The perimeter of a right triangle is 24 centimeters. Three times the length of the longer leg minus two times the shorter leg exceeds the hypothenuse by 2 centimeters. What are the lengths of all three legs?
We draw a triangle, with x = shorter leg, y = longer leg, z = hypotenuse
The perimeter of a triangle is 24 centimeters.
x + y + z = 24
Three times the length of the longer leg minus two times the shorter leg exceeds the hypothenuse by 2 centimeters.
3y - 2x = z + 2
Since it's a right triangle, the Pythagorean theorem holds:
x² + y² = z²
So the system of equation is
(1) x + y + z = 24
(2) 3y - 2x = z + 2
(3) x² + y² = z²
Solve (1) for z:
(4) z = 24 - x - y
Substitute 24 - x - y for z in (2)
3y - 2x = 24 - x - y + 2
Simplify and solve for x
4y - x = 26
(5) 4y - 26 = x
Use (4) to substitute 24 - x - y for z in (3)
x² + y² = (24 - x - y)²
Square out the right side:
x² + y² = 576 + x² + y² - 48x - 48y + 2xy
Simplify
48x + 48y - 2xy = 576
Divide through by 2
(6) 24x + 24y - xy = 288
Use (5) to substitute 4y - 26 for x in (6)
24(4y - 26) + 24y - (4y - 26)y = 288
96y - 624 + 24y - 4y² + 26y = 288
-4y² + 146y - 912 = 0
Divide through by -2
2y² - 73y + 456 = 0
That factors as
(2y - 57)(y - 8) = 0
2y - 57 = 0; y - 8 = 0
2y = 57 y = 8
y = 28.5
Substitute each in (5)
4(28.5) - 26 = x; 4(8) - 26 = x
88 = x; 6 = x
We assumed that y is greater than x, since y was the longer leg.
So we can ignore y=28.5 and x=88, since that would make y the
shorter leg.
So x = 6, y = 8
Substitute in (4)
z = 24 - 6 - 8
z = 10
What are the lengths of all three legs?
They are shorter leg = x = 6, longer leg = y = 8, hypotenuse = z = 10
Edwin
