Question

How do you find the area of an isosceles right angle triangle with a perimeter of 40 units?

Answer 1

`Area=400(3-2sqrt2)`

Explanation:

An isosceles triangle has two of its sides the same length.

This triangle is also a right angled triangle so Pythagoras's Theorem holds.

(1) find the hypotenuse

call the equal lengths `""x`

then we have by Pythagoras

`x^2+x^2=c^2`

`=>c=xsqrt2`

so the sides are:`" "x,x,xsqrt2`

(2) Find `x`

the perimeter is 40

`:. x+x+xsqrt2=40`

`2x+xsqrt2=40`

`x(2+sqrt2)=40`

`x=40/(2+sqrt2)=20(2-sqrt2) " on rationalising"`

(3) Find the area

`Area =1/2xx x xx x`

`Area =1/2(20(2-sqrt2) )^2`

`Area =1/2 xx 400xx(4-4sqrt2+2)`

`Area =200xx(6-4sqrt2)`

take out common factors

`Area=400(3-2sqrt2)`

`= 68.63`

Answer 2

`A = 68.63` square units.

Explanation:

Let the two equal sides of the triangle be `x`.

The hypotenuse is `40-2x` because the perimeter is `40` units.

By Pythagoras: The square on the hypotenuse is equal to the sum of the squares on the other two sides.

`(40-2x)^2 = x^2 + x^2`

`1600-160x+4x^2 = 2x^2`

`2x^2 -160x +1600 =0" "div 2`

`x^2 -80x +800 = 0" "larr` solve for `x`

`x = (-(-80) +-sqrt(6400 - 4(1)(800)))/2`

`x =(80+sqrt3200)/2 = 68.28" "larr` reject as too big

`x = (80-sqrt3200)/2 = 11.7157`

The equal sides of the right-angled isosceles triangle are also the base and height. Use `x` to find the area.

`A = (bxxh)/2`

`A = (11.7157 xx 11.7157)/2 =68.63` square units