Each side of a square $A B C D$ is 12 cm. A point $P$ lies on side $D C$ such that area of $△ A D P :$ area of trapezium $A B C P = 2 : 3$ . Find $D P$ .

9 \cdot 6

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5 \cdot 4

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9 \cdot 1

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7 \cdot 6

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Solution

The correct option is A $9 \cdot 6$ cm
ABCD is a square with each side $12$ cm square.
P is a point on side DC.
$\frac{A r e a o f △ A D P}{A r e a o f t r a p e z i u m A B C P} = \frac{2}{3}$
Let CP=x
$∴ D P = 12 - x$
Now,
$A r e a o f △ A D P = \frac{1}{2} \times A D \times D P$
= $\frac{1}{2} \times 12 \times (12 - x)$
= $6 (12 - x)$
Area of trapezium ABCP= $\frac{(A B + C P)}{2} C B$
= $\frac{(12 + x)}{2} 12$
= $(12 + x) 6$
Now,
$\frac{6 (12 - x)}{6 (12 + x)} = \frac{2}{3}$
$=> 36 - 3 x = 24 + 2 x$
$=> - 3 x - 2 x = 24 - 36$
$=> - 5 x = - 12$
$=> x = 2.4$
$∴$ Length of CP= $2.4$ cm
$∴$ Length of DP= $12 - 2.4$ cm
= $9.6$ cm

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Each side of a square ABCD is 12 cm. A point P lies on side DC such that area of △ADP: area of trapezium ABCP=2:3. Find DP.

Each side of a square $A B C D$ is 12 cm. A point $P$ lies on side $D C$ such that area of $△ A D P :$ area of trapezium $A B C P = 2 : 3$ . Find $D P$ .