Question

if abcd is a trapezoid with midsegment ef, find its perimeter.
the perimeter of this trapezoid is

Explanation:

Step1: Recall midsegment formula for trapezoid

The midsegment \( EF \) of a trapezoid is the average of the two bases \( BC \) and \( AD \). So, \( EF=\frac{BC + AD}{2} \).
Given \( BC = 3x-13 \), \( AD = 23 \), and \( EF = 2x - 3 \). Substitute into the formula:
\( 2x-3=\frac{(3x - 13)+23}{2} \)

Step2: Solve for \( x \)

Multiply both sides by 2 to eliminate the denominator:
\( 2(2x - 3)=(3x - 13)+23 \)
\( 4x-6 = 3x + 10 \)
Subtract \( 3x \) from both sides:
\( 4x-3x-6=3x - 3x+10 \)
\( x - 6 = 10 \)
Add 6 to both sides:
\( x=10 + 6=16 \)

Step3: Find length of \( BC \)

Substitute \( x = 16 \) into \( BC = 3x-13 \):
\( BC=3(16)-13=48 - 13 = 35 \)

Step4: Calculate the perimeter

The perimeter of trapezoid \( ABCD \) is \( AB + BC+CD + DA \). From the diagram, \( AB = 5 + 5=10 \) (since \( E \) is the midpoint of \( AB \), so \( AE = EB = 5 \)), \( BC = 35 \), \( CD = 9 + 9 = 18 \) (since \( F \) is the midpoint of \( CD \), so \( DF = FC = 9 \)), and \( DA = 23 \).
Perimeter \(=10 + 35+18 + 23\)
\(=10+35 = 45\); \( 45+18 = 63\); \( 63+23 = 86\)

Answer:

86