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 $12 m$ . Area of parallelogram is

204 m^{2}

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408 m^{2}

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816 m^{2}

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402 m^{2}

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Solution

The correct option is C $408 m^{2}$
Area of parallelogram $A B C D$
$=$ Area of $△ A D B +$ Area of $△ A B C$
$= \frac{1}{2} \times 34 \times 12 + \frac{1}{2} \times 34 \times 12$
$= 204 + 204 = 408 m^{2}$ .

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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. Area of parallelogram is

The correct option is C 408 m2Area of parallelogram ABCD= Area of △ADB+ Area of △ABC=12×34×12+12×34×12=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 $12 m$ . Area of parallelogram is

The correct option is C $408 m^{2}$
Area of parallelogram $A B C D$
$=$ Area of $△ A D B +$ Area of $△ A B C$
$= \frac{1}{2} \times 34 \times 12 + \frac{1}{2} \times 34 \times 12$
$= 204 + 204 = 408 m^{2}$ .