Question

The length of each side of an equilateral triangle is increased by 20%, resulting in triangle ABC. If the length of each side of the original equilateral is decreased by 20%, resulting in triangle DEF, how much greater is the area of triangle ABC than the area of triangle DEF?

Answer 1

Answer: Area of ΔABC is 2.25x the area of ΔDEF.

Step-by-step explanation: Because equilateral triangle has 3 equal sides, area is calculated as

$A=\frac{\sqrt{3} }{4} a^{2}$

with a as side of the triangle.

Triangle ABC is 20% bigger than the original, which means its side (a₁) measures, compared to the original:

a₁ = 1.2a

Then, its area is

$A_{1}=\frac{\sqrt{3} }{4}(1.2a)^{2}$

$A_{1}=\frac{\sqrt{3} }{4}1.44a^{2}$

Triangle DEF is 20% smaller than the original, which means its side is:

a₂ = 0.8a

So, area is

$A_{2}=\frac{\sqrt{3} }{4} (0.8a)^{2}$

$A_{2}=\frac{\sqrt{3} }{4} 0.64a^{2}$

Now, comparing areas:

$\frac{A_{1}}{A_{2}}= (\frac{\sqrt{3}.1.44a^{2} }{4})(\frac{4}{\sqrt{3}.0.64a^{2} } )$

$\frac{A_{1}}{A_{2}} =$ 2.25

The area of ΔABC is 2.25x greater than the area of ΔDEF.