Question

if $tu = 10x$, $ps = x + 23$, and $qr = x + 28$, what is the value of $x$?

Explanation:

Step1: Identify the midline theorem

In a trapezoid, the midline (or midsegment) length is the average of the two bases. Here, \( PS \) is the midline, \( TU \) and \( QR \) are the two bases. So the formula is \( PS=\frac{TU + QR}{2} \).

Step2: Substitute the given expressions

We know \( TU = 10x \), \( PS=x + 23 \), and \( QR=x + 28 \). Substitute into the formula:
\( x + 23=\frac{10x+(x + 28)}{2} \)

Step3: Multiply both sides by 2 to eliminate the denominator

\( 2(x + 23)=10x+x + 28 \)
Expand the left side: \( 2x+46 = 11x+28 \)

Step4: Solve for \( x \)

Subtract \( 2x \) from both sides: \( 46=9x + 28 \)
Subtract 28 from both sides: \( 9x=46 - 28=18 \)
Divide both sides by 9: \( x = 2 \)

Answer:

\( x = 2 \)