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.

Explanation:

Step1: Use isosceles - triangle property

Since \(\triangle CDE\) is isosceles with \(\angle D\cong\angle E\), then \(CD = DE\). So we set up the equation \(4x + 9=7x - 5\).

Step2: Solve the equation 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, we get \(x=\frac{14}{3}\).

Step3: Find the length of each side

For \(CD\): Substitute \(x = \frac{14}{3}\) into \(CD = 4x + 9\), \(CD=4\times\frac{14}{3}+9=\frac{56}{3}+9=\frac{56 + 27}{3}=\frac{83}{3}\).
For \(DE\): Substitute \(x=\frac{14}{3}\) into \(DE = 7x - 5\), \(DE=7\times\frac{14}{3}-5=\frac{98}{3}-5=\frac{98 - 15}{3}=\frac{83}{3}\).
For \(CE\): Substitute \(x=\frac{14}{3}\) into \(CE = 16x-27\), \(CE=16\times\frac{14}{3}-27=\frac{224}{3}-27=\frac{224 - 81}{3}=\frac{143}{3}\).

Answer:

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