Question 585308: A 60cm piece of wire is bent to form a right triangle. If the hypotenuse is 26cm long, what are legs of the right triangle?
Found 2 solutions by Schaman_Dempster, mananth:
Answer by Schaman_Dempster(26)   (Show Source):
You can put this solution on YOUR website! Let the other sides of the right triangle be x and y.
Since the length of the wire was 60 cm, the whole 60 cm has been bent to form a right triangle and that means 60 cm is as well the perimeter of the triangle.
Therefore, x + y + 26 = 60
or x + y = 34
or y = 34 -x ---------------- (1)
Using the Pythagoras theorem,
x^2 + y^2 = 26^2
x^2 + (34 - x)^2 = 26^2 {Using equation (1)}
x^2 + 34^2 + x^2 - 68x = 26^2
2x^2 - 68x + 34^2 - 26^2 = 0
2x^2 - 68x + (34 -26)(34+26) = 0 { a^2 - b^2 = (a+b)(a-b)}
2x^2 - 68x + 8*60 = 0
2x^2 - 68x + 480 = 0
x^2 - 34x + 240 = 0 { Dividing the equation throughout by 2}
x^2 - 24x - 10x + 240 = 0
x (x -24) - 10(x-24) =0
(x -10) (x-24) = 0
x =10, x =24
Plugging back the value of x in equation (1), we get:
When x =10, y = 34 -10 = 24
When x= 24, y = 34-24 =10
Hence, the two legs of the right triangle are 10 cm and 24 cm.
Answer by mananth(16946)  (Show Source):
You can put this solution on YOUR website! Wire length = 60 cm
right triangle is formed
perimeter = 60 cm
hypotenuse = 26
so sum of legs = 60 -26 = 34cm
let one leg be x
other leg = 34-x
(34-x)^2+x^2= 26^2
1156-68x+x^2= 676
2x^2-68x+1156-676=0
2x^2-68x+480=0
/2
x^2-34x+240=0
x^2-24x-10x+240=0
x(x-24)-10(x-24)=0
(x-24)(x-10)=0
The legs are 24 & 10 cm
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