Can someone please help me with this problem?

C 6

Step-by-step explanation:

Pythagoras

c² = a² + b²

117 = x² + (x+3)² = x² + x² + 6x + 9 = 2x² + 6x + 9

=>

108 = 2x² + 6x

54 = x² + 3x

0 = x² + 3x - 54 = (x-6)(x+9)

just think by what numbers can we divide 54 without any remainder (so that when multiplying we get 54) ? and which pair of these have a difference of 3 (so that with +/- combination we get "3x" in the multiplication)?

the solutions for x are the 2 numbers that would make one of the bracket expressions 0, as this turns the whole expression to 0.

x = 6 and x = -9

a negative number does not make any sense for a side length.

so, the only possible solution is x = 6

h/o/e/s mad

step-by-step explanation:

si si si

hi how are you doing today? if you ever need just tell me

step-by-step explanation: