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# Area of Circle – Singapore Mathematics Olympiad – 2013

Try this beautiful problem from Singapore Mathematics Olympiad based on Area of Circle. You may use sequential hints to solve the problem.

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

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).