Question

Consider a rhombus with diagonals of length $6$ and $8$ and the inscribed circle $C_1$. The vertices of the rhombus are the midpoints of the sides of a rectangle, inscribed in a circle $C_2$. The ratio of the radius of $C_1$ and the radius of $C_2$ is

• A. Less than $0.48$
• B. $0.48$
• C. $0.56$
• D. $0.64$
• E. Greater than $0.64$

Answer 1

The correct option is B $0.48$

ABCD is a rectangle and PQRS is the rhombus. Let radius of circle $c_1$ be r and that of circle $c_2$ be 'R'.

In rhombus PQRS, diagonals PR and QS are $6$ and $8$ respectively

We know that diagonals of a rhombus bisect each other at $90°$. Then in $△POQ$, using Pythagoras theorem,

${PQ}^2={PO}^2+{OQ}^2\Rightarrow PQ=\sqrt{{\left(\frac{6}{2}\right)}^2+{\left(\frac{8}{2}\right)}^2}=\sqrt{9+16}=\sqrt{25}=5㎝$

$ar△POQ=\frac{1}{2}\times PO\times OQ=\frac{1}{2}\times 3\times 4=6㎠$

$\Rightarrow \frac{1}{2}\times PQ\times r=6\Rightarrow r=\frac{6\times 2}{5}=2.4㎝$

Diameter of circle $c_2$ is equal to the diagonal of rectangle ABCD.

$\Rightarrow AC=\sqrt{{AD}^2+{DC}^2}=\sqrt{{PR}^2+{SQ}^2}=\sqrt{6^2+8^2}=\sqrt{100}=10㎝$

$\therefore$Radius $=\frac{10}{2}=5㎝=R$

$\therefore \frac{r}{R}=\frac{2.4}{5}=0.48$