consider parallelogram abcd below. use the information given in the figure to find x, m∠a, and m∠adb.

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Accepted Answer

# Explanation:
## Step1: Find $ x $
In a parallelogram, opposite sides are equal. So $ AD = BC $. Given $ AD = 5x $ and $ BC = 5 $, we set up the equation $ 5x = 5 $. Dividing both sides by 5, we get $ x = \frac{5}{5}=1 $.
## Step2: Find $ m\angle A $
In a parallelogram, consecutive angles are supplementary. So $ \angle A + \angle C = 180^\circ $. Given $ m\angle C = 114^\circ $, then $ m\angle A = 180^\circ - 114^\circ = 66^\circ $. Wait, no, wait. Wait, in the parallelogram $ ABCD $, $ AB \parallel CD $ and $ AD \parallel BC $. Wait, actually, $ \angle A $ and $ \angle C $ are opposite angles? No, no. Wait, consecutive angles: $ \angle A $ and $ \angle B $ are consecutive? Wait, no, let's correct. In parallelogram $ ABCD $, $ \angle A $ and $ \angle C $ are opposite angles? No, no, opposite angles are equal, consecutive angles are supplementary. Wait, $ AB \parallel CD $, so $ \angle A + \angle D = 180^\circ $, and $ \angle A = \angle C $? Wait, no, I made a mistake. Wait, the figure: $ ABCD $ is a parallelogram, so $ AD \parallel BC $, $ AB \parallel CD $. So $ \angle A $ and $ \angle C $: are they opposite? Wait, vertices are $ A, B, C, D $ in order, so $ AB $ is adjacent to $ BC $, $ BC $ to $ CD $, $ CD $ to $ DA $, $ DA $ to $ AB $. So opposite angles: $ \angle A $ and $ \angle C $, $ \angle B $ and $ \angle D $. Consecutive angles: $ \angle A $ and $ \angle B $, $ \angle B $ and $ \angle C $, $ \angle C $ and $ \angle D $, $ \angle D $ and $ \angle A $. So consecutive angles are supplementary. So $ \angle B + \angle C = 180^\circ $? Wait, no, let's look at the triangle $ BCD $. Wait, in the parallelogram, $ AD = BC = 5 $ (since we found $ x = 1 $, so $ AD = 5x = 5 $, so $ AD = BC = 5 $). Now, in triangle $ BCD $, we have $ BC = 5 $, $ \angle CBD = 32^\circ $, $ \angle C = 114^\circ $. Wait, maybe we need to use the triangle. Wait, no, first, $ x $: $ AD = BC $, $ AD = 5x $, $ BC = 5 $, so $ 5x = 5 $, so $ x = 1 $. Then, $ m\angle A $: in parallelogram, $ \angle A = \angle C $? Wait, no, opposite angles are equal. Wait, $ \angle C = 114^\circ $, so $ \angle A = 114^\circ $? But that contradicts the triangle. Wait, no, the triangle $ BCD $: $ BC = 5 $, $ CD $ is equal to $ AB $, $ AD = 5 $. Wait, maybe I messed up the angle. Wait, let's look at triangle $ BCD $. In triangle $ BCD $, angles sum to $ 180^\circ $. So $ \angle CBD = 32^\circ $, $ \angle C = 114^\circ $, so $ \angle CDB = 180^\circ - 32^\circ - 114^\circ = 34^\circ $. Then, since $ AD \parallel BC $, $ \angle ADB = \angle CBD = 32^\circ $? Wait, no, $ AD \parallel BC $, and $ BD $ is a transversal, so alternate interior angles: $ \angle ADB = \angle CBD = 32^\circ $. Wait, let's redo:

1. Find $ x $:
In parallelogram $ ABCD $, $ AD = BC $ (opposite sides of parallelogram are equal). Given $ AD = 5x $ and $ BC = 5 $, so:
$$
5x = 5 \implies x = 1
$$

