Question 814608
a 60 cm piece of wire is bent to form a right triangle.If the hypotenuse is 26 cm, find the length of the longer of the two legs of the right triangle.
:
Call the legs a & b
a + b = 60 - 26
a + b = 34
b = (34-a)
:
a^2 + b^2 = 26^2
replace b with (34-a)
a^2 + (34-a)^2 = 676
a^2 + 1156 - 68a + a^2
a^2 + a^2 - 68a + 1156 - 676 = 0
2a^2 - 68a + 480 = 0
simplify, divide by 2
a^2 - 34a + 240 = 0
Factors to 
(a-24)(a-10) = 0
a = 24 cm is the longer leg

:
See if that checks out on your calc (b=10)
Enter {{{sqrt(24^2 + 10^2)}}} results: 26