Question

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1) If the side of an equilateral triangle is increased to 2 times of its normal length then find the percentage of area increased ????



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Answer 1

Let the side of the equilateral triangle be a


We know that area of equilateral triangle = √3a²/4

Now original area = √3a²/4

Given that new length = 2 time of normal

=> 2a

So area = √3(2a)²/4

=> area = √3 × 4a²/4

=> area = √3a²


Area increased = √3a² - √3a²/4

=> area increased = √3a²( 1 - 1/4)

=> area increased = √3a²(4/4 - 1/4)

=> area increased = √3a²(3/4)

=> area increased = 3√3a²/4


So percentage increased = (3√3a²/4) ÷( √3a²/4) × 100

=> percentage increased = 3√3a²/4 × 4/√3a² × 100

=> percentage increased = 3 × 100

=> percentage increased = 300%


Hope it helps dear friend

Answer 2

Let the initial side  be  a₁  and  area  A₁ and  final side  be  a₂ and  area A₂

area of  Δ, A₁ = √3a₁²/4

Given, Increasing the side 2 times.

New  length, a₂ = 2a₁

                      a₂/a₁ = 2

New area, A₂ =  √3a₂²/4

$\frac{A_1}{A_2}=\frac{\frac{\sqrt{3}(a_1)^{2}}{4}}{\frac{\sqrt{3}(a_2)^{2} }{4}}\\\\\frac{A_1}{A_2}=(\frac{a_1}{a_2})^{2}\\\\\frac{A_2}{A_1}=(\frac{a_2}{a_1})^{2}\\\\\frac{A_2}{A_1}-1=(\frac{a_2}{a_1})^{2}-1\\\\\frac{A_2-A_1}{A_1}=(2)^{2}-1\\\\\frac{A_2-A_1}{A_1}=4-1\\\\\frac{A_2-A_1}{A_1}=3\\\\\frac{A_2-A_1}{A_1}\times100\%=3\times100\%\\\\\frac{A_2-A_1}{A_1}\times100\%=300\%$

Hence, Increase  in area   = 300%