Question

If the area of an isosceles triangle is 128cm^2 . Find its perimeter .

Answer 1

Step-by-step explanation:

Let third side of triangle x

⇒Perimeter =30 cm

semiparameter(s)=2perimeter

 

⇒2S=30,S=15

⇒12+12+x=30 ⇒ x=6

⇒A=s(s−a)(s−b)(s−c)   (Heron's formula)

⇒A=15(15−12)(15−12)(15−6)=15(3)(3)(9)

⇒A=915  cm2

Answer 2

key points :-

A triangle in which any two side equal then that triangle is considered as isosceles triangle .

let suppose in a triangle in which AB = BC then AB + BC > AC

or 2AB > AC .

Given :-

• area of isosceles triangle = 128 cm²

To find :-

• perimeter of isosceles triangle.

Formula used :-

$\sf \: area \: of \: isosceles \: triangle \: = \dfrac{ {a}^{2} }{2} \\ \\ \sf \: perimeter \: of \: isosceles \: triangle \: \\ \sf = 2a + a\sqrt{2}$

Solution :-

we have area of isosceles triangle .

$\sf \implies \: \dfrac{ {a}^{2} }{2} = 128 \\ \\ \sf \implies \: {a}^{2} = 256 \\ \\ \sf \implies \: a = \sqrt{256} \\ \\ \bf \implies \: a = 16cm$

now perimeter :-

$\sf \: perimeter \: = 2a + a\sqrt{2} \\ \\ \sf \: put \: the \: value \: of \: \bf a(side) \sf\: we \: get \: - \\ \\ \sf \implies \: 2 \times 16 + 16 \sqrt{2} \\ \\ \sf \implies \: 32 + 16 \sqrt{2} \\ \\ \sf \implies \: 4 \sqrt{2} + 16 \sqrt{2} \\ \\ \sf \implies \:20 \sqrt{2} cm \:$

hence, perimeter of isosceles triangle is 20√2 cm .