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RRB NTPC Mains or Stage 2 Model Paper 8 with solutions
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© examsnet.com
Question : 14
Total: 120
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
Δ
F
B
E
=
108
m
2
,
then the length of AC is:
63 m
36
√
2
m
63
√
2
m
72
√
2
m
Validate
Solution:
(b) Let the side of the square be x, then
B
E
=
x
3
a
n
d
B
F
=
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
A
C
2
=
A
D
2
+
D
C
2
=
x
2
+
x
2
=
2
x
2
=
2
×
1296
=
2592
or
A
C
=
√
2592
=
36
√
2
© examsnet.com
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