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8/12/2019 Maths Prev2

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Question 14

Sol:Given, points are collinear

Question 15


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Question 21

Question 22


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Question 23

Question 24


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Justification: Here, DE || CA

and

Steps of construction:

(i) Draw a triangle ABC with side BC = 6 cm , AB = 5 cm and ABC = 600

(ii) Draw any ray AX making an acute angle with BC on the side opposite to the vertex A.(iii) Along AX, mark off 4 points A1, A2, A3and A4on AX such that AA1 = A1A2=A2A3=A3A4.

(iv) Join A4C and draw a line through A3parallel to A4B intersecting AB to B

(v) Draw a line through B parallel to the line BC to intersect AC at C. Then ABC is the required triangle.

Justification:A4B || A3B [By construction]

[By Basic proportionality theorem]

But [By construction]

.. (i)

BC||BC [By construction]

ABC ~ABC [By AA similarity criterion]

(from (i))

(By Basic proportionality theorem)

Question 17


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17.Three consecutive vertices of a parallelogram are (2, 1); (1, 0) and (4, 3). Find the coordinates of the fourthvertex.

Sol:Let the vertices of the parallelogram be

A = (2, 1), B =(1, 0) , C = (4, 3) and D= (x, y)In parallelogram, diagonals bisect each other.

i.e., Mid point of AC = Mid point of BD

(1, 1) =

,

=2

Hence, the fourth vertex = (1, 2)

Question 18

18.If the point C (1, 2) divides internally the line segment AB in the ratio 3 : 4, where the coordinates of A are (2,5), find the coordinates of B.

Sol:Let the coordinates of B be (x, y)Given AC : CB = 3 : 4 (internally)

Co ordinates of C =

Hence, the coordinates of B are ( 5, 2)

Question 3

3.In an A.P., the sum of its first n terms is n2+2n. Find its 18thterm.Sol:

GivenReplacing n by (n 1), we get

Now

From (1) and (2)

To get AP, substituting n = 1, 2, 3,. . respectively in (3), we get

.(from (3))

Hence, AP is 3, 5, 7, 9,

Now 18thterm =Hence, 18thterm = 37.

Question 17

17.Three consecutive vertices of a parallelogram are (2, 1); (1, 0) and (4, 3). Find the coordinates of the fourthvertex.


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Sol:Let the vertices of the parallelogram be

A = (2, 1), B =(1, 0) , C = (4, 3) and D= (x, y)In parallelogram, diagonals bisect each other.i.e., Mid point of AC = Mid point of BD

(1, 1) =

,

=2

Hence, the fourth vertex = (1, 2)

Question 18

18.If the point C (1, 2) divides internally the line segment AB in the ratio 3 : 4, where the coordinates of A are (2,

5), find the coordinates of B.Sol:Let the coordinates of B be (x, y)Given AC : CB = 3 : 4 (internally)

Co ordinates of C =

Hence, the coordinates of B are ( 5, 2)

Question 23

23.A hemispherical bowl of internal diameter 36 cm is full of some liquid. This liquid is to be filled in cylindricalbottles of radius 3 cm and height 6 cm. Find the number of bottles needed to empty the bowl.

Sol:

Given 2r = 36

r = 18

Volume of the hemi-sphere =

Now volume of the cylinder = r2h

The number of cylinders that can be filled with cylinder


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= 72

Hence the number of bottles to empty the bowl = 72 bottles

(OR)

Water flows out through a circular pipe whose internal radius is 1 cm, at the rate of 80 cm/second into an emptycylindrical tank, the radius of whose base is 40 cm. By how much will the level of water rise in the tank in half anhour ?Sol:Rate of flow of water = 80 cm/secRadius of cylindrical PIPE = r = 1 cmRadius of cylindrical Tank = R = 40 cmTime = 30 minutes = 30 60 = 1800 secLength of the pipe = h = 80 30 60 = 144000 Let the height of the water level raised = HVolume of water flows from a pipe = Volume of water in a cylindrical tank

Hence, the level of water rise up to a height of 90 cm

Question 24

24.A pole 5 m high is fixed on the top of a tower. The angle of elevation of the top of the pole observed from apoint A on the ground is 60 and the angle of depression of point A from the top of the tower is 45. Find the

height of the tower. (Take = 1.732)

Sol:Let BC = height of the Tower = x

CD = height of the pole = 5 mAngle of elevation = BAD = 600Angle of depression = CAB = 450

Let AB = y

Consider BAD,

-----------(1)

ConsiderCAB

-------------(2)Substituting (2) in (1)


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Rationalising the denominator

'>

Hence, the height of the tower = 6.83 m

Question 3

3.In an A.P., the sum of its first n terms is n2+2n. Find its 18thterm.Sol:

GivenReplacing n by (n 1), we get

Now

From (1) and (2)

To get AP, substituting n = 1, 2, 3,. . respectively in (3), we get

.(from (3))

Hence, AP is 3, 5, 7, 9,

Now 18thterm =Hence, 18thterm = 37.

