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Given parallelogram ABCD, diagonals AC and BD intersect at point E. AE=2x, BE=y+10, CE=x+2 and DE=4y−8. Find the length of BD.A. 18B. 32C. 16D. 6

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For the given parallelogram, the diagonals AC and BD intersect at point E. That means;


\begin{gathered} BE=DE \\ \text{Also,} \\ AE=CE \end{gathered}

Substitute for the given values and we'll have;


\begin{gathered} BE=DE \\ y+10=4y-8 \\ \text{Collect all like terms and you'll have;} \\ y-4y=-8-10 \\ -3y=-18 \end{gathered}

Divide both sides by -3 and you'll have;


\begin{gathered} (-3y)/(-3)=(-18)/(-3) \\ y=6 \end{gathered}

Note that line segment BD is made up of;


\begin{gathered} BD=BE+DE \\ BD=(y+10)+(4y-8)_{} \\ BD=y+10+4y-8 \\ BD=4y+y+10-8 \\ BD=5y+2 \\ \text{When y=6, we now have} \\ BD=5(6)+2 \\ BD=30+2 \\ \text{BD}=32 \end{gathered}

Therefore, the answer is option B

User Maximilian Lindsey
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