Circles - Cheat Sheet Topic Overview

Center Radius Form, for circle centered at (h,k) with radius, r(x−h)2+( y−k )2=r2

Completing the square for center-radius form1. Move loose numbers to one side

2. Group x’s and y’s3. Divide middle term by 2 and square it – ADD TO BOTH SIDES!

4. put factors into Squared Form ( )2 ( remember the number will be half of the middle term)5. You’re in center-radius form!!!

We complete the square twice to put general form equations of circles into Center-Radius form, then graph! Recognize a circle by finding an x2

Systems with CirclesAny point of intersection is a solution to the system – solve graphically!

Systems with ParabolasAny point of intersection is a solution to the system – solve graphically!

Watch out for sneaky turning points and sneaky solutions- know how to manipulate your calculator!( 12-6 was a quiz!)Area of a Sector

Areaof a Sector=Area of a˚

( angle measure of sector360 )

Area of a circle = π r2

Arc Length of a Sector (In degrees)

Arc Length=Circumference( angle measure of sector360 )

Circumference of a circle = πd

Solving for arc length IN RADIANSs = rθ where: s = arc length; r = radius; θ = central angle

Radians – unit of angle measureAn angle is 1 radian when the length of the arc of the circle is equal to the radius

ConversionsSet up a proportion and solve for desired angle measure!

radiansdegrees

= π180

Formulas for area and circumference are on the reference sheet.

KNOW DIFFERENCE B/T AREA and LENGTH (CIRCUMFERENCE)

Part 1- Circles

Proving Circles Similar

1. Translate to get centers to coincide2. Dilate with center of dilation at one of the centers and scale factor using

the radii.

Circle Regents Type questions

1. The equation of a circle is . What are the coordinates of the center and the length of the radius of the circle?1) center and radius 42) center and radius 43) center and radius 164) center and radius 16

2. A circle with a radius of 5 was divided into 24 congruent sectors. The sectors were then rearranged, as shown in the diagram below.

To the nearest integer, the value of x is1) 312) 163) 124) 10

3. In the diagram below of circle O, the area of the shaded sector AOC is and the length of is 6 inches. Determine and state .

YOU TRY!

4. If is the equation of a circle, the length of the radius is1) 252) 163) 54) 4

5. Triangle FGH is inscribed in circle O, the length of radius is 6, and .

What is the area of the sector formed by angle FOH?1)2)

3)4)

6. In the diagram below of circle O, the area of the shaded sector LOM is .

If the length of is 6 cm, what is ?1) 10º2) 20º3) 40º4) 80º

Circle THEOREMS - Cheat SheetCENTRAL ANGLES

VERTEX MUST BE ON THE CENTER OF THE CIRCLE.

RULE: ANGLE = INTERCEPTED ARC

NSCRIBED ANGLESVERTEX MUST BE ON THE CIRCLE.

RULE: ANGLE =HALF THE INTERCEPTED ARC

ANGLES FORMED BY 2 CHORDS

“BOW –TIE ANGLES”VERTEX is NOT on the center and NOT

on the circle.

RULE: ARC¿1+ ARC ¿2 ¿

2

Cyclic Quadrilaterals

Quadrilateral inscribed in a circleOPPOSITE ANGLES ARE SUPPLEMETARY

(ADD UP tO 180 )

SPECIAL INSCRIBED ANGLESFormed by a Tangent and chord

Rule: ANGLE =HALF THE INTERCEPTED ARC

Sneaky AngleFormed by a secant and chord

Rule: Find the measure of inscribed adjacent angle and subtract from 180.

ANGLES FORMED BY TWO SECANTS, OR TWO TANGENTS OR A SECANT AND A TANGENT.

RULE: Outside ∡=FARC−NARC2

Thales Theorem

Watch out! This is not necessarily a parallelogram, so consecutive angles are not supplementary. ONLY OPPOSITE ANGLES.

SPECIAL CASEIf the chord is a

diameter, rt angles!

Special Case- “ Ice cream Cone”Narc + Outside angle = 180

1) Always solve for arcs first. (HIGLIGHT DIAMETERS!).KNOW HOW TO WORK WITH RATIOS!2) Solve the parts in the order they are presented in for the problem.3) GO SLOW.4) Highlight parts you're solving for (in different colors).5) Fill in/mark up the diagram

Parallel Chords

The arcs between two parallel chords are congruent

Congruent Chords

The arcs outside (subtended by) congruent chords are congruent

Segment Length INSIDE Circle (PP)

If a diameter/radius is perpendicular to a chord, then it bisects that chord.

Segment Length

OUTSIDE Circle

A right angle is formed at the point of tangency between a tangent and diameter/radius

Tangents that meet at one

exterior point are congruent:

Big Circles (Yesterday’s Review)!Similar Circles1. Translate (if necessary) first - STATE VECTOR2. Dilate at a center of dilation and a scale factor (image / pre-image)3. ConcludeCircle Proofs Tips

Mark up your diagram Come up with a plan Use your proof pieces

part x part = part x partor

pp = pp

3 common tangents

4 common tangents

whole x outer = whole x outer

orwo = wo

Look for congruent angles and congruent sides Always look for radii (always congruent) and Inscribed Angles!!!

