+(2) As we don't know exact intercept of line and X-axis we can not determine whether line intersects the circle or not. Not sufficient.

To elaborate more: we can draw infinitely many parallel lines with X-intercept more than 1 and slope -1/10, some will intersect the circle (for example line with X-intercept 1.1) and some not (for example line with X-intercept 1,000,000). Check the image below for two possible scenarios: blue line (with the slope of -1/10 and the x-Intercept greater than 1) intersects the circle while the red line (also with the slope of -1/10 and the x-Intercept greater than 1) does not.

Attachment:

graph.png [ 15.11 KiB | Viewed 561 times ]

Answer: E.Look at the diagrams.

Statement 1: If x intercept > 1, the line can be any of the following (and can be drawn in many more ways)

Attachment:

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Statement 2: If slope = -1/10, the line can be drawn in any of the following ways. (and many more)

Attachment:

Ques1.jpg [ 14.73 KiB | Viewed 1153 times ]

Using both together: Look at the diagrams above. Both have x intercepts greater than 1 and slope = -1/10. In one case, it will intersect the circle, in the other case, it will not. SO both statements together are not sufficient. Answer (E).

Draw some diagrams to figure it out. Even using both statements, you get two cases - one in which K intersects C and another in which it doesn't.Attachment:

Ques4.jpg [ 9.68 KiB | View

Line P is described by the equation 3x + 2y - 4 = 0. Which of the following lines is parallel to P and has the same x-intercept as the line 5x - y + 10 = 0?Choices

A)6x = -4y + 8

B)-y = (3/2)x + 5C)3y = (- 9/2)x - 9D)4x + 6y +8 = 0E)6x - 4y = -12

Eqn of line parallel to 3x + 2y - 4 = 0 will be of the form 3x + 2y + k = 0 where k is a constant. From the 5 options only options A and C satisfy the above. So eliminate the others from this condition itself.Also, x intercept of line 5x - y + 10 = 0 is -2. To find x intercept, substitute y = 0. Hence we get x intercept = -2.From A, on simplifying we get -> 3x + 2y - 8 = 0. x intercept = 8/3. Not what we want, so eliminate.From C, we get -> 3x + 2y + 6 = 0. Thus x intercept = -2. Bingo!

Line parallel to P will have same slope as P. Also it has same x-intercept as line 5x-y+10=0 i.e x=-2Slope of line P 2y=-3x+4 => slope = -3/2

Look for an answer with x-intercept =-2 and slope =-3/2A) x-intercept = 4/3. EliminateB) x-intercept = -10/3 EliminateC) x-intercept = -2. Also, slope = -3/2. We have the answer.OA C.

A right triangle ABC has to be constructed in the xy-plane so that the right angle is at A and AB is parallel to x axis. The coordinates of A, B and C are to satisfy the inequalities -3 ≤ x ≤ 5 and 2 ≤ y ≤11 and x and y are integers. The number of different triangles that can be constructed with these properties are?A. 100 B. 6480C. 2320D. 1500E. 9000

We have the rectangle with dimensions 9*10 (9 horizontal dots and 10 vertical). AB is parallel to x-axis and AC is parallel to y-axis.

Choose the (x,y) coordinates for vertex A: 9C1*10C1=90;Choose the x coordinate for vertex B (as y coordinate is fixed by A): 8C1, (9-1=8 as 1 horizontal dot is already occupied by A);Choose the y coordinate for vertex C (as x coordinate is fixed by A): 9C1, (10-1=9 as 1 vertical dot is already occupied by A).

9C1*10C1*8C1*9C1=6480.

For A, there are 90 possibilities on the plane.For each location of A, as the triangle has to have same properties, there are 8 possibilities on the x axis and 9 possibilities on y axis.So the answer is 90*8*9 = 6480

A right triangle ABC has to be constructed in the xy-plane so that the right angle is at A and AB is parallel to x axis. The coordiantes of A, B and C are to satisfy the inequalities -1 ≤ x ≤ 7 and 1 ≤ y ≤ 7. The numbers of different triangles that can be constructed with these properties are?

A63

B336C567D3024E5040

Question should specify that x- and y-coordinates of A, B, and C are to be integers.

We have the rectangle with dimensions 9*7 (9 horizontal dots and 7 vertical). AC is parallel to y-axis and AB is parallel to x-axis.

Choose the (x,y) coordinates for vertex A (right angle): 9C1*7C1;Choose the x coordinate for vertex B (as y coordinate is fixed by A): 8C1, (9-1=8 as 1 horizontal dot is already occupied by A);Choose the y coordinate for vertex C (as x coordinate is fixed by A): 6C1, (7-1=6 as 1 vertical dot is already occupied by A).

9C1*7C1*8C1*6C1=3024.Answer: D.

Note: A and C share the same x coordinate. A and B share the same y coordinate.

The x coordinates have 9 options (x = -1,0,1,2,3,4,5,6,7) pick 2: 1 for point B, and 1 for point A and C. 9P2The y coordinates have 7 options (y= 1,2,3,4,5,6,7) pick 2: 1 for point A and B, and 1 for point C. 7P2Number of triangles w/ A as the right angle = 9P2 * 7P2 = 3024

Right triangle PQR is to be constructed in the xy-plane so that the right angle is at P and PR is parallel to the x-axis. The x and Y coordinates of P,Q and R are to be integers that satisfy the inequalitites -4≤ X≤ 5 and 6≤ y≤ 16. How many different triangles with these properties could be constructed?

A)110B)1100C)9900D)10000E)12100

Thinking about this on the coordinate plane confuses me more than just thinking about it as a combination problem.P (A,B)R (C,B)Q (A,D)

So you need four values to solve the problem.

2 x-values {A and C}2 y-values {B and D}

When you approach the question this way the orientation of the triangle is irrelevant. Because by using the same x-value for point P and point Q along with the same y-value for point point P and point R. The right angle is at P and PR is parallel to the x-axis.

So I need 2 x values from -4≤ X≤ 5.

Pick the first one from 10 choicesPick the second one from 9 choices

10*9 = 90 and I need 2 y values from 6≤ y≤ 16Pick the first one from 11 choicesPick the second one from 10 choices

11*10 = 110So combine the 2 and we have 9900 (110*90) possible triangles.

I realized my error when I started to make a table of the possible coordinates for point P using the triangle in the upper right of the four choices

With point P and (4,6) I realized that I only have one possible length for PR*If P is (4,6)andthe triangle has the orientation in the upper right of my image.*Then R must be (5,6)

First a picture to summarize the discussion so far.

Attachment:

Triangles.JPG [ 8.26 KiB | Viewed 2208 times ]

or

We have the rectangle with dimensions 10*11 (10 horizontal dots and 11 vertical). PQ is parallel to y-axis and PR is parallel to x-axis.

Choose the (x,y) coordinates for vertex P (right angle): 10C1*11C1;Choose the x coordinate for vertex R (as y coordinate is fixed by A): 9C1, (10-1=9 as 1 horizontal dot is already occupied by A);

Choose the y coordinate for vertex Q (as x coordinate is fixed by A): 10C1, (11-1=10 as 1 vertical dot is already occupied by A).

10C1*11C1*9C1*10C1=9900.

Answer: C.

X can have 10 values & Y can 11 values.Just choose a random Point P. Now, P has both x & y values. So, P can have 11*10=110 values.PR is vertical. So, the y coordinates of P & R would be equal. Since PQR is a right triangle, QR would be perpendicular to the X axis & eventually parallel to the Y axis. So, the X coordinates of P & Q would be the same.

Since, we've fixed P. Q has the same value for x as that of P. So, there will be 10 (since 1 value of y is already taken by P) possible options for y coordinates for Q. Similarly, as P & R have the same value for y coordinates, R will have 9 possible values for x.So, the number of triangles would be 110*9*10=9,900.We're just calculating the number of variations after fixing a point.. P in this case. You'd get the solution if you fixed either Q or R.

Or

When y=6We know -4<=x<=+5, there are 10 integral points.The points are: {-4,6},{-3,6},{-2,6},{-1,6},{0,6},{1,6},{2,6},{3,6},{4,6},{5,6}So, number of distinct line segments that can be formed are 10C2=45.And these line segments PR as it should be parallel to x-axis.With every line segment there can be 20 triangles:

Let's see how:

consider line segment with points: P={-4,6} and R={1,6}We know there are 10 points above P that can serve us as Q. Q should always be vertically above or below point P because P is the right angle.So Q can be: {-4,7},{-4,8},{-4,9},{-4,10},{-4,11},{-4,12},{-4,13},{-4,14},{-4,15},{-4,16}See for yourself that now: for just one segment PQ we have 10 triangles.PQR.However, we can flip the points on the same line segment PR, so that P={1,6} and R={-4,6}We know there are 10 points above P that can serve us as Q. Q should always be vertically above or below point P because P is the right angle.So in this case Q can be:{1,7},{1,8},{1,9},{1,10},{1,11},{1,12},{1,13},{1,14},{1,15},{1,16}See for yourself that for just one segment PQ we have another 10 triangles.So, a total of 20 distinct triangles for just one line segment.

thus: a total of 20 * 10C2 = 20 * 45 = 900 triangles for all line segments where PR is on y=6.

Remember, we were talking only about PR that lies on y=6 line.There are 11 such lines between y=6 and y=16 i.e. 16-6+1. Thus; a total of 900 * 11 = 9900 triangles can be formed with given conditions.

I just plotted 10 horizontal points and 11 vertical points and solved it using the following formula:

(11-1)*(10-1)*11*10 = 10*9*11*10=9900Matched with Bunuel's approach and found the logic was pretty similar.

If you plot the points; you would see that you can make 90 right triangles at every point. Then, just multiply this figure with the total number of points.

In the rectangular coordinate system, points (4, 0) and (– 4, 0) both lie on circle C. What is themaximum possible value of the radius of C ?

(A) 2(B) 4(C) 8(D) 16(E) None of the above

It can be B, but the points mentioned can be a chord and that would make the radius larger. It takes 3 distinct points to define a circle. Only 2 are given here. The only thing we can conclude from the question that center lies on the Y-axis. But it could be ANY point on it, hence we can not determine maximum value of r.

The two points essentially identify a single chord of the circle C. Since no other information is provided, however, the radius of the circle can essentially be anything. All this information tell us is that the radius is greater than 4. It does not give us an upper limit.

Another way to look at it is that the two points, lets call them A and B, are equidistant to the centre of the circle, lets call that O. i.e. OA = OBHence the centre will lie on the Y axis (anywhere where x = 0).So not enough information to determine.

Yet another way to look at it is:Radius^2 = (Difference of X of O to A)^2 + (Difference of Y of O to A)^2From the question stem we know that A is (4,0). Using the above logic we also know that the centre lies on x=0. Using B would yield the same result as we are after distance it will always end up being positive anyway.This formula reduces to (4-0)^2 + (y-0)^2 = R^2Depending on the value of y, the length of the radius will keep growing.

You CAN calculate the distance between any two points with given coordinates on a plane (no matter in which quadrants they are). For example the distance between two points (4,0) and (-4,0) is simply 8.

Generally the formula to calculate the distance between two points and

is .

Next, the distance between (4,0) and (-4,0) won't necessarily be the DIAMETER of a circle. The minimum length of a diameter is indeed 8 (so min r=4) but as ANY point on the y-axis will be equidistant from the given points then any point on it can be the center of the circle thus the maximum length of the radius is not limited at all.

praveenvino wrote:

if we join the line connecting the the points (-4,0) and (4,0) to the center of the circle say (0,y), radius will be maximum at the point where the area formed by the above triangle is min. The area will be 0 if the height is 0, which means the center is in the line connecting two pnts (-4,0) and (4,0). Isn't?

The red part is not correct.

Again: The minimum length of a diameter is indeed 8 (so min r=4) but as ANY point on the y-axis will be equidistant from the given points then any point on it can be the center of the circle thus the maximum length of the radius is not limited at all.

Check 2 possible circles:Circle with min radius of 4 (equation x^2+y^2=4^2):

Attachment:

Circle with radius of 5 (equation x^2+(y-3)^2=5^2):

Generally circle passing through the points (4, 0) and (– 4, 0) will have an equation and

will have a radius of . As you can see min radius will be for , so and max radius is not limited at all (as can go to +infinity as well to -infinity).

Do lines and cross?

1. 2.

First of all: equations given ARE NOT linear equations. We know that if two different lines do not cross each other they are parallel. How can we tell, based on the equations, whether the lines are parallel? We can check the slopes of these lines: parallel lines will have the same slopes. NOT that the slopes of lines must be negative reciprocals of each other (as it was mentioned in the earlier posts): in this case they are perpendicular not parallel.

Second of all: we have quadratic equations. These equations when drawn give parabolas. The question is: do they cross? This CAN NOT be transformed to the question: "are they parallel?" In the wast majority of cases the word "parallel" is used for the lines. Well we can say that concentric circles are parallel, BUT GMAT, as far as I know, uses this word ONLY about the lines (tutors may correct me if I'm wrong). Next, the word "parallel" when used for curves (lines, ...) means that these curves remain a constant distance apart. So strictly speaking two parabolas to be parallel they need not only not to intersect but also to remain constant distance apart. In this case, I must say that this can not happen. If a curve is parallel (as we defined) to the parabola it won't be quadratic: so curve parallel to a parabola is not a parabola. So I think that at this point we can stop considering this concept in regard to the original question.

So in which cases parabolas do not cross? There are number of possibilities: We can shift the vertex: the

parabolas and will never intersect (note that they won't be exactly parallel but they will never intersect). We can consider downward and upward parabolas and in some cases they also never intersect... Of course there can be other cases as well.

