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.
2D Geometry
Area of Square
Answer: 1041
Singapore Mathematical Olympiad - 2013 - Junior Section - Problem Number 17
Challenges and Thrills -
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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