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Manipal Entrance Test 2013 Math Paper
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© examsnet.com
Question : 50
Total: 80
AB is a diameter of a circle and C is any point on the circumference of the circle, then
The area of ΔABC is maximum when it is isosceles
The area of ΔABC is minimum when it is isosceles
The perimeter of ΔABC is minimum when it is isosceles
None of these
Validate
Solution:
Area of
Δ
A
B
C
,
A
=
1
2
x
√
d
2
−
x
2
for
max
∕
min
d
A
d
x
=
0
⇒
1
2
√
d
2
−
x
2
+
1
2
×
{
−
2
x
2
√
d
2
−
x
2
}
=
0
⇒
d
2
−
x
2
−
x
2
2
√
d
2
−
x
2
=
0
⇒
x
=
d
√
2
Also for
x
<
d
√
2
,
d
A
d
x
>
0
and for
x
>
d
√
2
,
d
A
d
x
<
0
Hence;
x
=
d
√
2
is the point of maxima.
∴
A
is max at
x
=
d
√
2
,
the area is maximum
y
=
d
√
2
∴ when
Δ
is isosceles,
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