As for the solution. We can follow the way dzyubam proposed (and I think it's the fastest way, provided we can

identify correct examples) and consider two cases. First case: and (upward and downward parabolas), which satisfies both statements, and see that in this case answer is YES, they cross each other;

and the second case: and (also upward and downward parabolas), which also satisfies both statements, and see that in this case answer is NO, they do not cross each other. Two different answers to the question, hence not sufficient.

Answer: E.

We can solve the question algebraically as well:

Do lines and cross?

(1) --> and , now if they cross then for some ,

should be true --> which means that equation must have a solution, some real root(s), or in

other words discriminant of this quadratic equation must be --> ? --

> ? Now can we determine whether this is true? We know nothing about , , and , hence no. Not sufficient.

(2) --> the same steps: if and cross then for some ,

should be true --> which means that equation must have a solution or in other words

discriminant of this quadratic equation must be --> ? --

> ? Now can we determine whether this is true? We know that but what about ? Hence no. Not sufficient.

(1)+(2) and --> and --> same steps as above --

> --> and the same question remains: is true? but what about ? Not sufficient.Answer: E.

What is the equation of a circle of radius 6 units centered at (3, 2)?

A. x2 + y2 + 6x – 4y = 23B. x2 + y2 - 6x + 4y = 23C. x2 + y2 + 6x + 4y = 23D. x2 + y2 - 6x – 4y = - 23E. x2 + y2 - 6x – 4y = 23

The equation will be :

If you simplify this, you'll get (e)

Right triangle ABC is to be drawn in the xy-plane so that the right angle is at A and AB is parallel to the y-axis. If the x- and y-coordinates of A, B, and C are to be integers that are consistent with the inequalities -6 ≤ x ≤ 2 and 4 ≤ y ≤ 9 , then how many different triangles can be drawn that will meet these conditions?

A. 54B. 432C. 2,160D. 2,916E. 148,824

We have the rectangle with dimensions 9*6 (9 horizontal dots and 6 vertical). AB is parallel to y-axis and AC is parallel to x-axis.

Choose the (x,y) coordinates for vertex A: 9C1*6C1;Choose the x coordinate for vertex C (as y coordinate is fixed by A): 8C1, (9-1=8 as 1 horizontal dot is already occupied by A);Choose the y coordinate for vertex B (as x coordinate is fixed by A): 5C1, (6-1=5 as 1 vertical dot is already occupied by A).9C1*6C*8C1*5C1=2160.Answer: C.

If A is the center of the circle shown above (see attachment) and AB = BC = CD, What is the value of x?A. 15B. 30C. 45D. 60E. 75

Triangle abc is equilateral, therefore all angles are 60. abc = acb = bac = 60.Similarly, Triangle adc is equilateral, therefore all angles are 60. adc = acd = dac = 60.

Now, angle bad = bac + dac = 120, since each triangle is a reverse image of the other.

Now triangle bad is isosceles, since ab = ad.Therefore abd = adb = 30, since bad =120

Or

Look at the diagram below:

Given that AB = BC = CD, also since AB is the radius then AB = AC = AD = radius, so we have that: AB = BC = CD = AC = AD, so basically we have two equilateral triangles ABC and ACD with common base of AC (ABC and ACD are mirror images of each other). Line segment BD cuts the angle ABC in half and since all angles in equilateral triangle equal to 60 degrees then x=60/2=30 degrees.

how can we say that line segment BD cuts the angle ABC in half which property is this , can u expalin me?ABC and ACD are two equilateral triangles, which are mirror images of each other if you join vertices B and D, the segment BD must cut AC, the common base, in half. Which makes BD perpendicular bisector of AC --> so BO is a hight, median, and bisector of angle ABC (O being intersection point of BD and AC).

Or

you can solve the question easily by knowing two important properties of square and rhombus.

1. diagonals bisect each other.2. each diagonal forms a 90 degree angle with the other diagonal

This is a proper rhombus... and this is one of the property of rhombus... that they bisect the diagonals in half.. and the diagonals bisect the angles in half...

also Angle (DAC is 120 degree )and Triangle DAC is an isosceles triangle.. with base BD... hence angle ABD is 30===============================================

in the figure, ABC is an equilateral triangle, and DAB is a right triangle. What is the area of the circumscribed circle?

(1) DA = 4 (2) Angle ABD = 30 degrees

You should know the following properties to solve this question:• All inscribed angles that subtend the same arc are equal. The Central Angle Theorem states that the measure of inscribed angle is always half the measure of the central angle. Hence, all inscribed angles that subtend the same arc are equal.• A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s hypotenuse, then that triangle is a right triangle.• In a right triangle where the angles are 30°, 60°, and 90° the sides are always in the ratio . Notice that the smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°).For more check Circles Triangles and chapters of Math Book: math-circles-87957.html and math-triangles-87197.html

So, from above we'll have that as DAB=90 degrees then DB must be a diameter of the circle. Next, as angles ACB and ADB subtend the same arc AB then they must be equal and since ACB=60 (remeber ACB is an equilateral triangle) then ADB=60 too. Thus DAB is 30-60-90 triangle and its sides are in ratio .

(1) DA = 4 --> the side opposite 30 degrees is 4, then hypotenuse DB=diameter=4*2=8 --> radius=4 --> . Sufficient.

(2) Angle ABD = 30 degrees --> we knew this from the stem, so nothing new. Not sufficient.

Answer: A.

I don't get why the Central Angle Theorem applies. In the figure, there is no central angle for that to occur with. So why would ADB be the same as ACB?

All inscribed angles that subtend the same arc are equal. Since angles ACB and ADB subtend the same arc AB then they must be equal. Next, there is no central angle shown on the diagram, I just mentioned Central Angle Theorem to explain why is above property true: angles ACB and ADB have the same central angle AOB (O is the center of the circle) and since they both equal to half of it then they must be equal.

. How are ABCD and BCEF connected? There is only a common side BC. All other tree sides of ABCD can be of any size and the area of ABCD can be (0, inf). BCEF is a rhombus and its area can be (0, ] ( included).

2. Statement 1. In other words, area of ABCD is that can be greater or equal to the area of BCEF (0, ]

3. Statement 2. So, area of BCEF can't be and we have (0, ) possible range for the area. it's still not enough as the area ABCD can be (0, inf)

4. Statements 1&2. The area of ABCD is and the area of BCEF can be (0, ) or always lesser than the area of ABCD. Sufficient.

Quadrilateral ABCD is a rhombus and points C, D, and E are on the same line. Is quadrilateral ABDE a rhombus?

(1) The measure of angle BCD is 60 degrees. (2) AE is parallel to BD

Attachments:

With statement 2, we can conclude that since AE is parallel to BD, therefore triangle ABD is mirror image of AED (similar triangle). We have, AD is equal to AB. With all this, we can assert that ED is equal to AB and AE is equal to BD.

In other way, to cut the long story short:-

From st 2, we can come closer to only this much.. ..ABD and AED are two similar "Isosceles" triangles, Joined together. But, we need to prove that all four sides are equal.

I have drawn one such example here:

'B' meets only one condition for a rhombus which is Parellelism. but it does not prove that all sides of ABDE are equal. Please see the attachment which suggests that 'B' only is not correct. We also need 'A' to prove all sides are equal. Hope it clarifies.

Rhombus is a quadrilateral with all four sides equal in length. A rhombus is actually just a special type of parallelogram (just like square or rectangle).

So ABCD is a rhombus means AB=BC=CD=AD.ABDE to be a rhombus it must be true that AB=BD=DE=AE.

C

D

A

B

E

(1) The measure of angle BCD is 60 degrees --> diagonal BD equals to the sides of rhombus, so BD=AB. Know nothing about DE or/and AE. Not sufficient.

(2) AE is parallel to BD --> ABDE is a parallelogram (as AE||BD and BA||DE), hence opposite sides are equal: BD=AE and AB=DE. But we don't know whether all sides are equal (AB=BD=DE=AE). Not sufficient.

(1)+(2) From (1): BD=AB and from (2) BD=AE and AB=DE --> AB=BD=DE=AE --> ABDE is a rhombus. Sufficient.

Answer: C.

AE is parallel to BD --> ABDE is a parallelogram (as AE||BD and BA||DE), hence opposite sides are equal: BD=AE and AB=DE. But we don't know whether all sides are equal (AB=BD=DE=AE)

Bunuel, all sides have to be equal as the question stem states that C, D and E are on the same line. And it also states that BD is parallel to AE. Try drawing any kind of rhombus with the following conditions and all sides will be equal. So why do we need statement A? Am I missing something?

From your reasoning above it's not clear how you came to the conclusion that alls sides must be equal.

Actually I don't even need to try drawing, to state that there are infinite # of cases possible for AE to be parallel to BD and ABDE not to be a rhombus. Just try to increase or decrease diagonal BD and leave everything else the same (AE||BD): you'll always have a parallelogram but in only one case a rhombus, when BD=AB.

hat are the lengths of sides NO and OP in triangle NOP below

QR=24, NR=10, RP=40

Notice that triangles NOP and the smaller triangle within it (with the lengths of sides 24 and 40) are similar: they share the same angle P and both are right angled, so all their three angles are equal.

In similar triangles the ratio of the corresponding sides are equal (corresponding sides are the sides opposite the

same angles). So, --> --> .

Next, since then .

Because the triangles are similar :NO/PN = QR/PRNO/24 = 50/40NO = 30Now OP = sqrt(50^2 + 30^2)=> PQ = 10sqrt(34)

What is the greatest possible area of a triangular region with one vertex at the center of a circle of radius one and the other two vertices on the circle?

A.

B.

C. D. 1E.

Clearly two sides of the triangle will be equal to the radius of 1.

Now, fix one of the sides horizontally and consider it to be the base of the triangle.

.

So, to maximize the area we need to maximize the height. If you visualize it, you'll see that the height will be maximized when it's also equals to the radius thus coincides with the second side (just rotate the other side to see). which means to maximize the area we should have the right triangle with right angle at the center.

.

Answer: B.

Adding onto what Bunuel said, there is an important property about isosceles triangles that will help you understand and solve this question.

First though, let us see how this particular triangle must be isosceles.

If one vertex is at the centre of the circle and the other two are on the diameter, then the triangle must be isosceles since two of its sides will be = radius of circle = 1.

Now for an isosceles triangle, the area will be maximum when it is a right angled triangle. One way of proving this is through differentiation. However, since that is well out of GMAT scope, I will provide you with an easier approach.

An isosceles triangle can be considered as one half of a rhombus with side lengths 'b'. Now a rhombus of greatest area is a square, half of which is a right angled isosceles triangle. Thus for an isosceles triangle, the area will be greatest when it is a right angled triangle.

For an isosceles triangle with given length of equal sides right triangle (included angle) has the largest area.

And vise-versa:

Right triangle with a given hypotenuse has the largest area when it's an isosceles triangle.

Now for the right angled triangle in our case, b = 1 and h = 1

Thus area of triangle = =

The measure of an inscribed angle is half of the measure of the arc it intercepts.If angle PRO = 35 degrees then the arc PO = 2*35=70 degreesIf angle PRO=35 degrees, then the angle RPQ=35 degrees (since lines PQ and line OR are parallel). Thus the arc QR= 2*35=70 degrees. Arc OR = 180=arc OP + arc PQ +arc QR.we found that arc OP = arc QR=70then arc PQ= 180-70-70=40 degrees

Translating 40 degrees into Pi: 40/360*(2Pir)= 2*9*Pi/9=2*Pi

Or OCP will equal 70 so qcr = 70 and you know the part under the diameter = 180140+180 = 320PQ = 360-320/36040/360 = 1/9circumference = 18 pi1/9*18pi = 2pi

Two squares, each of side lengths 1 unit and having their centres at O, are rotated with respect to each other to generate octagon ABCDEFGH, as shown in the figure above. If AB = 43/99, find the area of the octagon.

Attachment:

a. 73/99 b. 86/99 c. 1287/9801 d. 76/99

First note that all the triangles will have the same dimensions. Let the legs of these triangles be and , also we

know that the hypotenuse equals to , so .

If we take one side of a square we'll see that --> --> square it --

> , as from above , then: --> ;

Now, the are of the octagon will be area of a square minus area of 4 triangles --

>

EASY WAY is as shown belowConnect AO and BO to get a AOB. We get totally 8 triangles like this.

Now. Lets calculate the AREA of AOB

AREA = .5*base*hiegt = 0.5*43/99 * 1/2 note: i/2 is the hieght as the length of a side of the given square is 1 and half of that is what is the hieght

hence AREA = 1/4 * 43/99

As told we have 8 similar triangles in this octagon.

Hence ANSWER = 8*1/4 * 43/99 = 86/99

Employees before July 1st: Total salary before july 1st:

Average before July 1st:

Employees after July 1st: Total salary after july 1st:

Average after July 1st:

Ans: "B""

Old number of employees = Old avg salary = So Old total salary =

New number of employees = New avg salary = So New total salary =

The opposite angle is equal to the sum of the two opposite exterior angles.So in this case Angle BDC = DAB + DBA2x = x + DBADBA = xHence DBA is a isoceles triangle.

Statement 1: AD = 6 AD = DB = BC = 6 because they are all isosceles triangles.Sufficient.

Statement 2: Gives you an angle but not how long any of the sides are - Insufficient.

A Circle inscribed in a equilateral triangle ABC so that point D lies on the circle and on line segment AC and point E lies on circle and on line segment AB. If line segment AB=6, what is the area of the figure created by line segments AD, AE and minot Arc DE.

A. 3√3 - (9/4) πB. 3√3 - πC. 6√3 - 3 πD. 9√3 - 3 π5. cannot be determined

Usually in GMAT questions when triangle is inscribed in circle it's either right triangle or equilateral. I've never seen the GMAT question involving the radius of the circle circumscribing scalene triangle.

For equilateral triangle inscribed in circle , where is the side of the triangle.

For right triangle inscribed in circle .

For scalene triangle inscribed in circle , where A is the area of the

triangle, are the three sides of the triangle and which is the semi perimeter of the triangle.

Since it is an equilateral triangle, the required area can be expressed as : [Area of Triangle - Area of Circle]/3

For an equilateral triangle with altitude 'h' :

Radius of inscribed circle = h/3Radius of circumscribed circle = 2h/3

Now we know that the side of the triangle = 6

Therefore, area of triangle = =

Radius of circle inscribed in an equilateral triangle (r) = = Therefore, area of circle = =

Thus, required area = = Answer : B

B - 3√3 - πArea of equilateral triangle - √3/4 (a^2) = √3/4 (6^2) = 9√3to find the inradius we have r = A/s where s is semiperimeter = 9√3/9 = √3

Area of the incircle = π r^2 = π (√3)^2 = 3πArea of Triangle - Area of Incircle = 9√3 -3πThis area will be equal from all three sides hence (9√3 -3π)/3 = 3√3 – π

The property given by sriharimurthy is absolutely correct. But I'd like to add couple of things to this:

1. In any triangle the three medians intersect at a single point, called centroid.2. In any triangle two-thirds of the length of each median is between the vertex and the centroid, while one-third is between the centroid and the midpoint of the opposite side.3. In any triangle the bisectors intersect at a single point, called incenter. The incenter is also the center of the triangle's incircle - the largest circle that will fit inside the triangle.4. In equilateral triangle altitude(height)=bisector=median.

From above:With an equilateral triangle, the radius of the incircle is exactly half the radius of the circumcircle --> R=2r.Radius of inscribed circle = height/3=bisector/3=median/3Radius of circumscribed circle = height*2/3=bisector*2/3=median*2/3

Bunuel, Correct me if i am wrong

Any Triangle in a circumscribed circle can be broken down into the median property (2/3rd, 1/3rd length). It does not have to restrict itself to equilateral triangles alone.

Any Triangle can have an inscribed circle. Here, rather than the medians, we have the bisectors drawn from each vertex, with all bisectors meeting at the center of the circle, also called incircle. However, if equilateral triangles have inscribed circles, the bisectors = medians and hence we can calculate the radius of the incircle

But how do we calculate the radius of a circle inscribed within a scalene triangle?

It will depend on what is given. Generally radius of a circle inscribed in a triangle is ,

OR , where , half of a perimeter and is an area of the triangle.

I should mention here that I've never seen the GMAT question requiring this formula. I wouldn't worry about this case at all.NOTE: The intersection of medians is not the center of circumscribed circle.

In the figure, each side of square ABCD has length 1, the length of line Segment CE is 1, and the length of line segment BE is equal to the length Of line segment DE. What is the area of the triangular region BCE?A.. 1/3 B. (2^-2 )/4 C. 1/2D. (2^-2)/2 E. 3/4

Note that as BE=DE the triangles BCE and CDE are congruent (all three sides are equal) --> . So if we continue the line segment CE it will meet with the crosspoint of the diagonals, let's call it point O. Also note that EO will be perpendicular to BD, as diagonals in square make a right angle.The area BCE=BOE-BOC.

Area of BOC is one fourth of the square's = .

Area BOE, . , half of the diagonal of a square. , CO is also half of the diagonal of a square.

So .

Area None of the answer choices shown is correct.

My answer is A.I split the graph in two right triangles. One triangle with angles 45-45-90, another with angles 60-30-90.

In triangle, 45-45-90, the length of the sides are in ratio 1:1:2^(1/2)In triangle, 30-60-90, the length of the sides are in ratio 1:3^(1/2):2.Using this I figured out area of each triangle and then the total.

The figure shown (please see attached) consists of three identical circles that are tangent to each other. If the area of the shaded region (center space where the three circles do not touch) is , what is the radius of each circle?a. 4b. 8c. 16d. 24e. 32

Always look at the numbers in the problem and at the answer choices. Do only as much math as needed.shaded region = triangle - 3 sectors = so 3 sectors = triangle is equilateral because each side = so each angle = 60, and each sector = 60/360 = 1/6 each circleso 3 sectors = 3*(1/6) = 1/2 each circle

so =

Or

I have a simpler solution to the above explanation your solution rocks but we wont have that much time in the GMAT exam to solve this question anywayshere it goes....64(3)^1/2 - 32pie is the area of the shaded region hence the area of triangle will be equal to 64(3)^1/2 and area of three corresponding circle is 32piei will take the area of triangle you can take the area of circles segments

Therefore, 64(3)^1/2 = area of Equilateral Triangle = ((3)^1/2)/4 * Side^2Hence (side)^2 = 64*4 = 2^8 ==> Side = 2^4 =16Remember this is the side of the Triangle Side of Triangle = 2(Radius of the Circle)therefore radius of circle = 8Answer B

Or

Let the radius of the circle be , then the side of equilateral triangle will be .Area of the shaded region equals to area of the equilateral triangle minus area of three 60 degrees sectors.Area of a 60 degree sector is 1/6 of the are of whole circle (as whole circle is 360 degrees and 60 is 1/6 of it), hence

are of 3 such sectors will be 3/6=1/2 of the area of whole circle, so (here if you could spot

that should correspond to then you can write --> );

Area of equilateral triangle equals to , where is the length of a side. So in our

case ;Area of the shaded region equals to ,

so --

> --> .Answer: B.

The square of a television screen is given as a length of the Diagonal. If the screens were flat, then the area of the 21in screen would be how many square inches greater than the area of a 19in screen?2416

3840

The diagonal of a square is always

and the side of a square is vice versa always

and --> --

> Answer: E.

Area of equilateral triangle is ;

Area of square is ;

As areas are equal, then --> --> .Answer: D.

Or

A(triangle) = A(square) =

Areas are equal.

Isolate t.

Take the square root of both sides.

Square root of a fraction:

Simplify.

Finally, the ratio.

one quick question where I am stumped. When you square root a square root is that where you are getting the 4th root?

Yes.

When you take the root again, you get which is equal to In other words, it the fourth root of 3.=========================================

Which of the following best approximates the percent by which the distance from A to C along a diagonal of square ABCD reduces the distance from A to C around the edge of square ABCD?a. 30%b. 43% c. 45%d. 50%e. 70%

Le the side of a square be .

Route from A to C along a diagonal AC is ;Route from A to C around the edge ABC is ;

Difference is --> .Answer: A.

Circle A is perfectly inscribed in a square, and the square is perfectly inscribed within circle B. The area of circle B is what percent greater than the area of circle A?

(A) 50%

(B) 100%

(C) 150%

(D) 200%

(E) 250%

UPDATE: Solution! One important note about inscribed shapes is that both squares and circles are perfectly symmetrical, meaning that if you know one length (radius, diameter, circumference, area; or side, diagonal, area, perimeter) you can solve for everything else.

In this case, is we call the radius of the smaller circle r, then the diameter of that circle is 2r and the area is pi*r^2.

If that circle is perfectly inscribed inside a square, then that means that the length of the diameter will perfectly fit within the boundaries of the square, making the side of the square also equal to 2r.

Now, if a circle is perfectly inscribed around that square, then the circle will hit each corner exactly once, making the diameter of the larger circle equal to the diagonal of the square. Using what we know about squares (or using Pythagorean Theorem), we know that the diagonal is equal to the side * √2, making the diameter of the larger circle equal to 2r*√2, and the radius of the larger circle then equal to half that: r*√2.

The area of the larger circle is the pi*(r*√2)^2, or 2*pi*r^2. Therefore, the area of the larger circle (2*pi*r^2) is twice that of the smaller circle (pi*r^2). But keep in mind that you must answer the right question! The question asks “how much GREATER is the larger circle than the smaller” not “the larger circle is what percent OF the smaller”. To get to twice the size, we only ADD 100%, so the correct answer is B.

The dimensions of a rectangular solid are 4 inches, 5 inches, and 8 inches. If a cube, a side of which is equal to one of the dimensions of the rectangular solid, is placed entirely within the sphere just large enough to hold the cube, what the ratio of the volume of the cube to the volume within the sphere that is not occupied by the cube?

(A) 10:17(B) 2:5(C) 5:16(D) 25:7(E) 32:25

The cube must have a side of 4 inches to fit in rectangular solid. Now as the cube is inscribed in sphere then the

radius of this sphere equals to half of the cube's diagonal --> , so radius of the sphere

is

Volume of the cube will be ;

Volume of the sphere will be .

The ratio of the volume of the cube to the volume within the sphere that is not occupied by the cube is , and after some calculations you'll get that this ratio is closer to 10/17 than to any other ratios in answer choices.Answer: A.P.S. I really doubt that GMAT would offer such a problem, as it requires lengthy calculations and bad approximations ( , ).

If 10, 12 and 'x' are sides of an acute angled triangle, how many integer values of 'x' are possible?(A) 7(B) 12(C) 9(D) 13(E) 11

The catch in the acute angled triangle.Consider the figures attached...

Case 1:

i.e.

Case 2:

i.e. Therefore x ranges from 7 to 15 = 9 Values..

6 < x < 16 is the same as values from 7 to 15 because only integral values are allowed.

Anyway, I am putting down the explanation. The question asks you for an acute triangle i.e. a triangle with all angles less than 90. An obtuse angled triangle has one angle more than 90. So the logic is that before one of the angles reaches 90, find out all the values that x can take.Starting from the first diagram where x is minimum and the angle is very close to 90, to the 2nd diagram where all angles are much less than 90 to the third diagram where the other angle is going towards 90.

Note: The remaining angle cannot be 90 because that will make 10 the hypotenuse but hypotenuse is always the greatest side.

In an obtuse angled triangle, only one angle can be more than 90 (because the sum of all angles has to be 180). The side opposite that angle is largest. The side opposite the smallest angle is the smallest side.Similarly, in a right angled triangle, the side opposite 90 degree angle is greatest (the hypotenuse). In our question, we have to consider only acute triangles. Therefore, we are working with right triangles. The values of x we obtain from them will be very close to the smallest/greatest that are acceptable.

In the first figure, we are trying to calculate the smallest value of x. In that case, 12 will be the hypotenuse of the triangle. x will be less than 12. The angle opposite x is the smallest possible (if you look at the second figure, the angle opposite x is getting greater)

In the third figure, angle opposite to x has become the greatest possible i.e. 90 degrees (actually it should be slightly less than 90). Here x will be the hypotenuse and hence the greatest side of the triangle. x can't be any greater because then the triangle will become obtuse.

In the leftmost diagram x = root 44 which is 6.somethingx should be greater than 6.something because the angle cannot be 90.

In the rightmost diagram, x = root 244 which is 15.something

x should be less than 15.something so that the angle is not 90.Values that x can take range from 7 to 15 which is 15 - 7 = 8 + 1 = 9 values.

A side of a triangle is less than the sum of other two sides.Also a side of a triangle is greater than the difference of other two sides. Since the other two sides are 10 and 12, x should be less than 22 and more than 2.

Now we only have to consider acute angled triangles. Look at the figures below.

In the first figure when x is very small, till the angle goes to 90, the triangle is obtuse, which is not allowed.

When x becomes = root 44 = 6.something, the angle is right angled. Now x greater than this value is allowed since we will get acute triangles. Values of 3, 4, 5 and 6 give obtuse angled triangles.Values of 7, 8, 9 ... give acute triangles so allowed.

Now look at the second figure. Finally x is = root 244 = 15.somethingSo values of x till 15.something make acute triangles. Therefore 7, 8, 9, 10, 11, 12, 13, 14 and 15 (9 values) are allowed.When x is even bigger (16, 17, 18, 19, 20, 21), it will again make an obtuse angled triangle.And, as we saw, there are two values of x (6.something and 15.something) when we get a right angled triangle.

If P and Q are two points on the line 3x + 4y = -15 such that OP and OQ= 9 units, The area of triangle POQ is given by

a) 18* sqrt(2)b) 3* sqrt(2)c) 6* sqrt(2)d) 15*sqrt(2)

OC is perpendicular to AB (PQ), so it's height of AOB and POQ;

and (points A and B are intersection of the line with the X and Y axis

respectively) --> ;

As AOB and OCB are similar then --> ;

;

;

.Answer: A.

a square is drawn by joining the midpoints of the sides of a given square . a third square is drawn inside the second square in the way and this process is continued indefinately . if a side of the first square is 4 cm. detemine the sum of areas of all squares?a 18b 32c 36d 64e none

Let the side of the first square be , so its area will be ;

Next square will have the diagonal equal to , so its area will be ;And so on.

So the areas of the squares will form infinite geometric progression: , , , , , ... with common ration equal

to .

For geometric progression with common ratio , the sum of the progression is , where is the first term.

So the sum of the areas will be .Answer: B.

A circle and a square have the same area. What is the ratio of the diameter of the circle to the diagonal of the square?(A) 2 : √(2)(B) 1 : 2√(C) 2√ : √2(D) 1 : √2(E) 1 : 2

--> --

> --> .

The shaded region in the figure above represents a frame shaped as a right triangle with sides 10 inches and 13 inches. The frame encloses a picture, shaped also as a right triangle, which has the same area as the frame itself. If the sum of the 2 legs of the picture is 17 inches, what is the length of the hypotenuse of the triangular picture?(A)sqrt(29) (B)sqrt(159) (C)13(D)sqrt(269)(E)17

Area of frame + picture = 65Since both have the same area, 65/2 = 32.5 - each are is 32.5.

The picture is a similar triangle to larger triangle. Therefore, the angles are the same and the sides are proportional.

Height must be less than 10. Base must be less than 13.

Area of picture = 1/2 b h = 32.5. bh = 65.

What two numbers add to 17 and multiply to 65, and are in proportion to h = 10, b = 13, and hypotenuse ?

You almost nailed it ...! Let me complete the rest for you.

Lets assume the sides of the picture as and and Hypotenuse as

Then we have and

We need

Square both the sides of equation

Take the square root of equation

Hence B.

A big cube is formed by rearranging the 160 coloured and 56 non-coloured similar cubes in such a way that the exposure of the coloured cubes to the outside is minimum. The percentage of exposed area THAT is coloured is:

1) 25.9%2) 44.44%3) 35%4) 61%5) None of these

We have 160+56=216=6^3 big cube.

Non exposed will be 4^3 little cubes, which will be "inside" the big cube, so exposed will be 216-4^3=152 little cubes out of which there will be 152-56=96 colored cube (as colored cubes are minimized on the surface).

Now surface area of big cube is 6*36=216 (exposed area) out of which the area of 96 is colored, so the percentage of

exposed area THAT is coloured is --> .

Answer: B.

Here, the solution is linked to the height of the triangle. Given, BC = 1, Radius = 1. ie OB = OC = 1. Therefore, we can say that /\ OBC is an equilateral triangle.Height of equilateral triangle is SQRT(3)/2 * (side)^2.Therefore, height = SQRT(3)/2.Therefore, area of /\ ABC = (1/2)*2 (Given) * SQRT(3)/2.Area = SQRT(3)/2.Ans : B.

We know that AC is a diameter. There is a property of a right triangle inscribed in circle:

A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s side, then that triangle is a right triangle and diameter is hypotenuse.

If the diagonal and the area of a rectangle are 25 in and 168 in2, what is the length of the rectangle?

A. 21B. 38C. 12D. 24E. 25

One way is:lw = 168l^2 + w^2 = 625(l-w)^2 = 289 so l-w = 17 or w = l - 17l(l-17) = 168solving so l = 24 or -7offcourse l cannot be (-)ve hence D

From common right triangle proportion we know 7:24:25just pluging these values in area formula we get correct value hence l=24 hence D.

let the length and breadth be l,b

sqrt(l^2+b^2)=25 => l^2+b^2=625lb = 168

eliminate B,E because 38^2 > 625 and 25^2 = 625

Option A: if l = 21, b = 168/21 = 8 , but 21^2 + 8^2 != 625Option C: if l = 12, b = 168/12 = 14 , but 12^2 + 14^2 != 625Option D: if l = 24, b = 168/24 = 7 , and 24^2 + 7^2 = 625

Circle O is inscribed in equilateral triangle ABC. If the area of ABC is , what is the area of circle O?(A) (B) (C) (D) (E) 1. Divide equilateral in 2 isosceles triangles2. s/2 : s/2 sqrt(3) : s3. area equilateral = s^2 sqrt(3) / 4

4. => s^2 sqrt(3) / 4 = 24 sqrt(3) => s = 4 sqrt(6)5. => height = s/2 sqrt(3) = 2 * sqrt(6) * sqrt(3)6. => diameter = 2/3 height = 4/3 * sqrt(6) * sqrt(3)7. => radius = 2/3 * sqrt(6) * sqrt(3)7. => area circle = Pi r^2 = Pi 4/9 * 6 * 3 = Pi 8

If 25 lines are drawn in a plane such that no two of them are parallel and no three are concurrent, then in how many points do they intersect?

A 2300 B 600 C 250 D 300 E none of these

let try draw lines one by one.

1st line - 0 points 2nd line - new 1 point 3td line - new 2 points + old 1 point 4th line - new 3 points + old 2+1 points 5th line - new 4 points + old 3+2+1 points n-th line - (n-1) points + (n-2) .... 3+2+1

therefore, S=1+2+3...(n-1)

S=(n-1)n/2=24*25/2=300

========

since any 2 lines has 1 intersect with each other, we need to find the number of ways to choose 2 out of 25: 25C2=25*24/2=300.

There are three spheres of dough with diameters of 2, 4, and 6 cm. If the three are combined into one large sphere, what is the radius of the large sphere?

Is the proper method to solving this problem: (1) find the volume of each sphere (2) add the volumes of the three spheres (3) calculate the radius from the new total volume?

Yes, R^3=(2/2)^3+(4/2)^3+(6/2)^3:

;

Volume of the large sphere: =====

--> .

The measures of the interior angles in a polygon are consecutive integers. The smallest angle measures 136 degrees. How many sides does this polygon have?

A) 8B) 9

C) 10D) 11E) 13

When I look at this question, I say, "I know how to find the Interior angle of a regular polygon. But this is not a regular polygon since it has angles 136, 137, 138, 139, 140, 141, 142 ....etc."Also, Interior angle of 8 sided regular polygon = 180*6/8 = 135Interior angle of 9 sided regular polygon = 180*7/9 = 140Interior angle of 10 sided regular polygon = 180*8/10 = 144etcThe average of the given angles can only match 140 (such that effectively, all the angles are 140) Hence, the polygon must have 9 sides.

Remember, capitalize on what you know.

Sum of interior angles of a polygon = (n-2)*180 (not necessarily regular polygon)

Why? See the figure below:

A 6 sided polygon can be split into 4 triangles each of which has a sum of interior angles 180 degrees.An n sided polygon can be split into n - 2 triangles.

When the polygon is regular, each angle is same so the sum is divided by the number of sides to get the measure of each angle e.g. in a regular hexagon, each interior angle = 4*180/6 = 120 degrees.

Now if I have a hexagon whose angles are 115, 117, 119, 121, 123 and x, what will be the angle x? We can see it in two ways - 1. The sum of all angles should be 4*180 = 720So 115 + 117 + 119 + 121 + 123 + x = 720 or x = 125

2. The average of the angles should be 120. (Since the sum of the angles is 720 and there are 6 sides) 119 and 121 average out as 120. (119 is 1 less than 120 and 121 is 1 more than 120)117 and 123 average out as 120. So 115 and x should average out as 120 too. Therefore, x should be 125.

In the question, the average of the given angles of the polygon can only be 140. So it must have 9 sides. To confirm, 136, 137, 138, 139, 140, 141, 142, 143, 144 - 9 angles with average 140. So the polygon must have 9 sides.

(It cannot be 144 or anything else because 10 angles (136, 137, 138, 139, 140, 141, 142, 143, 144, 145) will not give an average of 144)

Sum of Interior Angles of a polygon is where is the number of sides (so is the number of angles);

We are told that the smallest angle is 136 degrees, the next will be 136+1 degrees, ..., and the largest one,

angle, will be degrees. The sum of the consecutive integers (the sum of angles) is given

by ;

So we have that --> , now try the answer choices: in order RHS to end with zero (as LHS is because of 360) then , out of the options listed, could be either 10 or 9, fits.

If the diagonal of rectangle Z is d, and the perimeter of rectangle Z is p, what is the area of rectangle Z, in terms of d and p?(A) (d2 – p)/3(B) (2d2 – p)/2(C) (p – d2)/2(D) (12d2 – p2)/8(E) (p2 – 4d2)/8

Let the sides of rectangle be and .

Given: and . Question:

Square --> --> substitute by --> --

> --> .

Answer: E.

When answer is in terms of 1 or 2 variables, my suggestion would be to quickly take simple values. (If variables are more than that, keeping a track of their values becomes cumbersome) I would say let the sides be 3 and 4 to get diagonal, d = 5 (Pythagorean triple). Then p = 2x3 + 2x4 = 14. You are looking for an area of 3x4 = 12Put values and check.

There is a teeny-weeny chance that two options may give you the answer you are looking for. In that case, you might have to take different values and put in those two options to pick out the winner, but the risk is worth it, in my opinion.

A sphere is inscribed in a cube with an edge of 10. What is the shortest possible distance from one of the vertices of the cube to the surface of the sphere?

(A) (B)

(C)

(D)

(E) It would be easier if you visualize this problem.

As sphere is inscribed in cube then the edge of the cube equals to the diameter of a sphere --> .

Next, diagonal of a cube equals to .

Now half of Diagonal-Diameter is gap between the vertex of a cube and the surface of the sphere --

>

Shortest distance=(diagonal of cube-diameter of sphere)/2=

Answer: D.

In the above circle AB = 4, BC = 6, AC = 5 and AD = 6. What is the length of DE?

(A) 6 (B) 7.5(C) 8(D) 9(E) 10

Attachment:

As all inscribed angles that subtend the same arc are equal then <BCD=<BED (as these angles subtend the arc BD) and <CBE=<CDE (as these angles subtend the arc CE). Also <BAC=<DAE. So triangles ABC and ADE are similar: in similar triangles, corresponding sides are all in the same proportion.

So, DE/BC=AD/AB --> DE/6=6/4 --> DE=9.

Because triangles ABC and ADE are similar their corresponding sides are all in the same proportion. Corresponding sides are the sides opposite the angles which are equal. For example as <BCA=<AED then the sides opposite them BA and AD are corresponding. So, in similar triangles the RATIO of corresponding sides are the same: BA/AD=BC/DE.

Next triangles are not congruent they are similar, so AB doesn't equal to AD (by the way stem gives AB = 4 and AD = 6).==================

In the diagram, triangle PQR has a right angle at Q and a perimeter of 60. Line segment QS is perpendicular to PR and has a length of 12. PQ > QR. What is the ratio of the area of triangle PQS to the area of triangle RQS?A. 3/2 B. 7/4 C. 15/8D. 16/9 E. 2

Attachment:

Let , and .

Given: (i);

Equate the areas: (area of PQR can be calculated by 1/2*leg*leg and 1/2* perpendicular to hypotenuse*hypotenuse) --> (ii);

Aslo (iii);

So, we have:(i) ;(ii) ;

(iii) .

From (iii) --> as from (i) and from (ii) then (

--> --> --> ;

From (i) and from (ii) --> solving for and --> and (as given that ).

Next, perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle. So, PQR and SQR are similar. In two similar triangles, the ratio of their areas is the square of the ratio of their sides:

Perpendicular to hypotenuse divides it in half if and only the right triangle is isosceles, so when PQ=QR but it's given that PQ>QR, so it's not the case.

.

So, --> Answer: D.

Car B starts at point X and moves clockwise around a circular track at a constant rate of 2 mph. Ten hours later, Car A leaves from point X and travels counter-clockwise around the same circular track at a constant rate of 3 mph. If the radius of the track is 10 miles, for how many hours will Car B have been traveling when the cars have passed each

other for the first time and put another 12 miles between them (measured around the curve of the track)?

A. 4pi – 1.6B. 4pi + 8.4C. 4pi + 10.4D. 2pi – 1.6E. 2pi – 0.8

Step by step analyzes:

B speed: mph;A speed: mph (travelling in the opposite direction);Track distance: ;

What distance will cover B in 10h: milesDistance between B and A by the time, A starts to travel:

Time needed for A and B to meet distance between them divided by the relative

speed: , as they are travelling in opposite directions relative speed would be the sum of their rates;

Time needed for A to be 12 miles ahead of B: ;

So we have three period of times:Time before A started travelling: hours;Time for A and B to meet: hours;Time needed for A to be 12 miles ahead of B: hours;

Total time: hours.Answer: B.

I will consider Pi as 3. So the distance round the circle is 20pi or 60 mi approx. Out of this Car B has already travelled 20 mi in 10 hours. Thus we are left with 40 mi to be covered by A and B together but driving in opposite direction @ 3km/hr and 2km/hr respectively. Thus to cover balance distance plus 12km i.e. 52miles, I need 52/5=10.4 hours.

Thus answer is 10hours plus 10.4 hours i.e. 20.4 hours and answer choice B (4pi+8.4) is correct.

If each side of ΔACD above has length 3 and if AB has length 1, what is the area of region BCDE?

(A)

(B)

(C)

(D) (E)

Attachment:

Since each side of ΔACD has length 3 then ACD is an equilateral triangle and its each angle is 60°.

Now, the are of equilateral triangle is (for more check math-triangles-87197.html);

Next, since angle A is 60° then right triangle ABE is a 30°-60°-90° right triangle. Now, in a right triangle where the angles are 30°, 60°, and 90° the sides are always in the ratio , the leg opposite 30° (AB) corresponds with and the leg opposite 60° (BE) corresponds with (the hypotenuse AE corresponds with 2). So, since

then , and the area of ABE is ;

The area of of region BCDE is the area of ACD minus the area of ABE: .Answer: B.

In the figure above, triangles ABC and MNP are both isosceles. AB is parallel to MN, BC is parallel to NP, the length of AC is 7 and the length of BY is 4. If the area of the unshaded region is equal to the area of the shaded region, what is the length of MP?

A. B.

C.

D.

E.

Since the area of unshaded region is equal to the are of shaded region, then the area of the big triangle is twice the

area of the little triangle (unshaded region): Next, triangles ABC and MNP are similar. In two similar triangles, the ratio of their areas is the square of the ratio

of their sides: . Thus --> --> --

> .Answer: D.

In the diagram, what is the value of x?

A.

B.

C.

D.

E.

Rotate the diagram as shown below:

Attachment:

Triangle BDC is a 30°-60°-90° right triangle. Now, in a right triangle where the angles are 30°, 60°, and 90° the sides are always in the ratio , the leg opposite 30° (BD) corresponds with , thethe leg opposite 60° (DC)

corresponds with , and the hypotenuse corresponds with 2. So, since then

and .

