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-A- 6B- 2C- 3D- 4

The correct option is B 2 In the figure,ABQP is a parallelogram. Therefore, ar( PAB) = 1/2 ar(ABQP) nbsp; nbsp; (since diagonal of a parallelogram divides ...

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Byju's Answer

Standard VIII

Mathematics

Formula for Area of Parallelogram

In the given ...

Question

In the given figure, ABQP and ABCD are two parallelograms on the same base AB. $\frac{A r e a o f p a r a l l e l o g r a m A B C D}{A r e a o f t r i a n g l e P A B}$ is equal to:

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Solution

The correct option is A
2

In the figure,ABQP is a parallelogram.
Therefore,
$a r (△ P A B) = \frac{1}{2} a r (A B Q P)$ (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)
$∴ a r (A B C D) = 2 \times a r (△ P A B)$
Hence, $\frac{A r e a o f p a r a l l e l o g r a m A B C D}{A r e a o f t r i a n g l e P A B} = 2.$

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