Can you help please on the following question:

Question

"The sum of the lengths of the legs of a right triangle is 8 feet. If the length of the hypotenuse is 6 feet, find the length of the shorter leg to the nearest tenth of a foot."

Thank you, Cassandra

Answer 1

We know that a^2 + b^2 = 6^2 since the hypotenuse is 6 ft and this is pythagoras equation (^2 means squared or power of 2).

We also know that 6 + a + b = 8 if we name the unknown sides of the triangle a and b.

Therefore we can isolate either one, let's take a.
a = 8 - 6 - b so a = 2-b

if we then insert 2-b for a in the pythagoras equation we get: (2-b)^2 + b^2 = 6^2

Solve for b and you'll know a since we already established that a + b = 8

Answer 2

EDIT: a + b = 2 not 8, sorry if I confused you

Answer 3

No worries. I will review your answer today and let you know if I need further help. Thank you Mr. Amg.

Answer 4

leg #1 --- x
leg #2 --- 8-x

x^2 + (8-x)^2 = 6^2
x^2 + 64 - 16x + x^2 = 36
2x^2 - 16x + 28= 0
x^2 - 8x + 14 = 0
x^2 - 8x + 16 = -14 + 16 --- I am completing the square
(x-4)^2 = 2
x-4 = ± √2
x = 4+√2 or x=4-√2

x = appr 5.41 or x = appr 2.59

sides are 5.41 and 2.59
or
sides are 2.59 and 5.41, (this is called a symmetric quadratic solution)

the shorter side is 2.59 ft