Question 363260
The diagram shows an isosceles right triangle.
The two legs are "x," and the hypotenuse is x radical 2.
:
A. If its perimeter is 10, find x.
x + x + {{{x*sqrt(2)}}} = 10
2x + {{{x*sqrt(2)}}} = 10
{{{x*sqrt(2)}}} = -2x + 10
Square both sides
x^2(2) = (-2x+10)^2
FOIL (-2x+10)*(-2x+10)
2x^2 = 4x^2 - 20x - 20x + 100
2x^2 = 4x^2 - 40x + 100
:
Combine on the right
0 = 4x^2 - 2x^2 - 40x + 100
:
A quadratic equation
2x^2 - 40x + 100 = 0
:
Simplify, divide by 2
x^2 - 20x + 50 = 0
Use the quadratic formula to find x, a=1; b=-20; c=50
Two solutions
x = 17.06
and
x = 2.93 is the only reasonable solution
Check
2(2.93) + {{{2.93*sqrt(2)}}} = 10.00
:
:
B. If its area is 12, find x.
.5(x*x) = 12
multiply both sides by 2
x^2 = 24
x = {{{sqrt(24)}}}
x = 4.9 
:
Check
.5 * 4.9 * 4.9 = 12.0