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Question

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


A
9.6 cm
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B
8.5 cm
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C
9.6 m
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D
10.6 cm
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Solution

The correct option is A 9.6 cm
ABCD is a square with each side $$12$$cm square.
P is a point on side CD.

$$\dfrac{Area\  of\  \triangle \ ADP}{Area\  of\ trapezium\ ABCP}=\dfrac{2}{3}$$

Let $$CP=x$$
$$\therefore DP=12-x$$

Now,
$$Area  of \triangle ADP=\dfrac{1}{2}\times AD\times DP$$

                            =$$\dfrac{1}{2}\times 12\times (12-x)$$

                            =$$6(12-x)$$

Area  of trapezium ABCP=$$\dfrac{(AB+CP)}{2}BC$$

                                            =$$\dfrac{(12+x)}{2}12$$

                                            =$$(12+x)6$$
Now,
$$\dfrac{6(12-x)}{6(12+x)}=\dfrac{2}{3}$$

$$=>36-3x=24+2x\\$$
$$=>-3x-2x=24-36\\$$
$$=>-5x=-12\\$$
$$=>x=2.4$$

$$\therefore$$ Length of CP = $$2.4$$cm
$$\therefore$$ Length of DP = $$12-2.4$$cm
                          = $$9.6$$cm

359342_197279_ans_6703b04ee16e4d8f9cb44b96237a5fce.png

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