Question

sv is the midsegment of the trapezoid qrtu. if qr = 2y + 22, sv = 8y - 78, and tu = 8y - 64, what is the value of y?

Explanation:

Step1: Recall the midsegment formula of a trapezoid

The midsegment \(SV\) of a trapezoid \(QRTU\) is given by the formula \(SV=\frac{QR + TU}{2}\).

Step2: Substitute the given expressions into the formula

Substitute \(QR = 2y+22\), \(SV=8y - 78\), and \(TU = 8y-64\) into \(SV=\frac{QR + TU}{2}\).
We get \(8y-78=\frac{(2y + 22)+(8y-64)}{2}\).

Step3: Simplify the right - hand side of the equation

First, simplify the numerator of the right - hand side: \((2y + 22)+(8y-64)=2y+8y+22 - 64=10y-42\).
So the equation becomes \(8y-78=\frac{10y-42}{2}\).
Since \(\frac{10y-42}{2}=5y - 21\), the equation is \(8y-78=5y - 21\).

Step4: Solve for \(y\)

Subtract \(5y\) from both sides: \(8y-5y-78=5y-5y - 21\), which simplifies to \(3y-78=-21\).
Add \(78\) to both sides: \(3y-78 + 78=-21+78\), so \(3y=57\).
Divide both sides by \(3\): \(y=\frac{57}{3}\).

Answer:

\(y = 19\)