Question

find the perimeter of an isosceles triangle right angle triangle having area 200 CM square

Answer 1

Let the triangle be ABC
AB=BC=x
Area = 1/2 (x.x)
200 = 1/2 x^2
x^2 = 400
Square rooting both side
x = 20
Therefore AB=BC=20
By Pythagoras theorem
AC= 20root2
Hence, perimeter= 20+20+20(1.41) = 68.2

Answer 2

$\huge\underline\mathfrak{Answer:}$

Let each equal side be x of an isosceles right triangle ABC. Then

Area of an isosceles right triangle ABC

$\sf\frac{1}{2}$(x$\times$x)= $\sf\frac{1}{2}x^2$

But area of a triangle = $\sf{200\:cm^2}$

•°• $\sf\frac{1}{2}x^2=200$

$\implies$$\sf{x^2=400=20\times}$

$\implies$$\sf{x=20cm}$

•°• Hypotenuse AC of ∆ABC

= $\sf\sqrt{x^2+x^2}$ cm

= $\sf\sqrt{2x^2}$cm = $\sqrt{2}$x cm.

Now, the perimeter of an isosceles right triangle

= $\sf{(x+x+}$$\sf\sqrt{2x)}$cm

= $\sf{(20+20+20}$$\sf\sqrt{2)}$cm

= $\sf{(40+20}$$\sf\sqrt{2}$cm

20√2 (√2 + 1) cm.