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

## Check the Answer

Answer: 80.

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

Challenges and Thrills – Pre – College Mathematics

## Try with Hints

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 ………………..

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………

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

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

= \(80 \pi\) (Answer).

## Other useful links

- https://www.cheenta.com/area-of-square-singapore-mathematical-olympiad-2013-problem-no-17/
- https://www.youtube.com/watch?v=8tvcymMGKMI