Question

The perimeter of an isosceles right-angled triangle is 6(√2+1) cm . Find its area. Please solve it with explanation.

Answer 1

$\textsf{Let the length of the equal sides of the isoceles right triangle be x}$

$\textsf{Then, its hypotenuse is}$

$\mathsf{Hypotenuse^2=x^2+x^2=2x^2}$

$\implies\mathsf{Hypotenuse=\sqrt{2}x}$

$\textsf{\underline{Given:}}$

$\textsf{Perimeter of the triangle = 6(\sqrt{2}+1) cm}$

$\implies\mathsf{\sqrt{2}x+2x=6(\sqrt{2}+1)}$

$\implies\mathsf{\sqrt{2}x(1+\sqrt{2})=6(\sqrt{2}+1)}$

$\implies\mathsf{\sqrt{2}x=6}$

$\implies\mathsf{x=3\sqrt{2}}$

$\textsf{Then, the are of the triangle }$

$\textsf{ =\frac{1}{2} *base*height}$

$\mathsf{=\frac{1}{2}*3\sqrt{2}*3\sqrt{2}}$

$\textsf{=9 square cm}$

Answer 2

Answer:

9 sq.cm

Step-by-step explanation:

Given,

Perimeter = 6(√2+1) cm

Let the each side of isosceles right angled triangle be x cm

Again,

(BC)² = (AB)² + (AC)²

⇒ (BC) ² = x² + x²

⇒ (BC) ² = 2x²

⇒ (BC) = √2x² = √2 x

According to question,

√2 x + 2x = 6(√2+1)

⇒ √2 x (1+√2) = 6( √2 + 1)

⇒ √2 x = 6

⇒ x = 3√2

Therefore,

Area = ½ x base x height

= ½ x 3√2 x 3√2 sq. cm

= ½ x 9 x 2 sq.cm

= 9 sq.cm

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