2. Find $ m\angle A $:
In parallelogram $ ABCD $, consecutive angles are supplementary. $ \angle A $ and $ \angle C $ are not consecutive. Wait, $ \angle A $ and $ \angle B $ are consecutive? No, $ \angle A $ and $ \angle D $ are consecutive? Wait, vertices: $ A, B, C, D $, so $ AB \parallel CD $, $ AD \parallel BC $. So $ \angle A + \angle B = 180^\circ $, $ \angle B + \angle C = 180^\circ $, so $ \angle A = \angle C $? No, that's opposite angles. Wait, opposite angles in parallelogram are equal. So $ \angle A = \angle C $, $ \angle B = \angle D $. But in the figure, $ \angle C = 114^\circ $, so $ \angle A = 114^\circ $? But that can't be, because in triangle $ ABD $ or $ BCD $, let's check the angles. Wait, maybe I made a mistake. Wait, the triangle $ BCD $: $ BC = 5 $, $ CD $ is equal to $ AB $, $ AD = 5 $. Wait, $ \angle C = 114^\circ $, $ \angle CBD = 32^\circ $, so $ \angle CDB = 180 - 114 - 32 = 34^\circ $. Then, since $ AB \parallel CD $, $ \angle ABD = \angle CDB = 34^\circ $ (alternate interior angles). Then, in triangle $ ABD $, $ AD = 5 $, $ AB $ is some length, but maybe we need to find $ \angle A $. Wait, no, in parallelogram, $ \angle A + \angle C = 180^\circ $? No, that's not right. Wait, consecutive angles: $ \angle A $ and $ \angle B $ are consecutive, so $ \angle A + \angle B = 180^\circ $. But $ \angle B $ is split into $ \angle ABD $ and $ \angle CBD $. $ \angle CBD = 32^\circ $, and $ \angle ABD = \angle CDB = 34^\circ $ (from triangle $ BCD $, $ \angle CDB = 34^\circ $, so $ \angle ABD = 34^\circ $ because $ AB \parallel CD $). So $ \angle B = \angle ABD + \angle CBD = 34^\circ + 32^\circ = 66^\circ $. Then, since $ \angle A + \angle B = 180^\circ $ (consecutive angles in parallelogram), $ \angle A = 180^\circ - 66^\circ = 114^\circ $? Wait, no, that's conflicting. Wait, no, $ \angle C = 114^\circ $, and in parallelogram, $ \angle A = \angle C $, so $ \angle A = 114^\circ $. Then, $ \angle B = \angle D = 180^\circ - 114^\circ = 66^\circ $. Then, in triangle $ BCD $, angles: $ \angle C = 114^\circ $, $ \angle CBD = 32^\circ $, so $ \angle CDB = 180 - 114 - 32 = 34^\circ $. Then, since $ AD \parallel BC $, $ \angle ADB = \angle CBD = 32^\circ $ (alternate interior angles, because $ AD \parallel BC $ and $ BD $ is transversal). Wait, let's confirm:

- $ x $: $ AD = BC $, so $ 5x = 5 \implies x = 1 $. Correct.
- $ m\angle A $: In parallelogram, opposite angles are equal. $ \angle A = \angle C = 114^\circ $. Wait, but earlier when we calculated $ \angle B = 66^\circ $, then $ \angle A = 180 - 66 = 114 $, which matches $ \angle C = 114 $. So that's correct.
- $ m\angle ADB $: Since $ AD \parallel BC $, and $ BD $ is a transversal, alternate interior angles $ \angle ADB $ and $ \angle CBD $ are equal. $ \angle CBD = 32^\circ $, so $ \angle ADB = 32^\circ $. Wait, but in triangle $ BCD $, $ \angle CDB = 34^\circ $, and $ \angle ADB $ is part of $ \angle ADC $. Wait, $ \angle ADC = \angle B = 66^\circ $ (opposite angles in parallelogram). $ \angle ADC = \angle ADB + \angle CDB $, so $ 66^\circ = \angle ADB + 34^\circ \implies \angle ADB = 32^\circ $. Yes, that matches. So:

## Step3: Find $ m\angle ADB $
Since $ AD \parallel BC $ and $ BD $ is a transversal, alternate interior angles $ \angle ADB $ and $ \angle CBD $ are equal. Given $ \angle CBD = 32^\circ $, so $ m\angle ADB = 32^\circ $. Wait, no, wait, $ \angle CBD = 32^\circ $, $ \angle CDB = 34^\circ $, $ \angle ADC = 66^\circ $, so $ \angle ADB = 66^\circ - 34^\circ = 32^\circ $. Correct.

# Answer:
$ x = 1 $, $ m\angle A = 114^\circ $, $ m\angle ADB = 32^\circ $

So filling in the boxes:

$ x = \boxed{1} $

$ m\angle A = \boxed{114} $

$ m\angle ADB = \boxed{32} $

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QUESTION IMAGE

Question

Explanation:

Step1: Find \( x \)

Step2: Find \( m\angle A \)

Answer: