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

Name: 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
Uploaded: Oct. 22, 2024, 3:41 p.m.
Duration: 3 min 44 s
Description: 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

Solution

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 = $$\tan\ \theta\ \ =\ \ \frac{\ r}{\sqrt{\ 3}r}\ =\ 30$$
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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