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Question

If an isosceles triangle ABC, in which AB=AC=6 cm, is inscribed in a circle of radius 9 cm, find the area of the triangle.

A
42cm2
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B
83cm2
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C
4cm2
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D
8cm2
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Solution

The correct option is A 83cm2

Given: ΔABC is isosceles inscribed in a circle of radius 9 cm & center O.

AB=AC=6 cm


To find out-Area(ΔABC)
Solution-
We join AO which intersects BC at P.
In ΔABC it is given that AB=BC
ABC=ACB ....(i)
Now, in ΔABP & ΔACP we have,
AB=AC ...(given),
side AP common
And, ABC=ACB ...(by i)
ΔABP & ΔACP are congruent.
i.e APC=APB.
But they are adjacent angles.
So, APC=APB=90o.......(ii)
and BP=PCBP=12×BC ...........(iii).
BPAO .....(iv)
Also, ΔABP is a right one with hypotenuse AB .........(v)
Let AP=x cm and PB=PC=y cm then
From Pythagorean theorem, we have AB2=AP2+BP2
36=x2+y2.........(vi)
And also ΔBPO is also a right angled triangle, so BO2=PB2+OP2
81=y2+(9x)2
81=y2+8118x+x2........(vii)
Subtracting equation (vi) from equation(vii), we get
8136=y2+8118x+x2x2y2
818136=18x
36=18x
x=2 cm
Thus, y2=3622
=364
=32
y=42 cm
In ΔABC, Base=BC=2(PB)=2y=2(42)=82 cm
and Height=x=2 cm
Therefore, area =12×82×2
=82sq. cm.

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