CAT 2003 Question Paper (Leaked) Question 77
There are two concentric circles such that the area of the outer circle is four times the area of the inner circle. Let A, B and C be three distinct points on the perimeter of the outer circle such that AB and AC are tangents to the inner circle. If the area of the outer circle is 12 square centimeters then the area (in square centimeters) of the triangle ABC would be
Let R ,r be radius of big and small circles respectively. We know that R=2r.
And since area = 12 ;
$$R^2 = \frac{12}{\pi}$$.
$$r^2$$ = $$\frac{3}{\pi}$$
By pythagoras theorem in the small triangle with side 'x' we have x = $$\sqrt{\ 3}r$$.
Angle OAB =Â $$\tan\ \theta\ \ =\ \ \frac{\ r}{\sqrt{\ 3}r}\ =\ 30$$
Hence Angle CAB = 60.
So area = $$\ \frac{\ 1}{2}\ \times\ 2\sqrt{\ 3}r\times\ 2\sqrt{\ 3}r\times\ \sin\ 60$$.
On substituting the value of sin 60 and $$r^2$$, we get
Area = $$9\sqrt3/\pi$$.
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Achal Mukaty
2Â years, 8Â months ago
From where did sin60 came?
Andy Andy
2Â years, 11Â months ago
I think the angle you found out is OAB and not AOB, just a typo mistake but excellent explaination