Question 7

7.A box contains 5 red balls, 4 green balls and 7 white balls. A ball is drawn at random from the box. Find theprobability that the ball drawn is(a) white.(b) neither red nor white.

Sol:Number of balls in the box = 5 + 4 + 7 = 16

n(S) = 16(i) Let A be the event of drawing a white ball

n(A) = 7

Probability of drawing a white ball =(ii) Let B be the event of getting a ball which is neither red nor white.The selection can be made from green balls.

n(B) = 4

Probability of drawing neither white nor red ball


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.Question 13

13.A toy in the form of a cone mounted on a hemisphere with same radius. The diameter of the base of the

conical portion is 7 cm and the total height of the toy is 14.5 cm. Find the volume of the toy. (use = )

Sol:

Given AB = 7 cm

Total height = h = 14.5 cmVolume of the toy = Volume of the cone + volume of the hemi sphere

=

=

=

=

=

=

Question 5

5. The 5thterm of an Arithmetic Progression (A.P.) is 26 and the 10th term is 51. Determine the 15thterm of theA.P.Sol: Given, in an A.P, the 5thterm = 26 and 10th term = 51Let a is the first term and d is the common difference. Then,

----- (i)

----- (ii)

Subtract equation (i) from (ii)

Substitute d in equation (i)

15thterm


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Question 6

6. Find the sum of all the natural numbers less than 100 which are divisible by 6.Sol: Natural numbers less than 100 and divisible by 6 are 6, 12, 18, ..., 96.The sequence 6, 12, 18, ..., 96 is an A.P.

We know,

Hence, the sum of the natural numbers less than 100 which are divisible by 6 is 816.

Question 16

16. Find the co-ordinates of the point equidistant from these points

Sol: Let is equidistant from A, B and C then PA = PB = PCPA= PB

Hence, the required point is

OR

Show that the points are the vertices of a square.

Sol: Given points


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i.e., all the four sides are equal and diagonals are also equal.

Hence, the given points form a square.

Question 17

17. Find the value of p for which the points are collinear.

Sol: Given the points

Question 6

6.The 8 th term of an Arithmetic progression is zero. Prove that its term is triple its th term.

Sol:

Let t1be the 1stterm of the AP and d be the common difference,

be the nth term of this AP.

According to question,

i.e.

or, ..(1)

Now,

=

[using (1)]


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term of this AP is triple of its term

Question 13

13.The base radius and height of a right circular solid cone are and respectively. It is melted and

recast into spheres of diameter each. Find the number of spheres so formed.Sol:

Base radius = and

height =

Volume of this right circular solid cone is

The volume of the sphere having radius is


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Question 15

16.Prove that the points are vertices of a right isosceles triangle.

Sol:

Let ,

Now,

And

AOB is the right angled isosceles triangle.

(OR)

If the point P(x, y) is equidistant from the points and prove thatSol:

is equidistant from the points


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Question 17

17. The line joining the points and is trisected at the points P and Q. If point P lies on the line

, find the value of k.Sol:

Let

Let P be

Then,

P lies on the line

Satisfies the equation

Question 22

22.Two pillars of equal height stand on either side of a roadway which is wide. From a point on the

roadway between the pillars, the elevations of the top of the pillars are and Find the height of thepillars and the position of the point.

Sol:

AB and CD are pillars of equal height.

Width of the road

Let M be a point on the roadway AC straight the angles of elevation of the

pillar AB from M is and that of CD from M is

Let,


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Let the height of both the pillars AB and CD be .

According to question,

=

(1)

.. (2)Dividing (1) by (2) we get,

=

and

The point M is away from the foot of the perpendicular of the pillar

From (1) we get,

or,

height of the pillar will beOr

A man on the deck of a ship is 10 m above water level. He observes that the angle of elevation of the top of a hill

is and the angle of depression of the base of the hill is Calculate the distance of the hill from the shipand the height of the hill.Sol:

Let

height of the man from water level


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In right angled triangles BMD,

=

Distance of the hill from ship isAlso in the right angled triangle AMD,

=

Height of the hill

Question 23

23.A tent is in the form of a cylinder of diameter and height , surmounted by a cone of equal base

and height . Find the capacity of the tent and the cost of canvas for making the tent

at .Sol:

Capacity (volume)of the tent; ;

where, , ,

V

Again, the total surface area of the tent

Where , ,


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So,

Therefore the cost of canvas for making the tent at per sq m is

OR,

If the radii of the ends of a bucket, high, are and determine the capacity and total surfacearea of the bucket.Sol:

Volume (capacity)of the bucket which is the frustum of a cone is given by,

;

Where, ; ; and

Therefore,

Again, the total surface area of the bucket

Where,

=49.6 cm, and

Therefore,

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