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# Area of Square - Singapore Mathematical Olympiad - 2013 - Problem No.17

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

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).

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