## Reg. Geometry Unit 4 Test Review Solutions November 5, 2013

Posted by Ms. Miller in Geometry.

Unit 4 Test Review

Solutions

1. right isosceles 2. equiangular equilateral
3. $4x-18+8x-6+5x=180$
$17x=204$
$x=12$
4.$\mathrm{m}\angle 1=127-75=52\textdegree$
5. There is a typo on this problem. Replace the 7x with 8x.
$8x+6=5x-7+2x+21$
$8x+6=7x+14$
$x=8$
6. $\mathrm{m}\angle BAD=102\textdegree$
7. $\angle B \cong \angle C$ 8. $3x+7=4x-13$
$x=20$
base angles:
$3(20)+7=67\textdegree$
vertex angle:
$180-2(67)=46\textdegree$
B
9. $\angle BAC\cong\angle BCA$
$\angle 3\cong\angle 4$
$\angle 1\cong\angle 2$
10. $5x=2x+18$
$3x=18$
$x=6$
$\mathrm{m}\angle B=120\textdegree$ and $\mathrm{m}\angle BAG=150\textdegree$
11. A 12. C
13. B 14. $\overline{AB}\cong\overline{LF}$
15.

 $\angle J\cong\angle M$ $\overline{JK}\cong\overline{MN}$ $\angle K\cong\angle N$ $\overline{KL}\cong\overline{NO}$ $\angle L\cong\angle O$ $\overline{JL}\cong\overline{MO}$
16. $\overline{XY}$
17. Statement 1 should read $\angle D\cong\angle C$.
Reason 4: Vertical angles
Reason 5: AAS
18. The Given statement should read $\overline{HK}$ and $\overline{IL}$ bisect each other.

Statements Reasons
1. $\overline{HK}$ and $\overline{IL}$ bisect 1. Given
2. $\overline{HJ}\cong\overline{KJ}$ 2. Def. bisector
3. $\angle IJH\cong\angle LJK$ 3. Vertical angles
4. $\overline{IJ}\cong\overline{LJ}$ 4. Def. bisector
5. $\triangle HIJ\cong\triangle KLJ$ 5. SAS
19. Reason 4: ASA
Reason 5: CPCTC
20. If we prove ΔADB and ΔCDB are congruent, then we can use CPCTC to prove the sides congruent.

Statements Reasons
1. D is the midpoint of $\overline{AC}$ 1. Given
2. $\overline{AD}\cong\overline{CD}$ 2. Def. midpoint
3. $\angle ADB\cong\angle CDB$ 3. Given
4. $\overline{BD}\cong\overline{BD}$ 4. Refl. prop. ≅
5. $\triangle ADB\cong\triangle CDB$ 5. SAS
6. $\overline{AB}\cong\overline{CB}$ 6. CPCTC
7. ΔABC is isosceles 7. Def. isosceles
21.

Statements Reasons
1. M is midpt of $\overline{KP}$ and $\overline{LN}$ 1. Given
2. $\overline{KM}\cong\overline{PM}$ 2. Def. midpoint
3. $\angle KML\cong\angle PMN$ 3. Vertical angles
4. $\overline{NM}\cong\overline{LM}$ 4. Def. midpoint
5. $\triangle KLM\cong\triangle PNM$

5. SAS
6. $\angle K\cong\angle P$ 6. CPCTC
7. $\overline{KL}\parallel\overline{NP}$ 7. Conv. Alt. Int. ∠s
22. To prove ΔABC is right isosceles, prove the sides congruent and the angles perpendicular.
Sides:
$AB=\sqrt{(7-1)^2+(0-2)^2}=\sqrt{40}$
$BC=\sqrt{(3-7)^2+(-2-0)^2}=\sqrt{20}$
$AC=\sqrt{(3-1)^2+(-2-2)^2}=\sqrt{20}$
Slopes:
AB: $m=\frac{0-2}{1-7}=-\frac{1}{3}$
BC: $m=\frac{-2-0}{3-7}=\frac{1}{2}$
AC: $m=\frac{-2-2}{3-1}=-2$

Therefore, $AC=BC$, and ΔABC is isosceles.

The product of the slopes is -1, so the lines are perpendicular. Therefore, ∠ACB is right.