# Area of a triangle is x1 sq units. If its sides are doubled, the area of the new triangle formed is x2 sq units. What is the percentage increase in the area?

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Feb 9, 2018

 300%.

#### Explanation:

Suppose that, in $\Delta A B C , B C = a , C A = b , A B = c$.

Also, let $\left[A B C\right]$ denote the Area of $\Delta A B C$.

Then, by Heron's Formula,

[ABC]=sqrt(s(s-a)(s-b)(s-c)); s=(a+b+c)/2.

Consider $\Delta A ' B ' C '$ in which,

$B ' C ' = a ' = 2 B C = 2 a , C ' A ' = 2 b , \mathmr{and} A ' B ' = 2 c$.

Hence, $s ' = \frac{a ' + b ' + c '}{2} = \frac{2 a + 2 b + 2 c}{2} = a + b + c = 2 s$.

$s ' - a ' = 2 s - 2 a = 2 \left(s - a\right)$.

Similarly, $s ' - b ' = 2 \left(s - b\right) , \mathmr{and} s ' - c ' = 2 \left(s - c\right)$.

$\therefore \left[A ' B ' C '\right] = \sqrt{s ' \left(s ' - a '\right) \left(s ' - b '\right) \left(s ' - c '\right)}$,

$= \sqrt{2 s \cdot 2 \left(s - a\right) \cdot 2 \left(s - b\right) \cdot 2 \left(s - c\right)}$,

$= 4 \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$.

$\Rightarrow \left[A ' B ' C '\right] = 4 \left[A B C\right] , \mathmr{and}$

${x}_{2} = 4 {x}_{1}$.

$\therefore \text{ The increase in the area=} {x}_{2} - {x}_{1} = 3 {x}_{1}$.

$\therefore \text{ The % increase in the area=} \frac{{x}_{2} - {x}_{1}}{x} _ 1 \cdot 100$,

=(3x_1)/x_1*100=300%,

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