Question

Two circles are placed in an equilateral triangle ,one small and other big.the ratio of radius of smaller circle to larger one is

Answer 1

Radius of smaller circle : Radius of larger circle = 1:3

Step-by-step explanation:

In-radius of equilateral triangle of side $a =\frac{a}{2\sqrt{3}}$

Diameter of larger circle = $\frac{a}{2\sqrt{3}}$

Here common tangent PQ touches the two circle at R, center of smaller circle is I.

Now, PQ║ BC. AR ⊥ PQ. Also,Δ PQR is an equilateral triangle with  a straight line AORID,\.

AD = $\frac{\sqrt{3}}{2}a$

RD = $\frac{a}{\sqrt{3}}$

AR = $\frac{\sqrt{3}}{2}a - \frac{a}{\sqrt{3}}$

     =$\frac{3a -2a}{2\sqrt{3}} = \frac{a}{2\sqrt{3}}$

AR = $\frac{1}{3} AD$

Radius of smaller circle = $\frac{1}{3} radius\ of\ larger\ circle$

Radius of smaller circle : Radius of larger circle = 1:3

Answer 2

Step-by-step explanation:

In-radius of equilateral triangle of side a =\frac{a}{2\sqrt{3}}a=23a

Diameter of larger circle = \frac{a}{2\sqrt{3}}23a

Here common tangent PQ touches the two circle at R, center of smaller circle is I.

Now, PQ║ BC. AR ⊥ PQ. Also,Δ PQR is an equilateral triangle with  a straight line AORID,\.

AD = \frac{\sqrt{3}}{2}a23a

RD = \frac{a}{\sqrt{3}}3a

AR = \frac{\sqrt{3}}{2}a - \frac{a}{\sqrt{3}}23a−3a

     =\frac{3a -2a}{2\sqrt{3}} = \frac{a}{2\sqrt{3}}233a−2a=23a

AR = \frac{1}{3} AD31AD

Radius of smaller circle = \frac{1}{3} radius\ of\ larger\ circle31radius of larger circle

Radius of smaller circle : Radius of larger circle = 1:3