If the area of an isosceles right triangle is 8 cm^{2}, what is the perimeter of the triangle?

#### Options

8 + \[\sqrt{2}\] cm

^{2}8 + 4 \[\sqrt{2}\] cm

^{2}4+ 8 \[\sqrt{2}\] cm

^{2}12 \[\sqrt{2}\] cm

^{2}

#### Solution

We are given the area of an isosceles right triangle and we have to find its perimeter.

Two sides of isosceles right triangle are equal and we assume the equal sides to be the base and height of the triangle. We are asked to find the perimeter of the triangle

Let us take the base and height of the triangle be *x* cm.

Area of a isosceles right triangle, say *A* having base *x* cm and

height *x* cm is given by `A = 1/2 (" Base" xx "Height " )`

*A* = 8 cm^{2}; Base = Height = *x* cm

`8=1/2 (x xx x)`

8 × 2 = (x)^{2}

^{x = `sqrt(16)`}

^{x = 4 cm }

Using Pythagorean Theorem we have;

(Hypotenuse )^{2} = ( Base)^{2} + (Height )^{2 }

(Hypotenuse )^{2 }= (4)^{2} + (4)^{2}

(Hypotenuse )2 = 16 + 16

Hypotenuse = `sqrt(32)`

Hypotenuse = 4` sqrt(2)` cm

Let ABC be the given triangle

Perimeter of triangle ABC, say *P* is given by

p = AB + BC + AC

AB = 4 cm; BC = 4 cm; AC = `4 sqrt(2)`

`p = 4 + 4 + 4 sqrt(2)`

`p = 8 + 4 sqrt(2) ` cm