In a parallelogram $A B C D$ diagonal $A C$ measures $34 m$ and the perpendicular distance of $A C$ from either of the vertices $B$ and $D$ is $12 m$ . Find area of parallelogram.

204 m^{2}

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408 m^{2}

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816 m^{2}

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None

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Solution

The correct option is B. $408 m^{2}$

In $Δ A B C$
Base $= A C = 34 m$
Height $= B M = 12 m$
Area of $Δ A B C = \frac{1}{2} \times 34 \times 12$
$= 204 m^{2}$
Similarly the area of $Δ A D C = 204 m^{2}$ [Diagonal of a parallelogram divides it into two triangles of equal area]
Area of parallelogram $= 204 + 204$
$= 408 m^{2}$

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In a parallelogram ABCD diagonal AC measures 34 m and the perpendicular distance of AC from either of the vertices B and D is 12 m. Find area of parallelogram.

The correct option is B. 408m2In ΔABCBase =AC=34 m Height =BM=12 mArea of ΔABC=12×34×12=204 m2Similarly the area of ΔADC=204 m2 [Diagonal of a parallelogram divides it into two triangles of equal area]Area of parallelogram =204+204 =408 m2

In a parallelogram $A B C D$ diagonal $A C$ measures $34 m$ and the perpendicular distance of $A C$ from either of the vertices $B$ and $D$ is $12 m$ . Find area of parallelogram.