Also notice that .

Next, from Pythagoras theorem --> --

> --> --> -->

rationalize: --> --> .

Answer: B.

In the figure, point D divides side BC of triangle ABC into segments BD and DC of lengths 1 and 2 units respectively. Given that ÐADC = 60º and ÐABD = 45º, what is the measure of angle x in degrees?A. 55 B. 60 C. 70 D. 75 E. 90

Complete solution for all the angles is in the image below:

Attachment:

x=45+30=75.

Notes: Sides with one blue segment crossing them are equal and sides with two blues segments crossing them are equal too. CO is perpendicular to AD --> OD=1 (from 30-60-90 right triangle property as sides are in ratio ) --> as OD=BD=1 then ODB is an isosceles triangle.

<CDO and <BDO are supplementary to each other (supplementary angles are two angles that add up to 180°), so <BDO=120 --> <DAB=180-(120+45)=15. As ODB is an isosceles triangle --> <DOB=<DBO=30. <OBA=45-30=15 --> AOB is an isosceles triangle, so OA=OB also COB is an isosceles triangle, so CO=OB --> OA=CO=OB. So, AOC is an isosceles triangle --> <CAO=<OCA=45 (as <COA=90) --> x=45+30=75.

Answer: D.

In the triangle above, if , then what is the area of ΔABC?

A. B. C. D. E.

Look at the diagram below:

Attachment:

Since triangle BDC is 45-45-90 right triangle its sides are in ratio , hence ;

Since triangle ABD is 30-60-90 right triangle its sides are in ratio , hence AD (the side opposite 60 degrees) is ;

The area of ABC is .Answer: C.

The length of arc AXB is twice the length of arc BZC, and the length of arc AYC is three times the length of arc AXB. What is the measure of angle BCA?

A. 20B. 40C. 60D. 80E. 120

Attachment:

When we are told that "the length of arc AXB is twice the length of arc BZC" means that , where O is the center of the circle. Similarly "the length of arc AYC is three times the length of arc AXB." means

that

So if then --> --> . Now, according to The Central Angle Theorem the measure of inscribed angle is always half the measure of the

central angle so --> .

Answer: B.

To elaborate more on angles and arcs:You should know that when you measure the length of an arc in degrees then it's the measure of corresponding central angle. For example means that arcBZC is 40/360=1/9 th of the circumference. Next, , according to the central angle theorem, will be half the measure of the central angle,

so . As for : it'll be supplementary angle to (supplementary angles are two angles that add up to 180°), so it equals to .

If a 4 cm cube is cut into 1 cm cubes, then what is the percentage increase in the surface area of the resulting cubes?A. 4%B. 166%C. 266%D. 300%E. 400%

The big cube which dimensions 4*4*4 can "produce" 4*4*4 small cubes with dimensions of 1*1*1 (base layer can give 4*4 small cubes and as there are 4 layers than total of 4*4*4 small cubes)

Cube has 6 faces. The surface area of a cube which has a side of 4cm is 6*4^2=6*16 cm^2.

Now, when the cube is cut into the smaller cubes with side of 1cm we'll get 4*4*4=64 little cubes and each will have the surface area equal to 6*1^2=6 cm^2, so total surface are of these 64 little cubes will be 6*64 cm^2.

6*64 is 4 times more than 6*16 which corresponds to 300% increase.Answer: D.

Or: general formula for percent increase or decrease, (percent change):

So the percent increase will be: .Answer: D.

Consider a regular polygon of p sides. The number of values of p for which the polygon will have angles whose values in degrees can be expressed in integers?A. 24 B. 23C. 22 D. 20 E. 21

Couple of things:

1. Sum of interior angles of a polygon is given by where is the number of sides (for example the sum

of interior angles of a triangle is degrees and the sum of interior angles of a quadrilateral

is degrees).

Question below talks about a regular polygon, which is a polygon with all equal sides and equal interior angles. Thus

each interior angle of a regular polygon is given by: (for example each interior angle of an equilateral

triangle is degrees and each interior angles of a square is degrees).For more on polygons check: math-polygons-87336.html

2. Finding the Number of Factors of an Integer

First make prime factorization of an integer , where , , and are prime factors of and , , and are their powers.

The number of factors of will be expressed by the formula . NOTE: this will include 1 and n itself.Example: Finding the number of all factors of 450:

Total number of factors of 450 including 1 and 450 itself is factors.For more on number properties check: math-number-theory-88376.html

BACK TO THE ORIGINAL QUESTION:

The question is: for how many values of (where p is the # of sides of a regular polygon) is an integer (or how many sided regular polygons exist which have interior angles equal to an integer).

Now, to be an integer must be an integer, so must be a factor of 360. How many

different positive factors does 360 have? Since then # of factors is ,

including 1 and 360. Thus if is any of these 24 values (1, 2, 3, ... 360) then is an integer.

Finally, as polygon can not have 1 or 2 sides (p can not be 1 or 2) then only 24-2=22 regular polygons exist which have interior angles equal to an integer: 3 sided (equilateral triangle), 4 sided (square), 5 sided (regular pentagon), ..., 360 sided.Answer: C.

A welder received an order to make a 1 million liter cube-shaped tank. If he has only 4x2 meter sheets of metal that can be cut, how many metal sheets will be required for this order? (1 cubic meter = 1,000 liters)

9290827875

1 million liter cube-shaped tank = 1,000,000/1,000 = 1,000 cubic meter tank --> the cube-shaped tank with a side of 10m (10^3=1,000).Surface area of this tank would be 6 (# of faces of a cube) * 10^2 (area of each face) = 600 m^2. # of 4*2=8m^2 sheet needed is 600/8=75.

In the figure shown, ABCD is a square with side length 5 sqrt 2. Two quarter-circles BED and DFB are centered at A and C, respectively. What is the area of the shaded region?

A.50pi-25B.25pi−50 C.50pi/3 D.25pi/3 E.25pi

There are a lot of ways to do this. You could add the two areas of the quarter-circles ABD and BCD: when you do that, you'll be adding the area of ADFB once, the area of CDEB once, and the area of the shaded region twice. That's just the area of the entire square plus the area of the shaded region. So if you then subtract the area of the square, you'll have the area of the shaded region (this would be easier to demonstrate by drawing diagrams, but hopefully what I mean is clear - it's very much like a Venn diagram question, in which you add the numbers in each of two overlapping groups; in that case you're counting the overlap twice).

The area of each quarter-circle is (1/4)*Pi*(5sqrt(2))^2 = 25*Pi/2. Adding the two quarter-circles' areas gives 25*Pi. Then subtracting the area of the square to get the area of the shaded region gives 25*Pi - (5*sqrt(2))^2 = 25*Pi - 50.

You might also be able to pick the right answer here without really doing any work. The area of the square is 50, and the area of the shaded region looks to be about half the area of the square. That alone gets you to answers B or D immediately. Since you can pretty reliably guess that you'll need to subtract one area from another to get the answer, B almost has to be correct.

−

A farmer has a field that measures 1000 ft wide by 2000 ft long. There is an untillable strip 20 ft wide on the inside edge of the field, and a 30 ft wide untillable strip bisects the field into two squares (approximate). Approximately what percentage of the field is tillable?

A) 98%B) 93% C) 91%D) 90%E) 88%

Total Area = 1000*2000Tillable Square's side horizontally = ( 2000-20-30-20 )/2 = 1930/2 = 965 Tillable Square's side vertically = (1000-20-20) = 960 = 960Consider it as 960:

Why approximated to 93 and not 91 because we shortened one side from 965 to 960. Thus, in reality the squares are bigger.

Ans: "B"

Actually you can solve the problem pretty fast by using following approach:

1. one shorter inside strip with width of 20 ft takes 20/2000 = 1% of field2. There is 2 short strips, 2 long strips (twice as long as shorts ones) and one short but wider strip that equals 30/20 = 1.5 short strips.3. Approximately we have 2 + 2*2 + 1.5 = 7.5 short strips --> ~ 7.5% or 92.5%4. As we didn't take into account overlaps between strips it will be slightly higher than 92.5%.Or you can use calculations but I think it will take more time:

Two spherical balls lie on the ground touching. If one of the balls has a radius of 8 cm, and the point of contact is 10 cm above the ground, what is the radius of the other ball?

(1) 18 cm (2) 40/3 cm (3) 25/2 cm (4) 13 cm (5) none of the these

For the moment, just think of the two circles touching each other at point, say P, as shown in the diagram

Now, say, you draw a tangent at P, the point of intersection. This will be a common tangent to both the circles at the point P. Join the center of each circle with P. We know that the line joining the center to the point of tangency will be perpendicular to the tangent.

Hence they will form an angle of 180 degrees and will therefore be straight.

In the rectangular coordinate system above, the area of triangular region PQR is:

- 12.5- 14- 10[square_root]2- 16- 25

The answer is 12.5

If you look carefully, the triangle is enclosed within a rectangle with dimensions 7 x 4. See attached figure.

Area of the

= Area of rectangle - Area of yellow triangle - Area of blue triangle - Area of red triangle

Attachments:

If the product of slopes of two lines is -1, those two lines are perpendicular.

Here; PR is perpendicular to PQ.

Point

Point

Point

Slope of PR

Slope of PQ

Product of slopes

Hence, and PQR is a right angled triangle.

Formula of distance between two points

Distance between two points

Distance between point P and point Q Likewise,

Distance between point P and point R

Area of a triangle = 1/2*(PQ)*(PR) = 1/2*5*5=12.5

Or

Using Pythagorean theorem we can easily find that PQ and PR are 5 each. Now we have to determine is the angle PQR right angle.

Yes it is. How we know that. Using Pythagorean theorem again we can determine that QR is simply by

solving = = .

We know that PQ and QR are both 5 and their base is and that diagonale of the square is . So, triangle PQR must be half of the square with the base of 5 or 12,5.

In the xy plane, the point (-2,-3) is the center of a circle. The point (-2,1) lies inside the circle and the point (4,-3) lies outside the circle. If the radius r of the circle is an integer, then r=

A. 6B. 5C. 4D. 3E. 2

I'd quickly mark the points on a plane to SEE the whole picture:

You can see that the radius must be more than 4 (since the distance between (-2, -3) and (-2, 1) is 4) but less than 6 (since the distance between (-2,-3) and (4, -3) is 6). It's given that r is an integer therefore r=5.Answer: B.

Line m and n pass through point (1,2). Is the slope of m greater than the slope of n?

(1) The x-intercept of m is greater than 1 and that of n is less than 1 (2) The y-intercept of m = 4 and that of n = (-2)

Best is to draw graphically.S1: gives u a range for x intercept. But since you have (1,2) fixed, you would see.. Line m has a -ve slope (in the given range) and Line n has a positive slope. Hence SUFF.S2: Makes it much more clear by giving u two other points. Hence SUFF.

Equation for Line my = mx + bEquation for Line ny = nx + bQuestion: is m > n?(1)X-intercept of Line m > 1X-intercept of Line n < 1

Solve for X-intercept of Line my = mx + b(0) = mx + bmx = - b

x =

x > 1

> 1

> mSolve for X-intercept of Line yy = nx + b(0) = nx + bnx = -b

x =

x < 1

< 1

< n

m < < nTherefore, (1) is sufficient

(2)Y-intercept of Line m = 4Y-intercept of Line n = -2

Solve for my = mx + b(2) = (1)m + 4m = -2

Solve for ny = nx + b(2) = (1)n -2n = 4n > mTherefore, (2) is sufficientAnswer: D

++++++

The question asks: "Is the slope of m greater than the slope of n?" So, we should NOT compare the absolute value of the slopes.

Notice that a higher absolute value of a slope indicates a steeper incline. So:

If the slopes of lines k and l are positive and line k is steeper then it will have the greater slope. If the slopes of lines k and l are negative and line k is steeper then its slope is more negative then the slope of line l(the absolute value of k's slope is greater), which means that the slope of l is greater than the slope of k.

If the slopes of the line l1 and l2 are of the same sign, is the slope of the line l1 greater than that of line l2?

(1) Lines l1 and l2 intersect at point(a,b) where a and b are positive(2) The X-intercept of line l1 is greater than the X-intercept

If the slopes of the line l1 and l2 are of the same sign, is the slope of the line l1 greater than that of line l2?

(1) Lines l1 and l2 intersect at point (a,b) where a and b are positive --> clearly insufficient: we can rotate the lines on this point so that we get an YES, as well as NO answers. Not sufficient.

(2) The X-intercept of line l1 is greater than the X-intercept of line l2 --> also insufficient: we can draw numerous lines with the same sign slopes so that we get an YES, as well as NO answers. Not sufficient.

(1)+(2) We know that: the slopes of the lines are of the same sign, that they intersect in I quadrant and the X-intercept of line l1 is greater than the X-intercept of line l2. There can be two cases:

You can see that in both cases the slope of L1 is greater than the slope of L1 as a steeper incline indicates a higher absolute value of the slope.

In red case: both lines have positive slope, L1 is steeper thus its slope is greater than the slope of L2;In blue case: both lines have negative slope, L2 is steeper thus its slope is more negative (the absolute value of its slope is greater) than the slope of L1, which also means that the slope of L1 is greater than the slope of L2.

Answer: C.

Lines n and p lie in the xy-plane. Is the slope of line n less than the slope of line p ?

(1) Lines n and p intersect at the point (5 , 1).(2) The y-intercept of line n is greater than the y-intercept of line p.

Algebraic approach:

Lines n and p lie in the xy-plane. Is the slope of line n less than the slope of line p?

We have two lines: and . Q: true?

(1) Lines n and p intersect at the point (5,1) --> --> . Not sufficient.(2) The y-intercept of line is greater than the y-intercept of line --> y-intercept is value of for , so it's

the value of --> or . Not sufficient.

(1)+(2) , as from (2) (RHS), then LHS (left hand side) also is less than

zero --> --> . Sufficient.

Answer: C.

Graphic approach:

Lines n and p lie in the xy plane. Is the slope of line n less than the slope of line p?

(1) Lines n and p intersect at (5,1)(2) The y-intercept of line n is greater than y-intercept of line p

The two statements individually are not sufficient.

(1)+(2) Note that a higher absolute value of a slope indicates a steeper incline.

Now, if both lines have positive slopes then as the y-intercept of line n (blue) is greater than y-intercept of line p (red) then the line p is steeper hence its slope is greater than the slope of the line n:

If both lines have negative slopes then again as the y-intercept of line n (blue) is greater than y-intercept of line p (red) then the line n is steeper hence the absolute value of its slope is greater than the absolute value of the slope of the line p, so the slope of n is more negative than the slope of p, which means that the slope of p is greater than the slope of n:

So in both cases the slope of p is greater than the slope of n. Sufficient.

If line p has a negative y-intercept then its slope is positive and it will still be more than the slope of n, with positive y-intercept (if the slope of n will be positive than p will still be steeper than n, and if the slope of n is negative it obviously will be less than positive slope of p). Consider first image and rotate line n (blue) so that it to have positive y-intercept and you'll easily see the answer.

In the xy-plane, is the slope of line L greater than the slope of line K?

(1) L passes through (5, 0) and K passes through (-5, 0).(2) L and K intersect with each other in the 2nd quadrant.

Answer is E, not C as everyone stated above.Note that: a steeper incline indicates a higher absolute value of the slope.Obviously each statement alone is not sufficient. When taken together, we can have 2 cases:

You can see that the slope of K (blue) is greater than that of L (red): 2>-1;

In this case the slope of K (blue) is less than that of L (red): -8<-1/2. Here, both lines have negative slopes. Line K is steeper than line L, which indicates that the absolute value of its slope is greater than that of line L, since slopes are negative, then slope of K is "more negative" (so less) than slope of L.

Answer: E.

A cylinder has a base with a circumference of 20pi meters and an equilateral triangle inscribed on the interior side of the base. A marker is dropped into the tank with an equal probability of landing on any point on the base. If the probability of the marker landing inside the triangle is (Sqrt 3)/4 , what is the length of a side of the triangle?

A. 3(sqrt 2pi)B. 3(sqrt 3pi)C. 10*sqrt (pi)

D. 10(sqrt 3pi)E. 20 pi

Since the probability of the marker landing on the portion of the base inside the triangle is then the portion of

the base (circle) inside the triangle must be of the area of the base.

Next: --> --> ;

The area of the equilateral triangle is of the base: --> also the ares of the

equilateral triangle is , where is the length of a side --

> --> .Answer: C.

A cylindrical tank has a base with a circumference of 4(sqrt(pi sqrt(3)) meters and an equilateral triangle painted on the interior side of the base. A grain of sand is dropped into the tank, and has an equal probability of landing on any particular point on the base. If the probability of the grain of sand landing on the portion of the base outside the triangle is 3/4, what is the length of a side of the triangle?

A.

B. C. D. E.

Given: and Now, as the probability of the grain of sand landing on the portion of the base outside the triangle is 3/4 then the the portion of the base (circle) outside the triangle must be 3/4 of the are of the base and the triangle itself 1/4 of the are of the base.

Next: --> square both sides --> --> --

> ;

The area of the equilateral triangle is 1/4 of the base: --> also the ares of the

equilateral triangle is , where is the length of a side --

> --> .Answer: E.

In the figure shown, what is the value of x?

(1) The length of line segment QR is equal to the length of line segment RS

(2) The legnth of line segment ST is equal to the length of line segment TU

Attachment:

x+<QSR+<UST=180 (straight line =180) and <R+<T=90 (as PRT is right angle)

(1) The length of line segment QR is equal to the length of line segment RS --> QRS is isosceles --> <RQS=<QSR=(180-R)/2 (as <RQS+<QSR+<R=180 --> 2<QSR+<R=180 --> <QSR=(180-R)/2). Not sufficient.

(2) The legnth of line segment ST is equal to the length of line segment TU --> UST is isosceles --> <SUT=<UST=(180-T)/2. Not sufficient.

(1)+(2) x+<QSR+<UST=180 --> x+(180-R)/2+(180-T)/2=180 --> x+(360-(R+T))/2=180 --> as R+T=90 --> x+(360-90)/2=180 --> x=45. Sufficient.Answer: C.

It should be: QRS is isosceles --> <RQS=<QSR=(180-R)/2 (as <RQS+<QSR+<R=180 --> 2<QSR+<R=180 --> <QSR=(180-R)/2).

Is the perimeter of triangle ABC greater than 20?

(1) BC-AC=10.(2) The area of the triangle is 20.

Question Stem : Is AB + BC + AC > 20?

St. (1) : BC = AC + 10

Triangle Property : The sum of any two sides of a triangle is always greater than the third.

Since we are given that one of the sides is greater than 10, the sum of the other two sides must also be greater than 10.Hence the perimeter will always be greater than 20.Statement is sufficient.

St. (2) : A = 40

Triangle Property : For triangles with same area, the perimeter is smallest for an equilateral triangle.

Area of equilateral triangle with side x =

Therefore, = 40

Now, in order to speed up calculations, I will assume to be equal to 2.

If the condition is satisfied with equal to 2 then it will definitely be satisfied with the actual value of which is less than 2.

Therefore, = 80

This tells us that x is almost 9. More importantly, it tells us that x is greater than 8. Thus perimeter will be 3*x = 24.

Since this is the minimum perimeter possible (actually it is still less than what the actual minimum would be due to our approximations), we can conclude that the question stem will always be true.Answer : DAnother interesting triangle property : For triangles with same perimeter, the area is maximum for an equilateral triangle. (If you think about it, this property goes hand in hand with the one we used in St. 2).

Yes, the OA is D. It was the hard one.

This problem could be solved knowing the properties sriharimurthy mentioned. +1.

For (1): The length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides.

For (2):A. For a given perimeter equilateral triangle has the largest area.B. For a given area equilateral triangle has the smallest perimeter.

I would go backward.

Let's assume the perimeter is 20. The largest area with given perimeter will have the equilateral triangle, so side=20/3. Let's calculate the area and if the area will be less than 20 it'll mean that perimeter must be more than 20.

Think this way is easier. .

Every point in the xy plane satisfying the condition ax + by ≥ c is said to be in region R. If a, b

and c are real numbers, does any point of region R lie in the third quadrant?

1. Slope of the line represented by ax + by – c = 0 is 2.

2. The line represented by ax + by – c = 0 passes through (-3, 0).

First of all, notice that ax + by – c = 0 or ax + by = c is the equation of the same line. A line

divides the plane into two regions. One of them, where every point (x, y) satisfies ax + by ≥ c,

is region R. 

Statement 1: Slope of line is 2

the line will pass through third quadrant and hence both regions will lie in the third quadrant.

Sufficient.

Statement 2: The line passes through (-3, 0).

Attachment:

A line passing through (-3, 0) could be the blue line or the green line. In either case, the line

will pass through the third quadrant and hence, will have both regions in the third quadrant.

So it is sufficient too? What about the x axis? That is also a line passing through (-3, 0). It does

not pass through the third quadrant. We would need the equation of the line to find out

whether our region R lies in the third quadrant. The equation of x axis is y = 0. So the

required region is y ≥ 0 i.e. the first and second quadrant. Hence using just this information,

we cannot say whether a point of region R lies in the third quadrant or not. 

Answer (A)

Points L, M, and, N have xy-coordinates (2,0), (8,12), and (14,0), respectively. Points P, Q, and

R have xy-coordinates (6,0), (8,4), and (10,0), respectively. What fraction of the area of the

triangle LMN is the area of the triangle PQR?

A. 

B. 

C. 

D. 

E. 

Look at the diagram below:

Attachment:

The area of triangle LMN (red) is 1/2*base*height=1/2*12*12=72;

The area of triangle PQR (blue) is 1/2*base*height=1/2*4*4=8;

The area of PQR is 8/72=1/9 of the area of triangle LMN.

Answer: A.

Else you can notice that triangles LMN and PQR are similar. Now, in two similar triangles, the

ratio of their areas is the square of the ratio of their sides, therefore since the ratio of the

sides is 4/12=1/3 then the ratio of the areas is (1/3)^2=1/9.

Triangle ABC will be constructed in a xy-plane according to the following conditions: Angle

ABC is 90 degrees, and AB is parallel to the y-axis; for each of points A, B and C, both the x-

coordinate and the y-coordinate must be integers; the range of possible x-coordinates is

0=<x=<5 and the range of possible y-coordinates is -4=<y=<6. Any two triangles with non-

identical vertices are considered different. Given these constraints, how many different

triangles could be constructed?

A. 50

B. 66

C. 2500

D. 3300

E. 4356

Since give that 0=<x=<5 and -4=<y=<6 then we have a rectangle with dimensions 6*11 (6

horizontal and 11 vertical dots). AB is parallel to y-axis, BC is parallel to x-axis and the right

angle is at B.

Choose the (x,y) coordinates for vertex B: 6C1*11C1;

Choose the x coordinate for vertex C (as y coordinate is fixed by B): 5C1, (6-1=5 as 1

horizontal dot is already occupied by B);

Choose the y coordinate for vertex A (as x coordinate is fixed by B): 10C1, (11-1=10 as 1

vertical dot is already occupied by B).

6C1*11C1*5C1*10C1=3,300.

Answer: D.

In the xy-plane, line k passes through the point (1,1) and line m passes through the point (1,-

1). Are the lines k and m perpendicular to each other

(1) Lines k and m intersect at the point (1,-1)

(2) Line k intersects the x axis at the point (1,0)

For one line to be perpendicular to another, their slopes must be negative reciprocals of each

other (if slope of one line is   than the slope of the line perpendicular to this line is  ). In

other words, the two lines are perpendicular if and only the product of their slopes is  . 

So basically the question is can we somehow calculate the slopes of these lines.

From stem we have one point for each line.

(1) gives us the second point of line  , hence we can get the slope of this line, but we still

know only one point of line  . Not sufficient.

(2) again gives the second point of line  , hence we can get the slope of this line, but we still

know only one point of line  . Not sufficient.

(1)+(2) we can derive the slope of line   but for line   we still have only one point, hence we

can not calculate its slope. Not sufficient.

Answer: E.

In the rectangular coordinate system, the line y = x is the perpendicular bisector of segment AB (not shown), and the y-axis is the perpendicular bisector of segment BC (not shown). If the coordinates of point A are (3; 2), what are the coordinates of point C?

A. (-3;-2)B. (-2; 3)C. (3;-2)D. (2;-3)E. (2; 3)

Since the line y=x is the perpendicular bisector of segment AB, then point B is the mirror reflection of point A around the line y=x, so its coordinates are (2, 3). The same way, since the y-axis is the perpendicular bisector of segment BC then the point C is the mirror reflection of point B around the y-axis, so its coordinates are (-2, 3). Answer: B.

Now, you can simply see that options A, C, and D (blue dots) just can not be the right answers. As for option E: point (2, 3) coincides with point B, so it's also not the correct answer. Only answer choice B remains.

An equilateral triangle that has an area of   is inscribed in a circle. What is the area of the circle?

A. 6pi B. 9pi C. 12pi D. 9pi 3^1/2 E. 18pi 3^1/2

This is how I solved it area of equilateral triangle =Square root 3/4*a^2=9 square root 3we get a =6know to calculate radius of an equlateral triangle in an inscribed circle we can use formulae

r=a*square root 3/6with this I get r=square root 3 and then area =3 pi,  but its not in the answer choices 

I'd recommend to know the ways some basic formulas can be derived in geometry rather than memorizing them.

Anyway,  , where   is the length of a side --> as given

that   then  ;

Now, given that this triangle is inscribed in circle (not circle is inscribed in triangle). The

radius of the circumscribed circle is   (you used the formula for the radius of

the inscribed circle  ) -->  .

Answer: C.

In the xy-plane, if line k has negative slope and passes through the point (s;-2), is the x-intercept of line k positive?

(1) s = 0(2) The y-intercept of line k is negative.

(1) s = 0. Directly gives the y-intercept of line k: (0, -2). Since the y-intercept of line k is negative and the slope of line k is negative too then the x-intercept of line k must also be negative. Sufficient.

(2) The y-intercept of line k is negative. The same here: since the y-intercept of line k is negative and the slope of line k is negative too then the x-intercept of line k must also be negative.

Answer: D.

A rectangle is inscribed in a circle of radius r. If the rectangle is not a square, which of the following could be the perimeter of the rectangle?

A. 2r sqr3B. 2r (sqr3 + 1)C. 4r sqr2D. 4r sqr3E. 4r (sqr3 + 1)

Since the radius of the circle is r, the diameter must be 2r. Now imagine the rectangle. The diameter must be the hypotenuse of the right angled triangle of the rectangle. Say, if its sides are a and b,

So when you square a and b and sum them, you should get 4r^2The given options are the perimeter of the rectangle i.e. 2(a+b). So I ignore 2 of the options and try to split the leftover into a and b. The obvious first choices are options (B) and (E) since we can see that we can split the sum into 3 and 1.

