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

2 Answers

Answer:

#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

Answer:

#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