Question

quadrilateral pqrs is an isosceles trapezoid and m∠s = 4a + 100°. what is the value of a?

Explanation:

Step1: Recall isosceles trapezoid property

In an isosceles trapezoid, base - angles are equal. $\angle R$ and $\angle Q$ are one pair of base - angles, and $\angle S$ and $\angle P$ are the other pair. Also, adjacent angles along a non - parallel side are supplementary. So, $\angle R+\angle S = 180^{\circ}$.

Step2: Substitute the given angle values

Given $\angle R = 112^{\circ}$ and $\angle S=4a + 100^{\circ}$, we substitute into the equation $\angle R+\angle S = 180^{\circ}$. So, $112+(4a + 100)=180$.

Step3: Simplify the left - hand side

First, combine like terms: $112+100 + 4a=180$, which gives $212+4a = 180$.

Step4: Solve for a

Subtract 212 from both sides: $4a=180 - 212$, so $4a=- 32$. Then divide both sides by 4: $a=\frac{-32}{4}=-8$.

Answer:

$-8$