Now check:

That is what we wanted. Hence, the answer is (B)

A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. Since a rectangle is made of two right triangles then a rectangle inscribed in a circle must have its diagonal as the diameter of the circle.

Now, since each option has in it, then let's check a right triangle where the angles are 30°, 60°, and 90° (standard

triangle for the GMAT), because its sides are always in the ratio . Notice that the smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°).

The perimeter of the rectangle is .

Answer: B.

Square S is inscribed in circle T. If the perimeter of S is 24, what is the circumference of T?

A. 6π

B. 12π

C. 3√2 π

D. 6√2 π

E. 12√2 πsquare forms two right angled triangles.Any time we have a right angle triangle inside a circle, the hypotenuse is the diameter.hypotenuse here = diagonal of the square = 6 sqrt(2) = diameter=> radius = 3 sqrt(2)Circumference of the circle = 2pi r = 6 pi sqrt(2)Answer is D.

An integer greater than 1 that is not prime is called composite. If the two-digit integer n is greater than 20, is n composite?

(1) The tens digit of n is a factor of the units digit of n.

(2) The tens digit of n is 2.

the number be of the form AB= 10A + B (splitting the number)

now here we are told A is a factor of B and AB>20so the number 10A+B will be of the form A(10+integer).

This shows that AB can be written as integer*integer => non prime => composite.

For condition one - They have mentioned that ten's digit is a factor of one's digit.Implying that numbers are - 22 ( 2 is factor of 2 ) , 24 ( 2 is a factor of 4 ) , 26, 28, 33, 36, 39 etc. Thus these numbers

are composite as they are not prime.

Condition 2 , gives us 21,22,23,24....29. Of which 23, 29 are prime. Thus insufficient.

In triangle ABC, if BC = 3 and AC = 4, then what is the length of segment CD?

A. 3 B. 15/4C. 5 D. 16/3 E. 20/3

Attachment:

A perpendicular bisector is a line which cuts a line segment into two equal parts at 90°. So AC to be a perpendicular bisector of BD it must not only cut it at 90° (which it does) but also cut it into two equal parts. Now, in order AC to cut BD into two equal parts right triangle ABD must be isosceles, which, as it turns out after some math, it is not.

Attachment:

Important property: perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle.

Thus, the perpendicular AC divides right triangle ABD into two similar triangles ACB and DCA (which are also similar to big triangle ABD). Now, in these three triangles the ratio of the corresponding sides will be equal (corresponding sides are the sides opposite the same angles marked with red and blue on the diagram).

So, --> --> .

Answer: D.

In similar triangles the ratio of the corresponding sides are equal (corresponding sides are the sides opposite the same angles marked with red and blue on the diagram).

The ratios you are equating are not of corresponding sides. If you want to equate AD/AB then it should be AD/AB=AC/BC --> AD/5=4/3 --> AD=20/3. Also AD/AB=CD/AC --> (20/3)/5=CD/4 --> CD=16/3.

Hi Bunuel,

How did you infer that angle BAC = angle ADC. Could you please explain that part.. I know this can be proved thro similarity, however I wanted to understand from your concept, which you have mentioned below. Pls explain.

<B+<D+<A=180, since <A=90 then <B+<D=90;Similarly in triangle ABC: <B+<BAC=90 since <B=90-<D then (90-<D)+<BAC=90 --> <BAC=<D.

If 0<x<y, is y-x < 0.00005

(1) x>1/60,000(2) y<1/15,000

Notice that , and .

So, we can rewrite the question as:

If 0<x<y, is y-x<3

(1) x>1 --> if and then the answer is YES but if and the answer is NO. Not sufficient.(2) y<4 --> if and then the answer is YES but if and the answer is NO. Not sufficient.

(1)+(2) Remember we can subtract inequalities if their signs are in opposite directions --> subtract (1) from (2): --> . Sufficient.

Answer: C.

In the rectangular coordinate system shown above, which quadrant, if any, contains no point ( x, y ) that satisfies the inequality 2x-3y≤−6? (the quadrants are the standard quadrants in a co-ordinate system, I can't really draw it out here)

A. NoneB. ΙC. ΙID. ΙIIE. IV

--> . Thi inequality represents ALL points, the area, above the line . If you draw this line you'll see that the mentioned area is "above" IV quadrant, does not contains any point of this quadrant.

Else you can notice that if is positive, can not be negative to satisfy the inequality , so you can not

have positive , negative . But IV quadrant consists of such points for which is positive and negative. Thus answer must be E.Answer: E.

2x - 3y <= -6=> -2x + 3y >= 6 ---- (1)

equate the both sides of the inequality:-2x + 3y = 6=>x/(-3) + y/2 = 1

comparing with x/a + y/b = 1x intercept is -3 and y intercept is 2ie the line passes thru (-3,0) and (0,2)

This line obviously passes through the 1st, 2nd and 3rd quadrantand going back to the original inequality (1)we understand the points that satisfy this inequality would be above this line which means there is no chance that any of these points would lie in quadrant 4. So the ans is E

The sides of a rectangular region measured to the nearest centimeter are 8 cm and 6 cm. Out of the following which is the largest possible value of actual area of rectangle ?

A. 48 B. 55.25C. 52.29D. 44 E. 41.25

Given: and , where a and b are the sides of the rectangle.

So, the largest area the rectangle can possibly have is a bit less than . The correct answer would be the largest value among the options which is less than 55.25, so 52.29. For example the sides can be: 8.3 and 6.3 --> 8.3*6.3=52.29.Answer: C.

Is triangle ABC an isosceles?

(1) AB/BC = 2(2) x≠y

I thought it was C, because the only way for the triangle to be isosceles given both conditions is for BC = AC , and that's not possible because side AC + BC (assuming they are the same length) would equal AB. And the third side of the triangle must be greater than the sum of the other two sides or less than the difference of the other two sides. So given both conditions, should we be able to say for sure that the triangle is not isosceles?

1) AB/BC = 2 --> . Not sufficient on its own.

(2) x≠y --> . Not sufficient on its own.

(1)+(2) Since and , then the only way ABC to be isosceles is when . But in this case as given that AB=2BC then AB=BC+BC=BC+AC which is not possible because the length of any side of a triangle

must be smaller than the sum of the other two sides. So, , which means that ABC is not isosceles. Sufficient.

Answer: C.

n the xy coordinate plane, line L and line K intersect at the point (4,3). Is the product of their slopes negative?

(1) The product of the x-intersects of lines L and K is positive.

(2) The product of the y-intersects of lines L and K is negative. Sol:To draw a line with +ve slope through a point(x1, y1) on a 2D plane; draw two lines passing through the point perpendicular to each other with one line parallel to x-axis and the other parallel to y-axis(depicted by red lines in the images). These red-lines make 4-quadrants with origin(x1,y1). Any line that passes through this origin and lies on 1st and 3rd quadrant will have +ve slope. Any line passing through the point and lying on the 2nd and 4th quadrants will have negative slope.

Attached images show two possible scenarios for St1 and St2 making the statements insufficient individually. Last image shows the only possible scenario for the lines, making both the statements sufficient together.

St1:Insufficient.

St2:Insufficient.

Combined: Sufficient.

Ans: "C"

In the xy coordinate plane, line L and line K intersect at the point (4,3). Is the product of their slopes negative?

We have two lines: and . The question: is ?

Lines intersect at the point (4,3) --> and

(1) The product of the x-intersects of lines L and K is positive. Now, one of the lines can intersect x-axis at 0<x<4 (positive slope) and another also at 0<x<4 (positive slope), so product of slopes also will be positive BUT it's also possible one line to intersect x-axis at 0<x<4 (positive slope) and another at x>4 (negative slope) and in this case product of slopes will be negative. Two different answers, hence not sufficient.

But from this statement we can deduce the following: x-intersect is value of for and equals to

--> so --> .

(2) The product of the y-intersects of lines L and K is negative. Now, one of the lines can intersect y-axis at 0<y<3 (positive slope) and another at y<0 (positive slope), so product of slopes will also be positive BUT it's also possible one line to intersect y-axis at y<0 (positive slope) and another at y>3 (negative slope) and in this case product of slopes will be negative. Two different answers, hence not sufficient.

But from this statement we can deduce the following: y-intercept is value of for and equals to --

> .

(1)+(2) and . As numerator in is negative, then denominator must also

be negative. So . Sufficient.

Answer: C.

In fact we arrived to the answer C, without using the info about the intersection point of the lines. So this info is not needed to get C.

If BE || CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of trapezoid BEDC?

A. 12 B. 18 C. 24 D. 30 E. 48

Triangles ABE and ACD are similar (they share the same angle CAD and also as BE is parallel to CD then angles by BE and CD are equal, so all 3 angles of these triangles are equal so they are similar triangles).

Property of similar triangles: ratio of corresponding sides are the same: --> --> and .

So in triangle ABE sides are , and : we have 3-4-5 right triangle ABE (with right angle CAD, as hypotenuse is ) and 6-8-10 right angle triangle ACD.

Now, the --> .Answer: B.

Yes, any triangle whose sides are in the ratio 3:4:5 is a right triangle. Such triangles that have their sides in the ratio of whole numbers are called Pythagorean Triples. There are an infinite number of them, and this is just the smallest. If you multiply the sides by any number, the result will still be a right triangle whose sides are in the ratio 3:4:5. For example 6, 8, and 10.

A Pythagorean triple consists of three positive integers , , and , such that . Such a triple is

commonly written , and a well-known example is . If is a Pythagorean triple, then so

is for any positive integer .

Or

This is how I approached:

Since BE is II to CD AND B is mid point of AC therefore BE is Midsegment which means that BE = 1/2 (CD) = 5 and AE = ED = 4, Observing triangle ACD : its a 6,8,10 Right Triangle & therefore area is 1/2(b/h) = 24similarly traingle ABE : its a 3,4,5 triangle & therfore area is 1/2 (b/h) = 6

Therefore area of trapezoid is = 24 -6 = 18. Answer is B

If in triangle ABC point D lies on side AC, what is the value of angle BCA ?

(1) The value of angle ABD is half that of angle BDC(2) The value of angle DBC is 10 degrees

Let angle BCD = x

Evaluating statement 1 alone:

1) The value of angle ABD is half that of angle BDC

Let angle ABD = y. Then angle BDC = 2y.In triangle BDC,Measure of the exterior angle is equal to the sum of the interior opposite angles.(y+z) + x = 180 - y =>2y + z + x = 180

We need the values of y and z to determine x. Hence, not sufficient.We eliminate A and D.

Evaluating statement 2 alone:

2) The value of angle DBC is 10 degrees

z = 10But we don’t have the value of angle BDC.Hence we eliminate B.Combining both statements as well, we don’t have the value of y.Hence we eliminate C.Answer is E.

If distinct points A, B, C, and D form a right triangle ABC with a height BD, what is the value of AB times BC ?

(1) AB = 6. (2) The product of the non-hypotenuse sides is equal to 24.

BD is a height means that B is a right angle and AC is a hypotenuse (so BD is a height from right angle B to the hypotenuse AC). Question thus asks about the product of non-hypotenuse sides AB and BC.

(1) AB = 6. Clearly insufficient.(2) The product of the non-hypotenuse sides is equal to 24 --> directly gives us the value of AB*BC. Sufficient.Answer: B.

In triangle ABC, point X is the midpoint of side AC and point Y is the midpoint of side BC. If point R is the midpoint of line segment XC and if point S is the midpoint of line segment YC, what is the area of triangular region RCS ?

(1) The area of triangular region ABX is 32. (2) The length of one of the altitudes of triangle ABC is 8.

This question makes use of the Midpoint theorem in case of triangles. According to the theorem, the segment joining the midpoints of two sides of a triangle is half the length of the third side and the smaller triangle thus formed is similar to the original triangle. The ratio of sides of the smaller tr to the larger tr = 1/2

=> A(smaller tr) : A(Larger tr) = 1:4From the given info we have:A(CYX) : A(ABC) = 1:4A(CSR) : A(CYX) = 1:4=> A(CSR) = 1/16 * A(ABC)

ST 1:A(ABX) = 1/2 * A(ABC) ....... Since they have the same height and the base of ABX is half the base of ABCThus from A(ABX), we can calculate A(CSR) => A(ABX)/8 = 4=> SUFFICIENTST 2:We cannot deduce anything from the length of one of the heigths.=> NOT SUFFICIENTANS: A

Notice that XY is the midsegment of triangel ABC and RS is the midsemgent of triangle XYC (midsegment is a line segment joining the midpoints of two sides of a triangle).

Several important properties:

1. The midsegment is always half the length of the third side. So, and --> ;2. The midsegment always divides a triangle into two similar triangles. So, ABC is similar to XYC and XYC is similar to RSC --> ABC is similar to RSC, and according to above the ratio of their sides is 4:1;

3. If two similar triangles have sides in the ratio , then their areas are in the ratio (or in another way in two

similar triangles, the ratio of their areas is the square of the ratio of their sides: .). So, since ABC is

similar to RSC and the ratio of their sides is 4:1 then , so the area of ABC is 16 times as large as the area of RSC;

4. Each median divides the triangle into two smaller triangles which have the same area. So, since X is the midpoint of AC then BX is the median of ABC and the area of ABX is half of the area of ABC. From the above we have that the area of ABX is 16/2=8 times as large as the area of RSC.

So, to find the area of RSC we need to find the area of ABX.

(1) The area of triangular region ABX is 32 --> the area of RSC=32/8=4. Sufficient.(2) The length of one of the altitudes of triangle ABC is 8. Only altitude is no use to get the area. Not sufficient.Answer: A.

