Try this beautiful problem from Singapore Mathematical Olympiad. 2013 based on the **area of Square**.

## Problem – Area of Square

*Let ABCD be a square and X and Y be points such that the lengths of XY, AX, and AY are 6,8 and 10 respectively. The area of ABCD can be expressed as \(\frac{m}{n}\) units where m and n are positive integers without common factors. Find the value of m+n. *

- 1215
- 1041
- 2001
- 1001

**Key Concepts**

**2D Geometry**

**Area of Square**

## Check the Answer

Answer: 1041

Singapore Mathematical Olympiad – 2013 – Junior Section – Problem Number 17

Challenges and Thrills –

## Try with Hints

This can the very first hint to start this sum:

Assume the length of the side is a.

Now from the given data we can apply Pythagoras’ Theorem :

Since, \(6^2+8^2 = 10^2\)

so \(\angle AXY = 90^\circ\).

From this, we can understand that \(\triangle ABX \) is similar to \(\triangle XCY\)

Try to do the rest of the sum……………………

From the previous hint we find that :

\(\triangle ABX \sim \triangle XCY\)

From this we can find \(\frac {AX}{XY} = \frac {AB}{XC} \)

\(\frac {8}{6} = \frac {a}{a – BX}\)

Can you now solve this equation ?????????????

This is the very last part of this sum :

Solving the equation from last hint we get :

a = 4BX and from this we can compute :

\(8^2 = {AB}^2 +{BX}^2 = {16BX}^2 + {BX}^2 \)

so , \( BX = \frac {8}{\sqrt {17}} and \(a^2 = 16 \times \frac {64}{17} = \frac {1024}{17}\)

Thus m + n = 1041 (Answer).

## Other useful links

- https://www.cheenta.com/triangle-inequality-problem-amc-12b-2014-problem-13/
- https://www.youtube.com/watch?v=ST58GTF95t4&t=140s