Triangle `ABC` is isosceles with `AB=AC and BC=65cm.` P is a point on `BC` such that the perpendiculardistances from P to `AB and AC` are `24cm and 36cm,` respectively. The area of triangle `ABC` (in sq cm is)

Options

A. 1254
B. 1950
C. 2535
D. 5070

Answer

C

Explanation

`Delta = (1)/(2) b^(2) sin 2 theta = b^(2) sin theta cos theta`...(i)
Now, `(x)/(24) = (65 -x)/(36) " " ( :' DeltaPMB = ~= DeltaPNC)`
or `60x = (24) (65) " or " x = 26`
`:. sin theta = (12)/(13) and cos theta = (5)/(13)`
Now, `(b)/(sin theta) = (65)/(sin 2 theta)`
or `b = (65)/(2 cos theta) = ((65) (13))/((2)(5)) = (13^(2))/(2)`
From Eq. (i) we get
`Delta = (13^(4))/(4) xx (12)/(13) xx(5)/(13) = (169) (15) = 2535`