Question 1204789: Find the perimeter of a right triangle whose hypotenuse is 2 and whose area is 1. Found 3 solutions by MathLover1, mananth, ikleyn:Answer by MathLover1(20849) (Show Source):
You can put this solution on YOUR website! Area of triangle = 1
Area of triangle = 1/2 * base * height
2= b*h
b=2/h
h^2+(2/h)^2=2^2 (pythagoras theorem)
h^2 +4/h^2 = 4
h^2 +4/h^2 -4=0
h^4+4-4h^2=0
h^4-4h^2+4=0
(h^2-2)^2=0
h=sqrt(2)
b= 2/sqrt(2) = sqrt(2)
Perimeter=
Perimeter =
a^2+b^2=c^2
c=2 (hypotenuse)
You can put this solution on YOUR website! .
Find the perimeter of a right triangle whose hypotenuse is 2 and whose area is 1.
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Let x and y be the legs.
Then we have
x^2 + y^ 2 = 4 (1) (Pythagorean equation)
xy = 2 (2) (from the area equation)
Multiply equation (2) by 2 and subtract if from equation (1). Then you get
x^2 - 2xy + y^2 = 0,
or
(x-y)^2 = 0.
It implies x = y.
So, our triangle is isosceles right angled triangle.
Then from equation (2), we have
x^2 = 2,
which implies x = y = .
The perimeter then is x + y + 2 = + + 2 = 2 + units.