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

2 Answers

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

Apr 20, 2017

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