Question

n the polygon below, if diagonal AD = 16x - 4, and diagonal CB = 12x + 28, find x

Answer 1

Given:

• Diagonal AD = 16x - 4

,

• Diagonal CB = 12x + 28

Let's find the value of x.

The given polygon is an isosceles trapezoid.

The diagonals of am isosceles trapezoid are equal.

Thus, we have:

Diagonal AD = Diagonal CB

$16x-4=12x+28$

Let's solve for x.

Add 4 to both sides:

$\begin{gathered} 16x-4+4=12x+28+4 \\ \\ 16x=12x+32 \end{gathered}$

Subtract 12x from both sides:

$\begin{gathered} 16x-12x=12x-12x+32 \\ \\ 4x=32 \end{gathered}$

Divide both sides by 4:

$\begin{gathered} \frac{4x}{4}=\frac{32}{4} \\ \\ x=8 \end{gathered}$

Therefore, the value of x is 8.

• ANSWER:

a. 8