Question

Find the perimeter of an isosceles right triangle whose area is 200 cm square

Answer 1

Find the perimeter of an isoceles right triangle whose area is 200 cm^2.

since , we know that in isoceles triangle two sides are always equal. but here given that the isoceles triangle is right triangle . if the triangle is isoceles right triangle then its base and height will be equal.

why hypotenuse will not be equal either to height or base of isoceles right triangle ?

its because , according pythagoras theorem

if the triangle is right (90°) then the length of hypotenuse of that triangle will always be greater than the base and the height of that triangle.

let b be the hypotenuse and ( a , a ) be the base and height of isoceles right triangle.

from ∆ ABC

area of isoceles right triangle=200cm^2

( 1 / 2 ) × base × height = 200

( 1 × a × a ) / 2 = 200

a^2 = 400 => a = √400 => a = 20 cm

now in ∆ ABC ,

by Pythagoras theorm , we get

(AC)^2 = ( BC )^2 + ( AB )^2

b^2 = ( 20 )^2 + ( 20 )^2

b^2 = 400 + 400 = 800

b = √800 => 28.2 cm ( approx. )

since , we know that perimeter of triangle

= sum of all the sides of triangle.

therefore, perimeter of isoceles right triangle

= a + a + b = 20 + 20 + 28.2 = 68.2 cm

Your Answer : 68.2 cm
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Answer 2

Hello mate.

Thanks for asking this question.

Your answer is =>

In any isosceles triangle , let, $\triangle ABC$ , always two sides are equal along with respective opposite angles too.

SO ,

=> side AB = side BC = x cm.......(1)

In the given triangle , the angle between the congruent sides is taken as a right angle = 90°

so, it's clear that , we have to use Pythagoras Theorem to find the length of sides.

We know that , the area of any triangle is given by =>

$=> \boxed{\frac{1}{2} \: base\:\times\:height}$

but , it is given that , area of triangle is $200\:cm^{2}$

so , our equation becomes =>

$=> \frac{1}{2} \:base \:\times\:height\:\:=200cm^{2}$

but , in Isosceles triangle , as the one angle is given as right angle , then it's clear that the other remaining angles are the congruent angles.

so ,

side AB and BC will be the base and height of isosceles triangle between which the angle is 90°.

From equation (1) ,

=> AB = BC = x cm

=> $\frac{1}{2} \:(x)\times(x)\:=\:200\:cm^{2}$

=> $\frac{1}{2}\: x^{2}\:=\:200\:cm^{2}$

=> $x^{2} \:=\: (200\:\times\:2) cm^{2}$

=> $x^{2} \:=\:400 \:cm^{2}$

=> $x \:=\:20 \:cm$

hence ,

=> x = side AB = side BC = 20 cm.

now , as we talked , we will use Pythagoras Theorem to find the third side or the $Hypotenuse$ i.e side AC.

By using Pythagoras Theorem =>

=> $AC \:=\: \sqrt{\:AB^{2}\:+\:BC^{2}}$

=> $AC \:=\: \sqrt{x^{2}\:+\:x^{2}}$

=> $AC \:=\: \sqrt{\:2\:x^{2}}$

=> $AC\:=\: \sqrt{2 \:\times\:20^{2}}$

=> $AC\:=\: \sqrt{2\:\times\:400}$

=> $AC\:=\: \sqrt{800}$

=> $\boxed{AC\:=\: 28.2\:cm}$

NOW ,

perimeter of triangle = sum of side lengths.

Perimeter of given triangle =

=> side AB + side BC + side AC

=> 20 cm + 20 cm + 28.2 cm

=> 68.2 cm

therefore ,

$\underline{Perimeter\: of\: given\: isosceles}$ $\underline{triangle\: is\: 68.2 \:cm.}$



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