Question

ru is the midsegment of the trapezoid pqst. if st = -x + 51, ru = 2x - 30, and pq = 4x - 82, what is the value of x? t u p s r q x =

Explanation:

Step1: Recall midsegment formula for trapezoid

The midsegment (or median) of a trapezoid is the segment that connects the midpoints of the non - parallel sides. The length of the midsegment \(m\) of a trapezoid with bases \(b_1\) and \(b_2\) is given by the formula \(m=\frac{b_1 + b_2}{2}\). In trapezoid \(PQST\), the bases are \(ST\) and \(PQ\), and the midsegment is \(RU\). So we have the equation \(RU=\frac{ST + PQ}{2}\).

Step2: Substitute the given expressions into the formula

We know that \(ST=-x + 51\), \(RU = 2x-30\), and \(PQ=4x - 82\). Substituting these into the formula \(RU=\frac{ST + PQ}{2}\), we get:
\(2x-30=\frac{(-x + 51)+(4x - 82)}{2}\)

Step3: Simplify the right - hand side of the equation

First, simplify the numerator of the fraction on the right - hand side: \((-x + 51)+(4x - 82)=-x+4x + 51-82 = 3x-31\). So the equation becomes \(2x-30=\frac{3x - 31}{2}\)

Step4: Eliminate the fraction by multiplying both sides by 2

Multiply both sides of the equation \(2x-30=\frac{3x - 31}{2}\) by 2 to get rid of the denominator:
\(2(2x - 30)=3x-31\)
Using the distributive property \(a(b - c)=ab-ac\) on the left - hand side, we have \(4x-60 = 3x-31\)

Step5: Solve for x

Subtract \(3x\) from both sides of the equation \(4x-60 = 3x-31\):
\(4x-3x-60=3x-3x - 31\)
\(x-60=-31\)
Then add 60 to both sides:
\(x-60 + 60=-31 + 60\)
\(x = 29\)

Answer:

\(29\)