Geometry Final Review

Solve for the given problems from Unit 1: If TV = 14x – 8, find TU.

• 3

• 5

• 29

• 42

Find the Distance between A and B given A(-4,1) and B(3,-1).

• 3.1

• 6.4

• 7.3

• 10.8

Tags

CCSS.HSG.GPE.B.7

Match the correct property name to the demonstrated property: If m<P + m<Q=90, then <p and< Q are complementary.

• Angle Addition postulate

• Definition of complementary angles

• Definition of a right angle

• Definition of vertical angles

• Definition of supplementary angles

Tags

CCSS.7.G.B.5

• Definition of supplementary angles

• Definition of complimentary angles

• Definition of vertical angles

• Angle Addition postulate

• Definition of a right angle

Tags

CCSS.7.G.B.5

Perform the transformations in the given problems from Unit 3: Triangle CDE with vertices C(2,-1), D(7,-4), and E (4,-6) (x, y)→(x – 3, y + 8)

• C’(6, -9) D’(10, -12) E’(7, -14)

• C’(10, 2) D’(15, -1) E’(12, -3)

• C’(-1, 7) D’(4, 4) E’(1, 2)

• C’(10, -4) D’(15, -7) E’(12, -9)

Tags

CCSS.HSG.CO.A.2

Square PQRS with vertices P(2,6), Q (6,5), R(5,1) and S(1,2): rotated 180°

• P’(2,6), Q’(6,5), R’(5,1) and S’(1,2)

• P’(6, -2), Q’(5, -6), R’(1, -5) and S’(2, -1)

• P’(-2, -6), Q’(-6, -5), R’(-5, -1) and S’(-1, -2)

• P’(-6, 2), Q’(-5, 6), R’(-1, 5) and S’(-2, 1)

Tags

CCSS.8.G.A.3

CCSS.HSG.CO.A.5

Rectangle GHIJ with vertices G(2, -1), H(7, -1), I(7, -4), and J(2, -4): reflected across y=x

• G’(-1, 2), H’(-1, 7), I’(-4, 7), and J’(-4, 2)

• G’(2, -1), H’(7, -1), I(7, -4), and J’(2, -4)

• G’(1, -2), H’(1, -7), I(4, -7), and J’(4, -2)

• G’(2, -1), H’(-7, 1), I(-7, 4), and J’(-2, 4)

Tags

CCSS.8.G.A.3

CCSS.HSG.CO.A.5

Find the value of <C and BC.

• a. 60

• b. 120

• c. 9

• d. 180

Tags

CCSS.HSG.CO.C.10

Find the measure of the exterior angle.

• 180

• 4

• 77

• 18

Tags

CCSS.8.G.A.5

Line BT is a perpendicular bisector of triangle ABC. Solve for x.

• -2

• 2

• 10

• -10

Tags

CCSS.HSG.CO.C.9