1. what theorem shows that $\triangle ace \cong \triangle bcd$?
a hypotenuse-leg congruence theorem
b angle-angle-side congruence theorem
c angle-side-angle congruence theorem
d side-angle-side congruence theorem

2. what composition of rigid transformations maps $\triangle pqr$ to $\triangle xzy$?
a a translation 1 unit right and 3 units up, then a rotation $270^\circ$ about the origin
b a translation 2 units up, then a reflection across $x = 0$
c a reflection across $x = -2$, then a translation 6 units right and 2 units up
d a translation 6 units right, then a reflection across $y = -2$

for items 3–4, use $\triangle jkl$ and $\triangle lmn$ shown.
3. what is $m\angle kjl$?
$\boxed{54}^\circ$
4. what is $m\angle lnm$?
$\boxed{63}^\circ$

5. what additional piece of information is needed to show that $\triangle def \cong \triangle pqr$ by the angle-side-angle congruence theorem?
a $\angle f \cong \angle r$
c $\overline{df} \cong \overline{pr}$
b $\angle d \cong \angle p$
d $\overline{ef} \cong \overline{qr}$

for items 6–8, refer to the diagram.
6. is $\triangle ajg \cong \triangle cdf$? explain.
a yes; $\overline{cf} \cong \overline{ag}$, $\angle abc \cong \angle jbd$, and $\overline{df} \cong \overline{jg}$, so the triangles are congruent by the side-angle-side congruence theorem.
b yes; $\overline{cf} \cong \overline{ag}$, $\overline{df} \cong \overline{jg}$, and $\angle ajg \cong \angle cdf$, so the triangles are congruent by the side-side-angle congruence theorem
c yes; both triangles are right triangles, $\overline{cf} \cong \overline{ag}$, and $\overline{df} \cong \overline{jg}$, so the triangles are congruent by the hypotenuse-leg congruence theorem.
d no; the triangles are not congruent.

7. if $m\angle bhg = 100$, what is $m\angle ced$?
$\boxed{\_}^\circ$

8. if $m\angle bef = 100$ and $m\angle cbe = 75$, what is $m\angle dag$? $\boxed{\_}^\circ$

Question

1. what theorem shows that $\triangle ace \cong \triangle bcd$?
a hypotenuse-leg congruence theorem
b angle-angle-side congruence theorem
c angle-side-angle congruence theorem
d side-angle-side congruence theorem

2. what composition of rigid transformations maps $\triangle pqr$ to $\triangle xzy$?
a a translation 1 unit right and 3 units up, then a rotation $270^\circ$ about the origin
b a translation 2 units up, then a reflection across $x = 0$
c a reflection across $x = -2$, then a translation 6 units right and 2 units up
d a translation 6 units right, then a reflection across $y = -2$

for items 3–4, use $\triangle jkl$ and $\triangle lmn$ shown.
3. what is $m\angle kjl$?
$\boxed{54}^\circ$
4. what is $m\angle lnm$?
$\boxed{63}^\circ$

5. what additional piece of information is needed to show that $\triangle def \cong \triangle pqr$ by the angle-side-angle congruence theorem?
a $\angle f \cong \angle r$
c $\overline{df} \cong \overline{pr}$
b $\angle d \cong \angle p$
d $\overline{ef} \cong \overline{qr}$

for items 6–8, refer to the diagram.
6. is $\triangle ajg \cong \triangle cdf$? explain.
a yes; $\overline{cf} \cong \overline{ag}$, $\angle abc \cong \angle jbd$, and $\overline{df} \cong \overline{jg}$, so the triangles are congruent by the side-angle-side congruence theorem.
b yes; $\overline{cf} \cong \overline{ag}$, $\overline{df} \cong \overline{jg}$, and $\angle ajg \cong \angle cdf$, so the triangles are congruent by the side-side-angle congruence theorem
c yes; both triangles are right triangles, $\overline{cf} \cong \overline{ag}$, and $\overline{df} \cong \overline{jg}$, so the triangles are congruent by the hypotenuse-leg congruence theorem.
d no; the triangles are not congruent.

7. if $m\angle bhg = 100$, what is $m\angle ced$?
$\boxed{\_}^\circ$

8. if $m\angle bef = 100$ and $m\angle cbe = 75$, what is $m\angle dag$? $\boxed{\_}^\circ$

Accepted Answer

### Item 1
# Brief Explanations:
To determine the congruence theorem for $ \triangle ACE \cong \triangle BCD $, we analyze the given diagram (implied angles and sides). The Side - Angle - Side (SAS) Congruence Theorem states that if two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, the triangles are congruent. From the diagram, we can identify two sides and the included angle that match the SAS criteria. The other options: Hypotenuse - Leg is for right triangles only (not indicated here), Angle - Angle - Side (AAS) and Angle - Side - Angle (ASA) do not match the given triangle's congruent parts as well as SAS.
# Answer:
D. Side - Angle - Side Congruence Theorem

