SSC CHSL Model Paper 7

Question : 39 of 100

Marks: +1, -0

ABCD is a square, F is the mid-point of AB and E is a point on BC such that BE is one-third of BsC. If area of ΔFBE = 108 m

^{2}

, then the length of AC is:

Solution:

Let the side of the square be x, then

BE = x/3

and

BF = x/2

Area of Δ FEB =

1/2×x/3×x/2=x^2/12

Now,

x^2/12=108

⇒

x^2= 108× 12= 1296

In ΔADC, we have

AC^2 = AD^2 + DC^2

= x^2 + x^2=2x^2

=2× 1296 = 2592

or AC

=√2592 =36√2

Go to Question:

Prev Question

Next Question