Question

If the sides of a right triangle have lengths of x-7, x an x+1, then x=
(a) (1, 7)
(b) (7, -1)
(c) (7, 4)
(d) (7, 12)
(e) (4, 12)

Answer 1

first tell me which is the longest side

Answer 2

x-7,x,x+1....... what do u thik have the biggest value

Answer 3

Hint: Work with the dearest Pythagorean Theorem.

Answer 4

But beofre using Pythagorean Theorem.. u must find the longest side

Answer 5

Obviously, x + 1 would be the largest side and therefore the hypotenuse because of the + 1 while x is neutral and -7 is negative. THerefore, now use Pythagorean THeorem.
\[c^2 = a^2 + b^2\]\[(x + 1)^2 = x^2 + (x - 7)^2\]\[x^2 + 2x + 1 = x^2 + x^2 - 14x + 49\]\[x^2 + 2x + 1 = 2x^2 - 14x + 49\]\[0 = x^2 - 16x + 48\]Can you figure out how to solve the4 quadratic from here?

Answer 6

Yes sir! Yes sir!

Answer 7

i got x = 4, 12

Answer 8

so answer choice (e) is right?

Answer 9

You, sir, are totally correct!

Answer 10

Kinda ironic since 12 is the only true solution, but whatever works ^_^

Answer 11

thanks

Answer 12

np :)