Question

When each side of an equilateral triangle is increased by 4 cm, its area increases by 5√3 cm^2, then the length of the side of the original triangle is

Answer 1

Given

• Side of an equilateral triangle is increased by 4 cm
• Area increases by 5√3 cm²

Explanation:

Let the original side of the equilateral triangle be x and ${\sf{ A_1}}$

$\circ{\boxed{\underline{\sf{ Area_{(Equilateral \ Triangle)} = \dfrac{ \sqrt{3} }{4} a^2 }}}}$

Now, We can find desired results as:-

${\sf{ A_1 = \dfrac{ \sqrt{3} }{4} x^2 }}$

The new side of the Equilateral triangle is (x + 4)cm and ${\sf{ A_2 = \dfrac{ \sqrt{3} }{4} (x + 4)^2 \ cm^2 }}$

According to Question,

$\colon\implies{\sf{ A_2-A_1 = 5 \sqrt{3} \ cm }} \\ \\ \\ \colon\implies{\sf{ \dfrac{ \sqrt{3} }{4} (x+4)^2 - \dfrac{ \sqrt{3} }{4} (x)^2 = 5 \sqrt{3} }} \\ \\ \\ \colon\implies{\sf{ \dfrac{ \cancel{ \sqrt{3} } }{4} \left( (x+4)^2 - x^2 \right) = 5 \ \cancel{ \sqrt{3} } }} \\ \\ \\ \colon\implies{\sf{ \dfrac{1}{4} (x^2 +16+8x-x^2) = 5 }} \\ \\ \\ \colon\implies{\sf{ \cancel{x^2} + 16 + 8x - \cancel{x^2} = 20 }} \\ \\ \\ \colon\implies{\sf{ 16 + 8x = 20 }} \\ \\ \\ \colon\implies{\sf{ 8x = 20-16}} \\ \\ \\ \colon\implies{\sf{ \cancel{ 8} \ x = \cancel{4} }} \\ \\ \\ \colon\implies{\boxed{\mathfrak\pink{ x = \dfrac{1}{2} cm }}}$

Hence,

• The length of the side of the original triangle is ${\sf{ \dfrac{1}{2} cm }}$.