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# Problem on Area of Triangle | SMO, 2010 | Problem 32

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

Try this beautiful problem from Singapore Mathematics Olympiad based on area of triangle.

## Problem – Area of Triangle (SMO Entrance)

Given that ABCD is a square . Points E and F lie on the side BC and CD respectively, such that BE = $\frac {1}{3} AB$ = CF. G is the intersection of BF and DE . If

$\frac {Area of ABGD}{Area of ABCD} = \frac {m}{n}$ is in its lowest term find the value of m+n.

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• 19
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### Key Concepts

2D – Geometry

Area of Triangle

Challenges and Thrills – Pre – College Mathematics

## Try with Hints

Here is the 1st hint for them who got stuck in this problem:

At 1st we will join the points BD and CG .

Then to proceed with this sum we will assume the length of AB to be 1. The area of $\triangle {BGE}$ and $\triangle {FGC}$ are a and b respectively.

Try to find the area of $\triangle {EGC}$ and $\triangle {DGF}$……………………..

Now for the 2nd hint let’s start from the previous hint:

So the area of $\triangle {EGC}$ and $\triangle {DGF}$ are 2a and 2b .From the given value in the question we can say the area of $\triangle {BFC} = \frac {1}{3}$(as BE = CF = 1/3 AB\).

We can again write 3a + b = $\frac {1}{6}$

Similarly 3b + 2a = area of the $\triangle DEC = \frac {1}{3}$.

Now solve this two equation and find a and b…………..

In the last hint :

2a + 3b = 1/3……………………(1)

3a + b = 1/6……………(2)

so in $(1) \times 3$ and in $(2) \times 2$

6a + 9b = 1

6a + 2b = 1/3

so , 7 b = 2/3 (subtracting above 2 equation)

b = $\frac {2}{21}$ and a = $\frac {1}{42}$

So,$\frac {Area of ABGD}{Area of ABCD} = 1-3(a+b) = 1- \frac {15}{42} = \frac {9}{14}$

Comparing the given values from the question , $\frac {Area of ABGD}{Area of ABCD} = \frac {m}{n}$

m = 9 and n = 14