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
B
8.5 cm
C
9.6 m
D
10.6 cm

Solution

## The correct option is A 9.6 cmABCD 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                          = $$9.6$$cmMaths

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