In the given figure, ∠C = 90o. If p = area of triangle ACD and q = area of triangle BCD, then what is the value of q/p ?

Hello Aayush, let DC be h and BC be x
Then by Pythagoras in triangle ACD,
AD^2 = AC^2 + DC^2
==> 8^2 = (3+x)^2 + h^2 -----(1)
And in triangle BCD,
BD^2 = BC^2 + CD^2
==> 7^2 = x^2 + h^2 -----(2)
Now (1) - (2) ==>
8^2 - 7^2 = (3+x)^2 - x^2
==> 15 * 1 = (3 + 2x) * 3 [a^2 - b^2 = ( a +b) (a - b)]
===> 15 = 9 + 6x
OR 6x = 6
==> x = 1 i.e BC = 1
Thus AC = 3 + 1 = 4
Now area of ACD p = 1/2 * AC * DC
Area of BCD q = 1/2 * BC * DC
==> q/p = BC/AC = 1/4
 

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Height of ACD = h (DC)
Base of ACD = x (AC)
1/2 * h * x = p

Height of BCD = h (DC)
Base of BCD = y (BC)
1/2 * h * y = q

So q/p 
= 1/2 * h * y/ 1/2 * h * x  ( Cancel 1/2 * h as they are equal)
= y/x 

= BC/AC

 
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