RRB NTPC Mains or Stage 2 Model Paper 8 with solutions

Question : 14 of 120

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 BC. If area of

ΔFBE = 108 m^2,

then the length of AC is:

Solution:

(b) Let the side of the square be x, then

BE =x/3 and BF= x/2

Area of ΔFEB=

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

AC = √2592 = 36 √2

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