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) About Me  (Show Source):
You can put this solution on YOUR website!

the area of a right triangle is:
A=%281%2F2%29b%2Aa where a and b+are legs


if hypotenuse is c=2, then

a=sqrt%282%5E2-b%5E2%29
a=sqrt%284-b%5E2%29......eq.1

if area is 1, then

%281%2F2%29b%2Aa+=1...substitute+a
%281%2F2%29b%2Asqrt%284-b%5E2%29+=1
b%2Asqrt%284-b%5E2%29+=2....square both sides
%28b%2Asqrt%284-b%5E2%29%29%5E2+=2%5E2
b%5E2%2A%284-b%5E2%29+=4
4b%5E2-b%5E4+-4=0
-b%5E4+%2B4b%5E2-4=0
-%28b%5E4+-4b%5E2%2B4%29=0
-%28b%5E4+-2b%5E2-2b%5E2%2B4%29=0
-%28%28b%5E4+-2b%5E2%29-%282b%5E2-4%29%29=0
-%28b%5E2%28b%5E2+-2%29-2%28b%5E2-2%29%29=0
-%28%28b%5E2+-2%29%28b%5E2-2%29%29=0

solution
b%5E2-2=0
b%5E2=2
b=sqrt%282%29

then

a=sqrt%284-sqrt%282%29%5E2%29......eq.1
a=sqrt%284-2%29
a=sqrt%282%29


perimeter:
P=a%2Bb%2Bc
P=sqrt%282%29%2Bsqrt%282%29%2B2
P=2sqrt%282%29%2B2

Answer by mananth(16946) About Me  (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= sqrt%282%29%2Bsqrt%282%29%2B2
Perimeter =2%2B2%2Asqrt%282%29
a^2+b^2=c^2
c=2 (hypotenuse)
+sqrt%282%29%5E2%2Bsqrt%282%29%5E2=+4



Answer by ikleyn(52775) About Me  (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 = sqrt%282%29.


The perimeter then is  x + y + 2 = sqrt%282%29 + sqrt%282%29 + 2 = 2 + 2%2Asqrt%282%29  units.

Solved.