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24. in the figure shown, ef is the midsegment of trapezoid abcd. find e…

Question

  1. in the figure shown, ef is the midsegment of trapezoid abcd. find ef. 24.________

Explanation:

Step1: Use the mid - segment formula

The formula for the mid - segment \(EF\) of a trapezoid \(ABCD\) is \(EF=\frac{AB + CD}{2}\). Here, \(EF = 5x+4\), \(AB=4x + 1\), and \(CD=7x + 5\). So, \(5x+4=\frac{(4x + 1)+(7x + 5)}{2}\).

Step2: Simplify the right - hand side

First, simplify \(\frac{(4x + 1)+(7x + 5)}{2}\). Combine like terms in the numerator: \((4x+7x)+(1 + 5)=11x+6\). Then \(\frac{11x + 6}{2}\). So the equation becomes \(5x+4=\frac{11x + 6}{2}\).

Step3: Cross - multiply

Multiply both sides of the equation \(5x+4=\frac{11x + 6}{2}\) by \(2\) to get \(2(5x+4)=11x + 6\). Expand the left - hand side: \(10x+8 = 11x+6\).

Step4: Solve for \(x\)

Subtract \(10x\) from both sides: \(8=x + 6\). Then subtract \(6\) from both sides to find \(x=2\).

Step5: Find the length of \(EF\)

Substitute \(x = 2\) into the expression for \(EF\). Since \(EF=5x+4\), then \(EF=5\times2+4\). Calculate \(5\times2+4=10 + 4=14\).

Answer:

\(14\)