Question

If the area of an isolationist triangle is 8 cm^2 what is perimeter of the triangle

Answer 1

Answer:

$\huge \underline{ \sf{ \red{Given}}}$

Area of isosceles triangle is 8cm²

Find :- Perimeter of isosceles triangle

Solution:-

Area of isosceles triangle= 1/2×base×height

WKT... In isosceles triangle two sides are equal.

So, Base and height are equal

Base=Height=A

Area = 8cm

$\frac{1}{2} \times a \times a = 8$

$\frac{1}{2} \times {a}^{2} = 8$

${a}^{2} = 8 \times 2$

${a}^{2} = 16$

$a = \sqrt{16}$

$a = 4$

Therefore, the two sides of triangle are 4,4.

From Pythagoras theorm we can find perimeter:-

hyp²=opp²+adj²

${hyp}^{2} = {4}^{2} + {4}^{2}$

${hyp}^{2} = 16 + 16$

${hyp}^{2} = 32$

$hyp = \sqrt{32}$

$hyp = 4 \sqrt{2}$

The other side of isosceles triangle is 4√2.

Now, Perimeter= 4+4+4√2 = 8+4√2

The perimeter of Isosceles triangle is 8+4√2 cm