How 9 Cheenta students ranked in top 100 in ISI and CMI Entrances?

Learn MoreContent

[hide]

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

**2D Geometry**

**Area of Square**

But try the problem first...

Answer: 1041

Source

Suggested Reading

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

Challenges and Thrills -

First hint

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........................

Second Hint

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 ?????????????

Final Step

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).

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

Cheenta is a knowledge partner of Aditya Birla Education Academy

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.

JOIN TRIAL