If the perimeter of an isosceles right triang | vedclass

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(A) Let the two equal sides of the isosceles right triangle be $x$. Then,the hypotenuse is $\sqrt{x^2 + x^2} = x\sqrt{2}$. The perimeter is given by $

Vedclass · Accepted Answer

$4.5$

Vedclass · Answer

(A) Let the two equal sides of the isosceles right triangle be $x$. Then,the hypotenuse is $\sqrt{x^2 + x^2} = x\sqrt{2}$. The perimeter is given by $x + x + x\sqrt{2} = x(2 + \sqrt{2})$. Given that the perimeter is $(6 + 3\sqrt{2}) 	ext{ m}$,we have: $x(2 + \sqrt{2}) = 3(2 + \sqrt{2})$. Therefore,$x = 3 	ext{ m}$. The area of the triangle is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes x 	imes x = \frac{1}{2} 	imes 3 	imes 3 = 4.5 	ext{ m}^2$.

If the perimeter of an isosceles right triangle is $(6+3 \sqrt{2}) \text{ m}$,then the area of the triangle is ... $\text{m}^2$.

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