ABCD is a parallelogram. P is any point on CD. If ar $(△ D P A) = 25 c m^{2}$ and ar $(△ A P C) = 40 c m^{2}$ , then area $(△ A P B)$ is

$45 c m^{2}$

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$75 c m^{2}$

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$65 c m^{2}$

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$32 c m^{2}$

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Solution

The correct option is C
$65 c m^{2}$

$A r e a (Δ A C D) = A r e a (Δ A P D) + A r e a (Δ A C P) = 65 c m^{2} Now, A B C D is a parallelogram. ∴ D C = A B . ∴ Δ A P B and Δ A D C have equal bases (A P and D C) and they are between same parallels (A P and D C) Hence, A r e a (Δ A P B) = A r e a (Δ A C D) = 65 c m^{2}$

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Similar questions

ABCD is a parallelogram. P is any point on CD. If ar( $△$ DPA) = 35 $c m^{2}$ and ar( $△$ APC) = 15 $c m^{2}$ , then area of ( $△$ APB) is equal to

A B C D

p

C D

a r (D P A) = 15 c m^{2}

a r (A P C) = 20 c m^{2}

a r (A P B) =

△

c m^{2}

△

c m^{2}

△

ABCD is a parallelogram. P is any point on CD. If $a r (△ D P A) = 15 c m^{2}$ and $a r (△ A P C) = 20 c m^{2},$ then $a r (△ A P B)$ =

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