 Try this beautiful problem from Singapore Mathematics Olympiad based on Area of Circle.

## Problem – Area of Circle

On the xy – plane , let S denote the region consisting of all points

(x,y ) for which $|x+ \frac {1}{2} y | \leq 10$ and $|x|\leq 10$ and $|y|\leq 10$. The largest circle centered at ( 0,0 ) that can be fitted in the region S has area $k\pi$. Find the value of k.

• 81
• 80
• 79
• 84

### Key Concepts

Area of Circle

2 D Geometry

But try the problem first…

Source

Singapore Mathematics Olympiad – 2013 – Senior Section – Problem No. 14

Challenges and Thrills – Pre – College Mathematics

## Try with Hints

First Hint………….

We can start the sum from here :

The region S is the hexagon enclosed by lines

$x = \pm 10$ ; $y = \pm 10$ and $x + \frac {1}{2} y = \pm 10$.

So the largest circle contained in S must be tangent to $x + \frac {1}{2} y = \pm 10$

Try to find the radius from this ………………..

Second Hint ………….

We can start this hint by using a diagram:

So following the last hint we can understand that its radius is the distance from the origin (0,0) to $x + \frac {1}{2} y = 10$

r = $\frac {10}{ \sqrt {1+ ({\frac {1}{2}})^2}} = 4\sqrt {5}$.

Now it very easy to find the rest of the solutiion………

Final Hint …………..

So the area of a circle is $\pi {r}^2$

= $\pi {(4\sqrt 5)}^2$

= $80 \pi$ (Answer).