Circle Regents Type questions

1. In circle O shown below, diameter is perpendicular to at point C, and chords , , , and are drawn.

Which statement is not always true?1)2)3)4)2

2. In the diagram shown below, is tangent to circle O at A and to circle P at C, intersects at B, , , and .

What is the length of ?1) 6.42) 83) 12.54) 16

3. In the diagram below, quadrilateral ABCD is inscribed in circle P.

What is ?1) 70°2) 72°3) 108°4) 110°

You try!

4. In the diagram below, , , , , and are chords of circle O, is tangent at point D, and radius is drawn. Sam decides to apply this theorem to the diagram: “An angle inscribed in a semi-circle is a right angle.”

Which angle is Sam referring to?1)2)3)4)

In the diagram below of circle O with diameter and radius , chord is parallel to chord .

If , determine and state .

Part 2- QuadrilateralsQuadrilateral Properties Cheat Sheet – Sides and Diagonals

Parallelogram Properties

2 sets of opposite sides are parallel2 sets of opposite sides are congruentDiagonals bisect each otherOpposite angles are congruent, Consecutive angles are supplementary

Rectangle Properties2 sets of opposite sides are parallel2 sets of opposite sides are congruentDiagonals bisect each other*Diagonals are congruent*Adjacent sides are perpendicular

Rhombus Properties

2 sets of opposite sides are parallel2 sets of opposite sides are congruentDiagonals bisect each other*All sides are congruent*Diagonals are perpendicularDiagonals bisect vertex angles

Square Properties2 sets of opposite sides are parallel2 sets of opposite sides are congruentDiagonals bisect each otherDiagonals are congruentAdjacent sides are perpendicularAll sides are congruentDiagonals are perpendicular

Trapezoid Properties

ONLY 1 set of opposite sides are parallelAngles between bases are supplentary

Isosceles Trapezoid PropertiesONLY 1 set of opposite sides are parallel*ONLY 1 set of opposite sides are congruent (legs)*Diagonals are congruent, BASE ANGLES ARE CONGRUENT

Tools:Slope:

m= y2− y1

x2−x1

or counting method

Midpoint:

M=(x1+x2

2,

y1+ y2

2)

Distance:d= √(x2−x1)

2+( y2− y1)2

Quadrilateral Family “Geome-tree”

Another way to think about it….

How you read this: All rectangles, rhombuses, and squares are parallelograms, but not all parallelograms are rectangles, squares and/or rhombuses. All squares are rectangles but not all rectangles are squares etc….

Types of Quadrilateral Regents Questions:

Helps prove:Helps prove:

Helps prove:

o Property Multiple Choice Questionso Using a property to solveo Coordinate Geometry Proofo Two-Column Quadrilateral Proof

Regents type Questions

1. Quadrilateral ABCD has diagonals and . Which information is not sufficient to prove ABCD is a parallelogram?1) and bisect each other.2) and 3) and 4) and

2. The diagram below shows parallelogram LMNO with diagonal , , and .

Explain why is 40 degrees.

3.

1) Coordinate Geometry Proof

2) Which statement is not always true about a rhombus? 1)

The diagonals are perpendicular

2)

The opposite sides are congruent.

3)

The adjacent sides are perpendicular.

4)

The opposite sides are parallel.

3) In the accompanying diagram of rhombus ABCD, Sides AB and BC are adjacent sides, If AB=4 x , and BC=x+3What is the value of x? What is the length of AB?

1) In the diagram below, quadrilateral STAR is a rhombus with diagonals and intersecting at E. , , , , , , and .

a) Solve for SR

b) Solve for RT

c) Solve for

2) Given: ABCD is a parallelogram. ∆ ADB≅ ∆ BCA . Prove ABCD is a rectangle.

Solids and 3D Shapes

KEY CONCEPTS IMPORTANT NOTES

14-1: CALCULATING AREA OF REGULAR

POLYGONS

Need to “break up the figure” into triangles

Steps:1. Calculate the apothem!2. If not already there, draw in the Apothem

(mark the right angles) and bisect the central angleto find the vertex angle of the small RIGHT triangle.

3. Bisect the base of the isosceles triangle to find the length of one side of the right triangle.

4. Use SOH CAH TOA to calculate the length of the apothem.5. Find the area of each triangle6. Multiply the area by the # of triangles in the regular polygon.

