Similar Triangles. To solve a proportions Cross multiply Solve

GEOMETRY FINAL REVIEW

Similar Triangles

PROPORTIONS To solve a proportions

Cross multiplySolve

EXAMPLE

SIMILAR FIGURES1. Write the proportions for

corresponding sides

2. Solve the proportion

EXAMPLE

EXAMPLE

EXAMPLE

GEOMETRIC MEANThe geometric mean of two numbers is the square root of the product of the numbers.

a b

EXAMPLEFind the geometric mean of the two numbers. Simplify your answer.1. 7 and 35

GEOMETRY FINAL REVIEW

Right Triangles

PYTHAGOREAN THEOREMa2 + b2 = c2

ca

b

EXAMPLE

TRIGONOMETRIC RATIOSSin A =

Cos A =

Tan A =

hypotenuse

opposite

hypotenuse

adjacent

adjacent

opposite

EXAMPLEFind the value of sine, cosine, and tangent ratios to the nearest hundredth.

SOLVING TRIANGLES FOR ALL MISSING PARTSIf you know two sides and one angle: To find the missing side use Pythagorean

theorem:a2 + b2 = c2

To find the two angles use inverse trigonometric functions:

Angle = Sin -1

Angle = Cos -1

Angle = Tan -1

hypotenuse

opposite

hypotenuse

adjacent

adjacent

opposite

If you know two angles and one side: To find the missing angle:

Add the two angles and subtract from 1800

To find the two missing sides use the trigonometric ratios.

Sin angle =

Cos angle =

Tan angle =

hypotenuse

opposite

hypotenuse

adjacent

adjacent

opposite

EXAMPLESolve each triangle for all the missing information. Round your answer to the nearest tenth.

EXAMPLESolve each triangle for all the missing information. Round your answer to the nearest tenth.

GEOMETRY FINAL REVIEW

Circles

ARCS & ANGLES Central Angle

Angle = Arc

Inscribed AngleAngle = ½ Arc

Inside the circleAngle = ½ (sum of the arcs)

Outside the circleAngle = ½ (difference of the arcs)

EXAMPLEFind the measure of the missing angle or

arc.

EXAMPLEFind the measure of the missing angle or

arc.

EXAMPLEFind the measure of the missing angle or

arc.

CHORDS & SECANTS Inside the circle

Part ∙ Part = Part ∙ Part

Outside the circle Outside ∙ Whole = Outside ∙

Whole

E

C

AB

D

B

EC

D

A

A

EC

D

EXAMPLESolve for x.

EXAMPLESolve for x.

ARC LENGTH/CIRCUMFERENCE C = 2pr

B

A

03602

mAB

r

ABoflengthArc

EXAMPLE:Find the length of arc BC. Leave your answer in terms of .p

7 in

40o

SECTOR AREA/AREA A = p r2

P

D

C

02 360

mAB

r

A

EXAMPLE:Find the area of the shaded region. Leave your answer in terms of .p

7 in300

GEOMETRY FINAL REVIEW

Area of Polygons

AREA/PERIMETERY OF POLYGONS Circle

C = 2 p rA = p r2

SquaresP = 4sA = s2

RectanglesP = 2L + 2WA = L W

r

s

L

W

AREA/PERIMETERY OF POLYGONS Parallelograms

P = 2b + 2lA = b h

TrapezoidsP = add up all 4 sidesA = ½ (b1 + b2) h

TrianglesP = a + b + cA = ½ b h

b

lh

b2

b1

h

a

c

b

AREA OF EQUILATERAL TRIANGLES

P = 3sA = ¼ s2 √3

s

AREA OF REGULAR POLYGONS Given side length and apothem

P = n sA = n [ ½ (s)(a)]

a s

AREA OF REGULAR POLYGONS Given side length only

P = n s To find a

1. = 360/2b n2. x = s/23. a = x/tan b4. A = n [ ½ (s)(a)] sb

EXAMPLE:Find the perimeter and area of each

polygon.1.

2.

Find the area of the regular polygons.1.

2 ft

GEOMETRY FINAL REVIEW

Surface Area & Volume

IMPORTANT TERMS FOR SURFACE AREA & VOLUME B = Area of the Base

Base is the shape not like the others Base does not mean the bottom shape Base is not one number it is an area (use the

previous chapter)

P = Perimeter of the Base

h = height of the polyhedron

l = slant height of the polyhedron

SURFACE AREA & VOLUME FORMULAS Prism

SA = 2B + Ph

V = Bh

PyramidSA = B + ½ P l

V = 1/3 B hB

h l

SURFACE AREA & VOLUME FORMULAS Cylinder

SA = 2 p r2 + 2 p r h

V = p r2 h

ConeSA = p r2 + p r l

V = 1/3 p r2 h r

h l

h

r

SURFACE AREA & VOLUME FORMULAS Sphere

SA = 4 p r2

V = 4/3 p r3

r

GEOMETRY WARM-UPAPRIL 29, 2014Find the surface area and volume of the right prism.1.

5 in9 in

2 in

GEOMETRY WARM-UPAPRIL 29, 2014Find the surface area and volume of the right pyramid.1.

7 in7 in

9 in