In the given figure- ABQP and ABCD are two parallelograms on the same base AB- Area of parallelogram ABCD - Area of triangle PAB is equal to-

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Standard IX

Mathematics

Theorem 1: Parallelograms

In the given ...

Question

In the given figure, ABQP and ABCD are two parallelograms on the same base AB. $\frac{Area of parallelogram ABCD}{Area of triangle PAB}$ is equal to:

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Solution

ABQP is a parallelogram. Therefore,
ar( $△ P A B) = \frac{1}{2} a r (A B Q P)$ (since diagonal of a parallelogram divides it into 2 equal areas)
But, ar(ABQP) = ar(ABCD) (parallelograms on the same base and between the same parallels)
Therefore ar(ABCD) = $2 \times a r (△ P A B)$
Hence, $\frac{Area of parallelogram ABCD}{Area of triangle PAB} = 2$

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In the given figure, ABQP and ABCD are two parallelograms on the same base AB. Area of parallelogram ABCDArea of triangle PAB is equal to: __

In the given figure, ABQP and ABCD are two parallelograms on the same base AB. $\frac{Area of parallelogram ABCD}{Area of triangle PAB}$ is equal to:

__