triangle congruence review
name sophia aroyo
period 1
part 1: name the congruent triangles.
1. δogd ≅ δ___
2. δbac ≅ δ___
3. δinj ≅ δ___
4. δtox ≅ δ___
5. δkgh ≅ δ___
10. δpqr ≅ δmnr find x
11. is δabc ≅ δadc? why?
15. find a, b, side c
part 2: name the congruent triangle and the congruent parts.
(a) δefi ≅ δa___
(b) fg ≅ ___
(c) ∠g ≅ ∠___
(d) gh ≅ ___
(e) fh ≅ ___
(f) ∠h ≅ ∠___
part 3: use the congruency statement to fill in the corresponding parts.
δefh ≅ δkgh
8. ∠e ≅ ∠___
9. ∠efi ≅ ∠___
10. fe ≅ ___
11. fh ≅ ___
12. te ≅ ___ (possible typo)
13. ∠efh ≅ ∠___

Question

Accepted Answer

To solve the problem of finding $ x $ in the triangle congruence (specifically the vertical angles and right triangles with congruent angles), we can use the properties of congruent triangles and angle - angle similarity or congruence.

### Step 1: Identify the congruent triangles and corresponding angles
We know that $ \triangle QOR\cong\triangle MNR $ (given). So, the corresponding angles are equal. Also, $ \angle QRO $ and $ \angle MRN $ are vertical angles, so $ \angle QRO=\angle MRN $. The triangles $ \triangle QOR $ and $ \triangle MNR $ are right - angled triangles ( $ \angle O = \angle N=90^{\circ} $ ).
Since the triangles are congruent, the non - right angles are equal. We are given that one of the non - right angles in $ \triangle MNR $ is $ 39^{\circ} $. Let's assume that the angle in $ \triangle QOR $ corresponding to the $ 39^{\circ} $ angle in $ \triangle MNR $ is equal. Also, since the sum of angles in a triangle is $ 180^{\circ} $, and we have a right angle ($ 90^{\circ} $) and another angle related to $ x $.

But a simpler way: In the right - angled triangles $ \triangle QOR $ and $ \triangle MNR $, if $ \triangle QOR\cong\triangle MNR $, then the acute angles are equal. If we consider the angle adjacent to $ x $ in $ \triangle QOR $ and the $ 39^{\circ} $ angle in $ \triangle MNR $, and since the triangles are congruent, the angle opposite to the right angle (the other acute angle) should be equal. Also, we can use the fact that in the intersecting lines, the vertical angles are equal and the triangles are right - angled.

Wait, actually, if we look at the diagram, $ \triangle QOR $ and $ \triangle MNR $ are congruent right - triangles. So, the angle at $ Q $ (let's say) in $ \triangle QOR $ is equal to the angle at $ M $ in $ \triangle MNR $. But more straightforward: In a right - triangle, the two acute angles are complementary (sum to $ 90^{\circ} $). But since the triangles are congruent, the angle marked $ x $ and the $ 39^{\circ} $ angle are related. Wait, actually, if $ \triangle QOR\cong\triangle MNR $, then $ \angle Q=\angle M $ and $ \angle O=\angle N = 90^{\circ} $, and $ \angle QRO=\angle MRN $ (vertical angles). So, the angle $ x $ and the $ 39^{\circ} $ angle are equal? No, wait, maybe the triangles are similar (by AA similarity, since two angles are equal: right angle and vertical angle). So, if $ \triangle QOR\sim\triangle MNR $ (by AA, as $ \angle O=\angle N = 90^{\circ} $ and $ \angle QRO=\angle MRN $), then the corresponding angles are equal. So, if one of the acute angles in $ \triangle MNR $ is $ 39^{\circ} $, then the corresponding acute angle in $ \triangle QOR $ is also $ 39^{\circ} $. But wait, maybe $ x = 39^{\circ} $? Wait, no, maybe I made a mistake. Wait, let's re - examine.

Wait, the problem says $ \triangle QOR\cong\triangle MNR $. So, corresponding parts are equal. Let's assume that $ \angle Q=\angle M $, $ \angle O=\angle N = 90^{\circ} $, and $ \angle QRO=\angle MRN $. If in $ \triangle MNR $, the angle at $ M $ is $ 39^{\circ} $, then the angle at $ Q $ is also $ 39^{\circ} $. But in $ \triangle QOR $, $ \angle O = 90^{\circ} $, so $ \angle Q+\angle QRO=90^{\circ} $. But $ \angle QRO=x $? Wait, no, maybe the angle labeled $ x $ is equal to $ 39^{\circ} $ because of congruence. Wait, maybe the answer is $ x = 39^{\circ} $. Wait, let's think again.

If $ \triangle QOR\cong\triangle MNR $, then the corresponding angles are equal. So, the angle in $ \triangle QOR $ that is equal to the $ 39^{\circ} $ angle in $ \triangle MNR $ is $ x $. So, $ x=39^{\circ} $.

### Step 2: Conclusion
Since the triangles $ \triangle QOR $ and $ \triangle MNR $ are congruent, the corresponding acute angles are equal. So, the value of $ x $ is equal to $ 39^{\circ} $.

Answer: $ \boldsymbol{39^{\circ}} $

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Step 1: Identify the congruent triangles and corresponding angles

Answer: