Question

1. △cde is an isosceles triangle with ∠d≅∠e. if cd = 4x + 9, de = 7x - 5, and ce = 16x - 27, find x and the measure of each side.

ce =
de =
cd =
x =
©gina wilson (all things algebra®, llc), 2014 - 2020

Explanation:

Step1: Set equal sides

In an isosceles triangle $\triangle CDE$ with $\angle D\cong\angle E$, the sides opposite these angles are equal, so $CD = DE$. Set up the equation $4x + 9=7x - 5$.

Step2: Solve for $x$

Subtract $4x$ from both sides: $9 = 7x-4x - 5$, which simplifies to $9 = 3x - 5$. Then add 5 to both sides: $9 + 5=3x$, so $14 = 3x$. Divide both sides by 3 to get $x=\frac{14}{3}$.

Step3: Find length of $CD$

Substitute $x = \frac{14}{3}$ into the expression for $CD$: $CD=4x + 9=4\times\frac{14}{3}+9=\frac{56}{3}+9=\frac{56 + 27}{3}=\frac{83}{3}$.

Step4: Find length of $DE$

Since $DE = CD$, $DE=\frac{83}{3}$.

Step5: Find length of $CE$

Substitute $x=\frac{14}{3}$ into the expression for $CE$: $CE = 16x-27=16\times\frac{14}{3}-27=\frac{224}{3}-27=\frac{224 - 81}{3}=\frac{143}{3}$.

Answer:

$x=\frac{14}{3}$, $CD=\frac{83}{3}$, $DE=\frac{83}{3}$, $CE=\frac{143}{3}$