the area of an isosceles rith triangle is 128cm^2 fond its perimeter [root 2=1.414
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let ABC is an isosceles triangle whose AB= BC
let AB=BC= x cm
therefore Area of right isosceles triangle = x^2/2
so, 128= x^2/2
x^2 = 256
x= 16
Now by using phythogorian Theorem of right isosceles triangle ,
h^2 = 2x^2
h^2 = 256×2
h = √512
h= 16√2
h= 16× 1.414
h= 22.624
so the perimeter of the given isosceles triangle = AB+BC+AC
= 16+ 16 + 22.624
= 54.624cm
let AB=BC= x cm
therefore Area of right isosceles triangle = x^2/2
so, 128= x^2/2
x^2 = 256
x= 16
Now by using phythogorian Theorem of right isosceles triangle ,
h^2 = 2x^2
h^2 = 256×2
h = √512
h= 16√2
h= 16× 1.414
h= 22.624
so the perimeter of the given isosceles triangle = AB+BC+AC
= 16+ 16 + 22.624
= 54.624cm
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