14-2: TRANSLATIONS

FORMING SOLIDS, PROPERTIES AND CROSS SECTIONS

Cylinders Formed by translation Base shape: circle (parallel and congruent shapes in the 3D figure) Lateral View: Rectangle (cross section perpendicular to the base) Base View: Circle (cross section parallel to the base)

Prisms Formed by translation Polygonal bases Named by their base shapes Lateral View: Rectangle (cross section perpendicular to the base) Base View: same as base shape (cross section parallel to the base)

Pyramid Formed by translation and dilation One base that is a polygon Named by base shape Lateral edges congruent

14-3: ROTATIONS FORMING SOLIDS, PROPERTIES AND CROSS SECTIONS

*Cylinders can be formed by rotations

Cone Formed by rotation of triangle Slant height: height from edge of base to top Base shape: circle (parallel and congruent shapes in the 3D figure) Lateral View: Triangle (cross section perpendicular to the base) Base View: Circle (cross section parallel to the base)

Sphere Formed by rotation of circle or semi-cirlce Lateral View: Circle (cross section perpendicular to the base) Base View: Circle (cross section parallel to the base) The great circle: largest circle within a sphere ; same diameter as sphere

14-4: VOLUME OF PRISMS AND

CYLINDERS, EQUAL VOLUMES,

CAVALIERI’S PRINCIPLE

Volume of Prisms and Cylinders: V = Bh

(B = area of the base; h is the height/depth of the prism/distance between the bases)

Cavalieri’s Principle: two of the same solid that have thesame base area and the same height, also have the same volume.

14-5: VOLUME OF PYRAMIDS, CONES

AND SPHERES

V(pyramid/cone) = 13

Bh

V(sphere) = 43

πr3

14-6: DENSITY AND SURFACE AREA

Population Density A ratio of the amount of a population that exists over a given area

Population Density = populationtotal area

Density of a 3D Solid A ratio that compares an object’s weight (mass) to the amount of space (volume) it takes up

Density = Mass

Volume

3 DIMENSIONAL SOLIDS PROPERTIESFeatures: Prism Cylinder Cone Pyramids SphereBase View Always

polygon Circle Circle Any Polygon Circle

Lateral ViewRectangle Rectangle Isosceles

TriangleIsosceles Triangle Circle

Formation Translating a polygon into 3 Dimensions

Translating a circle into 3 Dimensions

OrRotating a

Rotating a Triangle

Translation and Dilation of a polygon

Rotate a semi-circle

RectangleCross Section Parallel to the Base

Same 2D shape as base view

Cross Section Perpendicular to the Base

Same 2D shape as lateral view

Number of Bases

2 2 1 1 n/a

Example

Formulas (Volume)

V = (area of the base) height

V = (area of the base) height V =

13

(area of the

base) height

V =13

(area of the base)

height

Helpful Tips For Solids Be prepared for application problems Careful typing into your calculator (especially fractions) Don’t Round until the end of the problem Leave in terms of Pi until the end of the problem (sometimes pi will cancel) Use appropriate formulas If you see equal height or equal volume, that’s Cavalieri’s Principle! Be prepared to describe how solids are formed through transformation Be prepared to describe what cross sections look like

SOLIDS PRACICEA. When a circle is translated a __________________________ is the solid formed

B. When a cylinder is cut by a plane, parallel to it’s base, the cross section is a ____________________

C. All cross sections of a sphere are circles a) True b) False

D. When a right triangle is rotated, the solid formed is a _______________________

E. Find the volume of the following right circular cylinder. Round to the nearest tenth.

F. The volume of a right circular cone is 80π and a radius of 4. Find the altitude.

G. The radius of a sphere is 2 feet. Find the volume of the sphere in terms of π.

4. In a solid hemisphere, a cone is removed as shown. Calculate the volume of the resulting solid to the nearest hunderedth. In addition to your solution, provide an explanation of the strategy you used in your solution.

6. Find the area of a regular nonagon below.

1. Calculate the Surface area and volume of the following solid:

2.a) The surface area of the prism below is 102 cm2. Find x.

b) If 1 can of paint covers 30cm2, how many cans of paint do you need to buy to cover the prism above?

The water tower in the picture below is modeled by the two-dimensional figure beside it. The water tower is composed of a hemisphere, a cylinder, and a cone. Let C be the center of the hemisphere and let D be the center of the base of the cone.

If feet, feet, and , determine and state, to the nearest cubic foot, the volume of the water tower.

3. Two containers show below hold candies of the same size. Container A holds 75 candies and container B holds 160 candies. Given the dimensions below, which container has a smaller population density?

A contractor needs to purchase 500 bricks. The dimensions of each brick are 5.1 cm by 10.2 cm by 20.3 cm, and the density of each brick is . The maximum capacity of the contractor’s trailer is 900 kg. Can the trailer hold the weight of 500 bricks? Justify your answer.a) Notice how density is given in kilograms per meters cubed? 1st convert dimensions of the brick so it’s in meters too!

b) Now volume of a brick.

c) Fill into density formula; find the mass (weight) of one brick.

d) Can the trailer hold 500 bricks? Justify your answer!