An equilateral triangle that has an area of is inscribed in a circle. What is the area of the circle?A. 6pi B. 9pi C. 12pi D. 9pi 3^1/2 E. 18pi 3^1/2I'd recommend to know the ways some basic formulas can be derived in geometry rather than memorizing them.

Anyway, , where is the length of a side --> as given

that then ;Now, given that this triangle is inscribed in circle (not circle is inscribed in triangle). The radius of the circumscribed

circle is (you used the formula for the radius of the inscribed circle ) --

> .Answer: C.

=========

A thin piece of wire 40 meters long is cut into two pieces. One piece is used to form a circle with radius r, and the other is used to form a square. No wire is left over. Which of the following represents the total area, in square meters, of the circular and the square regions in terms of r?

A.

B.

C.

D.

E.

The area of a circle will be - and meters of wire will be used;

There will be meters of wire left for a square. Side of this square will be , hence the

area of the square will be .

The total area will be - .Answer: E.

Very minimal calculations are required if you assume a convenient value for R e.g. R = 7/2 since

Circumference in this case So length of leftover wire = 18Side of square = 18/4

Area of square =

Now put R = 7/2 in the second half of the options you want to check.i.e. say you want to check option (E)

Option (E) is correct since the area of the square is (9/2)^2 when R = 7/2 (as shown above)

-- The length of each piece = 20;

Circumference of the circle is --> --> ;Perimeter of the square --> --> ;

The total area is . Now, you should substitute the value of in the answer choices and see which one

gives . Answer choice E works.

The area bounded by the curves |x + y| = 1, |x - y| = 1 is A. 3B. 4 C. 2D. 1E. None

|x+y|=1 represents two lines: x+y=1 and x+y=-1 --> y=1-x and y=-1-x. Find the x and y intercept of these lines to plot;|x-y|=1 represents two lines: x-y=1 and x-y=-1 --> y=x-1 and y=x+1. Find the x and y intercept of these lines to plot;

Notice that these lines are mirror images of each other. Here is a square you get when you plot them:

Notice that the diagonal of this square is equal to 2 (the difference between x intercepts). Area of a square is diagonal^2/2=2^2/2=2.

Answer: C.

If equation |X| + |Y|= 5 encloses a certain region on the coordinate plane, what is the area of that region?

5102550100

Below is the the region we get by joining X and Y intercepts which are (0, 5), (0, -5), (5, 0), (-5, 0):

Now, diagonals of the rectangle are equal (10 and 10), and also are perpendicular bisectors of each other (as the are on X and Y axises), so it must be a square. Area of a square equals to diagonal^2/2=10^2/2=50.

Answer: D.

You can find the area of a square in either way: .

The difference between the square given by |x+y|+|x-y|=4 and the square given by |x|+|y| =5 is that the first one is "horizontal" square and the second one is the square turned by 45 degrees.|x+y|+|x-y|=4:

|x|+|y| =5:

We are not assuming the values for x and y. We are expanding |x| + |y| = 5 to get the figure.

If x>0 and y>0 we'll have x+y=5;If x>0 and y<0 we'll have x-y=5;If x<0 and y>0 we'll have -x+y=5;If x<0 and y<0 we'll have -x-y=5;

So we have 4 lines (4 linear equations which will give 4 segments) when drawn these lines give the square shown in my original post. But there is an easier way to get this figure: get X and Y intercepts (which are (0, 5), (0, -5), (5, 0), (-5, 0)) and join them.

Well it cannot be a circle because it's a square. The reason why it's a square is given in previous posts (you can also see a diagram there).

Next, in an xy plane, the circle with center (a, b) and radius r is the set of all points (x, y) such that:

. There is no resemblance whatsoever between this equation and the equation given in the question.

In the x-y plane, the area of the region bounded by the graph of |x+y| + |x-y| = 4 is

A. 8B 12C. 16D. 20E. 24

Yes, it can be done by graphing. |x+y| + |x-y| = 4 can expand in four different wasy:

A. x+y+x-y = 4 --> x=2B. x+y-x+y = 4 --> y=2C. -x-y +x-y= 4 --> y=-2D. -x-y-x+y=4 --> x=-2

So you can draw all these four lines x=2, x=-2, y=2, y=-2 to get a square with the side of 4:

Attachment:

Why cant we consider (4,0) and (0,4) as points on graph ? then area would be different... , right?

First of all we are not considering points separately, as we have X-Y plane and roots of equation will represent lines, we'll get the figure bounded by this 4 lines. The equations for the lines are:x=2x=-2y=2y=-2

This lines will make a square with the side 4, hence area 4*4=16.Second: points (4,0) or (0,4) doesn't work for |x+y| + |x-y| = 4.

Need help in solving equations involving Mod...... help?

OK there can be 4 cases:|x+y| + |x-y| = 4

A. x+y+x-y = 4 --> x=2B. x+y-x+y = 4 --> y=2C. -x-y +x-y= 4 --> y=-2D. -x-y-x+y=4 --> x=-2

The area bounded by 4 graphs x=2, x=-2, y=2, y=-2 will be square with the side of 4 so the area will be 4*4=16.

In our original question when we were solving the equation |x+y| + |x-y| = 4 each time x or y were cancelling out so we get equations of a type x=some value twice and y=some value twice. And these equations give the lines which are parallel to the Y or X axis respectively so the figure bounded by them is a "horizontal" square (in your question it's "diagonal" square).

If equation |x/2|+|y/2| = 5 encloses a certain region on the coordinate plane, what is the area of this region?A. 20B. 50C. 100D. 200E. 400

First of all to simplify the given expression a little bit let's multiply it be 2: --> .Now, find x and y intercepts of the region (x-intercept is a value(s) of x for y=0 and similarly y-intercept is a value(s) of y for x=0):

--> --> and ;

--> --> and .

So we have 4 points: (10, 0), (-10, 0), (0, 10) and (-10, 0).

When you join them you'll get the region enclosed by :

You can see that it's a square. Why a square? Because diagonals of the rectangle are equal (20 and 20), and also are perpendicular bisectors of each other (as they are on X and Y axis), so it must be a square. As this square has a

diagonal equal to 20, so the .

Or the --> .Answer: D.

Given:

You will have 4 case:

and --> --> ; and --> --> ;

and --> --> ; and --> --> ;

A circular mat with diameter 20 inches is placed on a square tabletop, each of whose sides is 24 inches long. Which of the following is closest to the fraction of the tabletop covered by the mat?

A. 5/12B. 2/5

C. 1/2D. 3/4E. 5/6

( --> );

;

.

Now this fraction is obviously between 1/2 and 3/4 --> and --> 157 is closer to 144 than to 216.Answer: C.

In the coordinate plane, a circle has center (2, -3) and passes through the point (5, 0). What is the area of the circle?

A. 3πB. 3√2πC. 3√3πD. 9πE. 18π

The point is that the radius does not equal to 3, it equals to . You can find the length of the radius either with

the distance formula (the formula to calculate the distance between two points and

is ) or with Pythagoras theorem. Look at the diagram below:

Attachment:

The radius of the circle is the hypotenuse of a right isosceles triangle with the legs equal to 3: --> .

Answer: E.

In the figure above, if x and y are each less than 90 and PS||QR, is the length of segment PQ less than the length of segment SR ?

(1) x>y(2) x+y>90

So, the lines PS & RQ are parallel, angles X & Y are each less than 90 degrees.

Statement #1 tells us angle X is greater than angle Y. In the diagram, I showed an exaggerated example of this ---- if Y is a much smaller angle, it follows a less steep diagonal, which travels a longer distance between the two lines, as shown in the diagram. Therefore, if (angle X) > (angle Y), then segment RS is longer than segment PQ. Statement #1 is sufficient.

(Notice that this logic depends on both angles staying less than 90 degrees. If X had a value greater than 90 degrees, it would start making a longer, less steep, line segment on the left side.)

Statement #2 says only that the sum (x + y) is greater than 90. The trouble with that is: it doesn't give us any way to distinguish x vs. y --- either one could be much bigger than the other, or they both could be equal. No way to distinguish x vs. y ==> no way to distinguish RS vs. PQ. Insufficient.

Correct answer = A

(1) x>y --> if the angles x and y were equal then the length of segment PQ would be equal to the length of segment SR (as PS||QR). Now, as x>y it means that point R is to the left of the position it would be if x and y were equal (previous case), or in other words, we should drag point R to the left to the position of R2 to make angle y less than x, thus making the length of segment SR bigger than the length of segment PQ. So as x>y than SR>PQ. Sufficient.

(2) x+y>90 --> clearly insufficient: if then the length of segment PQ would be equal to the length of segment SR but if and then the length of segment PQ would be less than the length of segment SR. Not sufficient.

Note that the shortest distance between two parallel lines is the perpendicular distance. As the angle keeps decreasing, the length of the line keeps increasing. So if we know that x>y, then PQ < SR. Stmnt 1 Sufficient.

Attachment:

Since statement 2 doesn't give any information about relative size of x and y, nothing can be said about PQ and SR. Not sufficient.

You have 6 sticks of lengths 10, 20, 30, 40, 50, and 60 centimeters. The number of non-congruent triangles that can be formed by choosing three of the sticks to make the sides is

A. 3B. 6C. 7D. 10E. 12

The length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides.

Based on this there can be only 7 triangles formed: (20, 30, 40), (20, 40, 50), (20, 50, 60), (30, 40, 50), (30, 40, 60), (30, 50, 60), (40, 50, 60).Answer; C

The points R, T, and U lie on a circle that has radius 4. If the length of arc RTU is 4*PI/3, what is the length of line segment RU?

A. 4/3B. 8/3C. 3D. 4E. 6

The circumference of a circle= , . --> Angle degrees (C center of the circle).

RCU is isosceles triangle as and degrees. Hence .Answer: D.

Let's call the center of the circle O; we then have a triangle ORU. Notice that OR and OU are both a radius, so they are equal in length, so this triangle must be isosceles, and the angles at R and U must be equal.

The circumference of the circle is 8*Pi. So if arc RTU is 4*Pi/3, then arc RTU is 1/6th of the circle. Thus the angle ORU is 1/6th of 360 degrees, so is 60 degrees. Now the angles at R and U must be equal, and since the angles in this triangle must add to 180, the angles at R and U must both be 60 degrees. So ORU is in fact equilateral, and every side is 4 long.

In the figure above, the area of square region PRTV is 81, and the ratio of the area of square region XSTU to the area of square region PQXW is 1 to 4. What is the length of segment RS?A. 5B. 5.5C. 6D. 6.5E. 7As the area of square region PRTV is 81 then --> ;

The area of square region and the area of square

region ;

Given: --> --> .Answer: C.

Or: as the ratio of the areas of squares is the square of the ratio of their sides then -->

As the area of square region PRTV is 81 then --> ;

The area of square region and the area of square

region ;

Given: --> --> .Answer: C.

Or: as the ratio of the areas of squares is the square of the ratio of their sides then -->

In the xy-plane, if line k has negative slope and passes through the point (−5,r ), is the x-intercept of line k positive?

(1) The slope of line k is –5.(2) r > 0

This question can be done with graphic approach (just by drawing the lines) or with algebraic approach. 

Algebraic approach:

Equation of a line in point intercept form is  , where:   is the slope of the line,   is the y-intercept of the line (the value of   for  ), and   is the independent variable of the function  .

We are told that slope of line   is negative ( ) and it passes through the point (-

5,r):   -->  .

Question: is x-intercept of line   positive? x-intercep is the value of   for   --

>   --> is  ? As we know that  , then the question basically becomes: is  ?.

(1) The slope of line   is -5 -->  . We've already known that slope was negative and there is no info about  , hence this statement is insufficient.

(2)   -->   -->  , as   we have that   is more than some negative number ( ), hence insufficient, to say whether  .

(1)+(2) From (1)   and from (2)   -->   -->  . Not sufficient to say whether  .

Answer: E.

Graphic approach:If the slope of a line is negative, the line WILL intersect quadrants II and IV. X and Y intersects of the line with negative slope have the same sign. Therefore if X and Y intersects are positive, the line intersects quadrant I; if negative, quadrant III.

When we take both statement together all we know is that slope is negative and that it crosses some point in II quadrant (-5, r>0) (this info is redundant as we know that if the slope of the line is negative, the line WILL intersect quadrants II). Basically we just know that the slope is negative - that's all. We can not say whether x-intercept is positive or negative from this info.

Below are two graphs with positive and negative x-intercepts. Statements that the slope=-5 and that the line crosses (-5, r>0) are satisfied.

-------------

:

A ladder of a fire truck is elevated to an angle of 60° and extended to a length of 70 feet. If the base of the ladder is 7 feet above the ground, how many feet above the ground does the ladder reach?

A. 35B. 42C. 35 root 3

D. 7 + 35 root 3E. 7 + 42 root 3

Triangle ABC is a 30°-60°-90° triangle. Now, in a right triangle where the angles are 30°, 60°, and 90° the sides are always in the ratio , the leg opposite 30° (AC) corresponds with , the leg opposite 60° (BC) corresponds

with and the hypotenuse AC corresponds with 2. So, --> --> .

Hence the leader reaches above the ground.

Answer: D.

Does the equation y = (x – p)(x – q) intercept the x-axis at the point (2,0)?

(1) pq = -8(2) -2 – p = q

x-intercepts of the graph is the values of for which . So, the x-intercepts

are and . The question basically asks whether either or equals 2.

(1) pq = -8. Not sufficient to say whether p or q equals 2.(2) -2 – p = q. Not sufficient to say whether p or q equals 2.

(1)+(2) Solving and gives us that either and OR and . In

either case one of the unknowns is 2, so intercepts the x-axis at the point (2,0). Sufficient.Answer: C.