The area of an isosceles triangle …
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Hans Raj 6 years, 7 months ago
let us consider a right Ld iscosceles triangle ABC rt Ld at B
AB = BC=x (Given as it is iscosceles triangle)
area = 1/2 x base x height
= 1/2 x x = 1/2 x2 = 5000
x2=10000
x=100 m
AB = 100 m
BC = 100 m
AC2=AB2 + BC2
= (100)2 + (100)2
= 10000 + 10000
= 20000
AC = 100{tex} \sqrt{2}{/tex} m
thus perimeter = AB + BC +AC = 100 + 100 + 100 {tex} \sqrt{2}{/tex}
= 100(2 + {tex} \sqrt{2}{/tex}) m
1Thank You