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Question 35
ABCD is a rectangle. P is a point on the side AB as shown in the given figure. If DP = 13, CP = 10 and BP = 6, then what is the value of AP ?
Solution
We have :
Now CP =10
BP=6
We know Angle B is a right angle
Now therefore CP^2=BC^2+BP^2
So we get BC=8
Now BC=AD=8
Now in triangle ADP
We get DP^2=AD^2+AP^2
So we get 169=64+AP^2
we get AP^2=105
AP=$$\sqrt{\ 105}$$
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