SOLUTION: Find the perimeter of a right triangle whose hypotenuse is 2 and whose area is 1.

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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!

the area of a right triangle is:
where and are legs


if hypotenuse is , then


......eq.1

if area is , then

...substitute

....square both sides










solution




then

......eq.1




perimeter:




Answer by mananth(16946)   (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)




Answer by ikleyn(52775)   (Show Source): You can put this solution on YOUR website!
.
Find the perimeter of a right triangle whose hypotenuse is 2 and whose area is 1.
~~~~~~~~~~~~~~~~~~~~~ 

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.

Solved.



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