Question

A circle is inscribed in a rhombus with diagonals 12 cm and 16 cm. The ratio of the area of circle to the area of rhombus is

• A. \(\frac{5π}{18}\)
• B. \(\frac{6π}{25}\)
• C. \(\frac{3π}{25}\)
• D. \(\frac{2π}{15}\)

Answer 1

The Correct Option is B

Given the circle is inscribed in the rhombus of diagonals \(12\) and \(16\) . 
Let O be the point of intersection of the diagonals of the rhombus. Then, \(OE\) (radius) ⊥ \(DC\).
Also \(DC = \sqrt{6^2+8^2} = 10\)
As area of \(ΔODC\) should be the same, we have,\(\frac{1}{2}×6×8=\frac{1}{2}×OE×10\)
\(⇒ OE = 4.8\)
\(∴\)  Required ratio of areas = \(\frac{π(4.8)^2}{\frac{1}{2}×12×16} = \frac{6π}{25}\)

So, the correct answer is (B): \(\frac{6π}{25}\)

Answer 2

Let the radius be r.
Apply Pythagoras Theorem,
\(x^2+r^2=6^2\)        \(……… (1)\)
\((10-x)^2+r^2=8^2\)    \(……. (2)\)
On solving both equations
\(x=3.6\)
\(r^2=36-(3.6)^2\)
\(r^2 = 23.04\)
Area of the circle,
\(=\pi r2\)
\(= \pi \times 23.04\)
The area of rhombus,
\(= \frac 12 d_1 d_2\)

\(= \frac 12 \times 12 \times 16\)
\(= 6 \times 16\)
\(= 96\)

Now, the ratio of area of cirele to the area of rhombus,
\(=\frac { \pi \times 23.04}{ 96}\)

\(= \frac {6\pi}{25}\)

So, the correct option is (B): \(\frac {6\pi}{25}\)