Question

The perimeter of an equilateral triangle is 126 cm . Find the area of incircle of the traingle

Answer 1

$\large\underline{\underline{Question:}}$

The perimeter of an equilateral triangle is 126 cm . Find the area of incircle of the traingle.

$\large\underline{\underline{To\:Find:}}$

$\underline{Area\:of\:the\:incircle}$

$\large\underline{\underline{We\:Know:}}$

• Perimeter of an equilateral triangle is :

☞ $perimeter = length × 3$

$\therefore length = \dfrac{perimeter}{3}$

• Area of a circle :

☞$Area = πr^{2}$

• Pythagoras theorem:

$a^{2} = b^{2} + c^{2}$

$\large\underline{\underline{Concept:}}$

$\therefore Radius\:of\:a\:circle = </p><p>\dfrac{Height}{3} of\:a\:triangle$

$\large\underline{\underline{Given:}}$

• Perimeter of the triangle = 126cm.

$\large\underline{\underline{Solution:}}$

First we have to find the side of the equilateral triangle.i.e.

$length = \dfrac{126}{3}cm$

$length = 42cm$

Hence ,the sides formed as:

• $Hypotenuse = 42 cm$
• $base = \dfrac{42}{2}cm = 21cm$
• Let the height be x cm.

By Pythagoras theorem :

${\boxed{a^{2} = b^{2} + c^{2}}}$

Where;

• a = hypotenuse
• b = base
• c = height.

Using the Pythagoras theorem ,we can find the height of the triangle.i.e.

${\boxed{42^{2} = 21^{2} + x^{2}}}$

$\Rightarrow 42^{2} - 21^{2} = x^{2}$

$\Rightarrow \sqrt{42^{2} - 21^{2}} = x$

$\Rightarrow \sqrt{42^{2} - 21^{2}} = x$

$\Rightarrow 21\sqrt{3} = x$

$\therefore Height = 21\sqrt{3}cm$

______________________________________

we know that,

$Radius\:of\:a\:circle = </p><p>\dfrac{Height}{3} of\:a\:triangle$

$Radius\:of\:a\:circle =</p><p>\dfrac{21\sqrt{3}}{3}$

${\boxed{\therefore Radius = \dfrac{21\sqrt{3}cm}{3}}}$

______________________________________

Area of the Circle = $Area = πr^{2}$

$π = \dfrac{22}{7}$

$\Rightarrow Area = \dfrac{22}{7} × \bigg(\dfrac{21\sqrt{3}}{3}\bigg)^{2}$

$\Rightarrow Area = \dfrac{22}{7} × \dfrac{1323}{9}$

$\Rightarrow Area = 22 × \dfrac{189}{9}$

$\Rightarrow Area = 462cm^{2}.$

${\boxed{\therefore Area = 462cm^{2}}}$