Question

math 3 unit 1 test reasoning with geometry version 2
name jim

1. in the parallelogram shown, find the measure of ∠a.

a. 25° b. 40° c. 140° d. -8.33°

1. solve for x.

a. 21.25 b. 20 c. 32 d. 40.4

1. in parallelogram abcd, find m∠bdc.

a. 90° b. 70° c. 25° d. 35°

1. given m∠1 = 2x + 16 and m∠2 = x + 14, solve for x.

a. -50 b. -2 c. 50 d. 2

Explanation:

1.

Step1: Use property of parallelogram

Adjacent angles in a parallelogram are supplementary, so $\angle A+\angle D = 180^{\circ}$, and also opposite - angles are equal. Let's first find $x$ using the fact that opposite - angles are equal. So $5x + 15=2x - 10$.
$5x-2x=-10 - 15$
$3x=-25$ (This is wrong. We should use the adjacent - angle property. Let's start over. Adjacent angles of a parallelogram are supplementary, so $(2x - 10)+(5x + 15)=180$).

Step2: Solve the equation for $x$

Combine like - terms: $2x+5x-10 + 15=180$, $7x+5 = 180$, $7x=180 - 5=175$, $x = 25$.

Step3: Find $\angle A$

Substitute $x = 25$ into the expression for $\angle A$: $\angle A=2x-10=2\times25-10=50 - 10 = 40^{\circ}$.

Step1: Use the property of corresponding angles

If two parallel lines are cut by a transversal, then corresponding angles are equal. So $3x + 42=5x-22$.

Step2: Solve the equation for $x$

Subtract $3x$ from both sides: $42=5x-3x-22$, $42 = 2x-22$.
Add 22 to both sides: $42 + 22=2x$, $64 = 2x$.
Divide both sides by 2: $x = 32$.

Step1: Use the property of parallelogram

In parallelogram $ABCD$, $AB\parallel CD$. $\angle ABD=\angle BDC$ (alternate - interior angles). Given $\angle ABD = 25^{\circ}$, so $\angle BDC=25^{\circ}$.

Answer:

B. $40^{\circ}$

2.