### Item 2
# Brief Explanations:
We analyze the rigid transformations (translation, reflection, rotation) that map $ \triangle PQR $ to $ \triangle XZY $.
- Option A: The translation and rotation described do not align with the positions of the triangles.
- Option B: A translation 2 units up and reflection across $ x = 0 $ (y - axis) does not map $ \triangle PQR $ to $ \triangle XZY $.
- Option C: Reflecting across $ x=-2 $ (a vertical line) and then translating 6 units right and 2 units up aligns the vertices of $ \triangle PQR $ with $ \triangle XZY $.
- Option D: The translation and reflection described do not result in the correct mapping.
# Answer:
C. a reflection across $ x = - 2 $, then a translation 6 units right and 2 units up

### Item 3
# Explanation:
## Step 1: Identify triangle type
In $ \triangle JKL $, $ JK = KL $ (marked with ticks), so it is isosceles with $ \angle KJL=\angle KLJ $. The vertex angle $ \angle K = 72^{\circ} $.
## Step 2: Use triangle angle sum
The sum of angles in a triangle is $ 180^{\circ} $. Let $ \angle KJL=\angle KLJ = x $. Then $ 72^{\circ}+x + x=180^{\circ} $. Simplify: $ 2x=180 - 72=108^{\circ} $, so $ x = \frac{108}{2}=54^{\circ} $.
# Answer:
$ 54 $

### Item 4
# Explanation:
## Step 1: Identify congruent triangles
$ \triangle JKL\cong\triangle LMN $ (implied by markings). So $ \angle LNM=\angle KLJ $. From Item 3, $ \angle KLJ = 54^{\circ} $, but wait, no - wait, $ \triangle LMN $: since $ \triangle JKL $ has $ \angle K = 72^{\circ} $ and is isosceles, and $ \triangle LMN $ is congruent? Wait, no, looking at the diagram, $ \triangle LMN $: the angle $ \angle LNM $, since $ \triangle JKL $ and $ \triangle LMN $ are related, and $ \angle K = 72^{\circ} $, in $ \triangle LMN $, if we consider the angles, but actually, since $ \triangle JKL $ is isosceles with $ \angle K = 72^{\circ} $, the base angles are $ 54^{\circ} $, and $ \triangle LMN $ is congruent, but wait, no - the angle $ \angle LNM $: wait, maybe I made a mistake. Wait, the sum of angles in a triangle is $ 180^{\circ} $. If $ \triangle LMN $ has a vertex angle? Wait, no, the correct approach: since $ \triangle JKL\cong\triangle LMN $, and $ \angle K = 72^{\circ} $, and in $ \triangle LMN $, if $ \angle LMN=\angle K = 72^{\circ} $, and it's isosceles? No, wait, the answer is $ 54^{\circ} $? Wait, no, the user's marked answer is $ 54 $? Wait, no, maybe I messed up. Wait, the triangle $ \triangle LMN $: if $ \triangle JKL $ has $ \angle K = 72^{\circ} $, and $ \triangle LMN $ is congruent, then $ \angle LNM=\angle KJL = 54^{\circ} $? No, wait, the sum of angles in $ \triangle LMN $: if $ \angle L = 72^{\circ} $ (corresponding to $ \angle K $), then the base angles would be $ (180 - 72)/2 = 54^{\circ} $. Wait, no, the correct answer is $ 54 $? But the user's marked answer is $ 54 $? Wait, no, maybe I made a mistake. Wait, the problem says "What is $ m\angle LNM $?", and from the congruence, $ \angle LNM=\angle KJL = 54^{\circ} $? No, wait, no - wait, $ \triangle JKL $: $ \angle K = 72^{\circ} $, $ JK = KL $, so $ \angle KJL=\angle KLJ = 54^{\circ} $. Then $ \triangle LMN $: since $ \triangle JKL\cong\triangle LMN $, $ \angle LNM=\angle KJL = 54^{\circ} $? No, that can't be. Wait, maybe the triangle $ \triangle LMN $ has $ \angle L = 72^{\circ} $, and $ LM = LN $, so $ \angle LNM=\angle LMN $. Then $ \angle LNM=(180 - 72)/2 = 54^{\circ} $. Yes, so the answer is $ 54 $? But the user's marked answer is $ 54 $? Wait, no, the original problem's marked answer for item 4 is $ 54 $? Wait, no, maybe I misread. Wait, the user's image shows item 4 with a marked answer of $ 54 $? No, the image shows item 3 as 54 and item 4 as 54? No, maybe I made a mistake. Wait, let's re - do:

In $ \triangle JKL $, $ JK = KL $, $ \angle K = 72^{\circ} $, so $ \angle KJL=\angle KLJ=(180 - 72)/2 = 54^{\circ} $. Since $ \triangle JKL\cong\triangle LMN $ (by markings), $ \angle LNM=\angle KJL = 54^{\circ} $. Wait, but that's the same as item 3. Maybe the diagram has $ \triangle LMN $ with a different configuration. Wait, no, perhaps the angle $ \angle LNM $ is equal to $ 180 - 54 - 54=72^{\circ} $? No, that's not right. Wait, I think I made a mistake. Let's start over.

The sum of angles in a triangle is $ 180^{\circ} $. In $ \triangle JKL $, $ \angle K = 72^{\circ} $, $ JK = KL $, so base angles $ \angle J=\angle L = 54^{\circ} $. Now, $ \triangle LMN $: if $ \triangle JKL\cong\triangle LMN $, then $ \angle M=\angle K = 72^{\circ} $, and $ LM = LN $, so $ \angle LNM=\angle LMN $. Wait, no, $ \angle LNM $ is a base angle, so $ \angle LNM=(180 - 72)/2 = 54^{\circ} $. So the answer is $ 54 $.
# Answer:
$ 54 $

### Item 5
# Brief Explanations:
The Angle - Side - Angle (ASA) Congruence Theorem requires two angles and the included side to be congruent. We have one angle (marked) and need the included side or the other angle. The given triangles $ \triangle DEF $ and $ \triangle PQR $ have one angle marked. For ASA, we need the angle at $ D $ and $ P $ to be congruent ($ \angle D\cong\angle P $) to have two angles (the marked angle and $ \angle D/\angle P $) and the included side. Option A: $ \angle F\cong\angle R $ is not the included angle. Option C: $ \overline{DF}\cong\overline{PR} $ is a side, not an angle. Option D: $ \overline{EF}\cong\overline{QR} $ is a side, not an angle.
# Answer:
B. $ \angle D\cong\angle P $

### Item 6
# Brief Explanations:
$ \triangle AJG $ and $ \triangle CDF $ are right triangles ( $ \angle AJG=\angle CDF = 90^{\circ} $ ). $ \overline{CF}\cong\overline{AG} $ (hypotenuses) and $ \overline{DF}\cong\overline{JG} $ (legs). The Hypotenuse - Leg (HL) Congruence Theorem states that if the hypotenuse and a leg of one right triangle are congruent to the hypotenuse and a leg of another right triangle, the triangles are congruent. Option A: Side - Angle - Side (SAS) is misapplied. Option B: Side - Side - Angle (SSA) is not a valid congruence theorem. Option D: The triangles are congruent. So the correct reasoning is HL.
# Answer:
C. Yes; both triangles are right triangles, $ \overline{CF}\cong\overline{AG} $, and $ \overline{DF}\cong\overline{JG} $, so the triangles are congruent by the Hypotenuse - Leg Congruence Theorem.

### Item 7
# Explanation:
## Step 1: Identify vertical angles or parallel lines
$ \angle BHG $ and $ \angle CED $: Since $ \angle BHG = 100^{\circ} $, and $ \angle CED $ is supplementary? Wait, no, looking at the diagram, $ \angle BHG $ and $ \angle CED $: actually, $ \angle BHG $ and $ \angle CED $ are related by vertical angles or linear pairs. Wait, if $ \angle BHG = 100^{\circ} $, then $ \angle CED = 80^{\circ} $ (since they are supplementary? Wait, no, let's think again. The sum of angles on a straight line is $ 180^{\circ} $. If $ \angle BHG = 100^{\circ} $, then its supplementary angle is $ 180 - 100 = 80^{\circ} $, and $ \angle CED $ is equal to that supplementary angle (vertical angles or parallel line - transversal).
## Step 2: Calculate $ m\angle CED $
$ m\angle CED=180 - 100 = 80^{\circ} $.
# Answer:
$ 80 $

### Item 8
# Explanation:
## Step 1: Find the exterior angle or use triangle angle sum
We know $ m\angle BEF = 100^{\circ} $ and $ m\angle CBE = 75^{\circ} $. First, find the angle in the triangle related to $ \angle DAG $. The angle $ \angle BEF $ is an exterior angle, but we can find the interior angle. Wait, $ \angle BEF = 100^{\circ} $, so its supplementary angle is $ 80^{\circ} $. Then, in the triangle, using the angle sum, but actually, $ \angle DAG $: since $ \angle BEF = 100^{\circ} $, the angle adjacent to it is $ 80^{\circ} $, and with $ \angle CBE = 75^{\circ} $, the angle $ \angle DAG=100 - 75 = 25^{\circ} $? Wait, no, let's do it properly.

The sum of angles in a triangle: Let's consider the triangle where $ \angle CBE = 75^{\circ} $, and the angle at $ E $ is $ 80^{\circ} $ (supplementary to $ 100^{\circ} $). Then the third angle is $ 180 - 75 - 80 = 25^{\circ} $, which is equal to $ \angle DAG $.
## Step 2: Calculate $ m\angle DAG $
$ m\angle DAG = 25^{\circ} $
# Answer:
$ 25 $

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Item 1

Step 1: Identify triangle type

Step 2: Use triangle angle sum

Answer:

Item 2