Geometry - 1st Semester Review

RS is the segment bisector.  If OQ = 32 and PO = 4x, what does x equal?

• 8
• 28
• 128
• 36

Tags

TEKS.MATH.G.6D

D is the midpoint of EF.
ED = 4x + 6
DF = 7x -9
 Find EF.

• 5
• 26
• 52
• 22

Tags

CCSS.HSG.CO.C.10

Find the midpoint of the line segment with endpoints (-2,0) & (9,-1).

• (1, 3)

• (3.5, -.5)

• (-.5, 3.5)

• (6, -.5)

Tags

CCSS.HSG.GPE.B.6

Find the midpoint of the segment with the endpoints:(4, 12) and (-6, 14)

• (-1, 13)
• (13, 1)
• (13, -1)
• (5, 13)

Tags

CCSS.HSG.GPE.B.6

How is the distance formula correctly written:

• $d\ =\ \sqrt{\left(y_{1\ }-\ y_2\right)^2\ -\left(x_{2\ }-\ x_1\right)^2}\$

• $d\ =\ \sqrt{\left(x_{2\ }-\ x_1\right)^2\ +\ \left(y_{2_{\ }}-\ y_1\ \right)2}$

• $d\ =\ \sqrt{\left(y_{1\ }-\ y\ _2\right)^2\ +\ \left(x_{2\ }-\ x_1\right)}^2$

• d = $\sqrt{\left(x\right)^2-\ \left(y\right)^2}$

Tags

CCSS.HSG.GPE.B.7

Find the distance between (-12, 1) and (12, -1).

• 24.05
• 24.06
• 24.07
• 24.08

Tags

CCSS.HSG.GPE.B.7

Solve

• 50°
• 58°
• 115°
• 226°

Tags

CCSS.7.G.B.5

Find the value of ∠A.

• 25°
• 30°
• 35°
• 90°

Tags

CCSS.7.G.B.5

Solve for x

• x = 8
• x = 6
• x = 4
• x = 2

Tags

CCSS.4.MD.A.2

Angle JHG measures 161.

• m∠GHK=80.5,

  m∠KHJ=161

• m∠GHK=161,

  m∠KHJ=80.5

• m∠GHK=80.5,

  m∠KHJ=80.5

• m∠GHK=161,

  m∠KHJ=161

Tags

CCSS.7.G.B.5