Question

(d) problem 20: (first taught in lesson 29) △def is isosceles and the length of base df is 8. find the other two side lengths if de = 5z - 4 and ef = 3z + 2. after you enter your answer press go. de = ef =

Explanation:

Step1: Set up the equation for an isosceles triangle

In an isosceles triangle, two sides are equal. So, \(DE = EF\).
\(5z - 4=3z + 2\)

Step2: Solve for \(z\)

Subtract \(3z\) from both sides:
\(5z-3z - 4=3z-3z + 2\)
\(2z-4 = 2\)
Add \(4\) to both sides:
\(2z-4 + 4=2 + 4\)
\(2z=6\)
Divide both sides by \(2\):
\(z=\frac{6}{2}=3\)

Step3: Find the length of \(DE\)

Substitute \(z = 3\) into \(DE=5z - 4\)
\(DE=5\times3-4=15 - 4=11\)

Step4: Find the length of \(EF\)

Substitute \(z = 3\) into \(EF=3z + 2\)
\(EF=3\times3+2=9 + 2=11\)

Answer:

\(DE = 11\), \(EF = 11\)