The given figure shows a trapezium in which AB || DC and diagonal AC and BD intersect at point P. If AP : CP = 3:5 and area of $△ A P B$ is $27 m^{2}$ , Find the area of $△ C P D$ .

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Solution

In $Δ A P B$ and $Δ C P D$ ,

$∠ A P B = ∠ C P D$ (vertically opposite angles)

$∠ P D C = ∠ P B A$ (alternate angles)

$∴ Δ A P B \sim Δ C P D$ (by AA similarity criterion)

$\frac{A P^{2}}{C P^{2}} = \frac{a r (Δ A P B)}{a r (Δ C P D)}$

(Since the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.)

${(\frac{A P}{C P})}^{2} = \frac{a r (Δ A P B)}{a r (Δ C P D)}$

$\Rightarrow {(\frac{3}{5})}^{2} = \frac{27 m^{2}}{a r (Δ C P D)}$

$\Rightarrow a r (Δ C P D) = \frac{27 \times 25}{9} = 75 m^{2}$

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The given figure shows a trapezium in which AB || DC and diagonal AC and BD intersect at point P. If AP : CP = 3:5 and area of $△ A P B$ is $27 m^{2}$ , then the area of $△ C P D$ is ___.

The given figure shows a trapezium in which AB is parallel to DC and diagonals AC and BD intersect at point P. If AP : CP = 2 : 3,

$A r (Δ C P D) : A r (Δ A P B) = ?$

Q. In the given figure, the diagonals AC and BD intersect at point O. If OB = OD and AB//DC, show that :

A r e a (Δ D C B) = A r e a (Δ A C B)

Q. In the figure, diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that

a r e a (△ A O D) = a r e a (△ B O C)

State true or false:

In a trapezium ABCD, side AB is parallel to side DC; and the diagonals AC and BD intersect each other at point P, then

Δ A P B

is similar to

Δ C P D .

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The given figure shows a trapezium in which AB || DC and diagonal AC and BD intersect at point P. If AP : CP = 3:5 and area of △APB is 27 m2, Find the area of △CPD .

The given figure shows a trapezium in which AB || DC and diagonal AC and BD intersect at point P. If AP : CP = 3:5 and area of $△ A P B$ is $27 m^{2}$ , Find the area of $△ C P D$ .