If c is the measure of the hypotenuse of a right triangle, how do you find each missing measure given a=x-47, b=x, c=x+2?

1 Answer
Feb 8, 2017

Answer:

#a = 2 + 9sqrt(57)#
#b = 49 + 9sqrt(57)#
#c = 51 + 9sqrt(57)#

Explanation:

#a = x - 47#
#b = x#
#c = x + 2#

and we know that the pythagoras theorem is:
#a^2 + b^2 = c^2#

so if we fill that in:
#(x-47)^2 + x^2 = (x+2)^2#

from here we can see that:
#a^2 = x^2 - 94x - 2209#
#b^2 = x^2#
#c^2 = x^2 + 4x + 4#

this leads to:
#2x^2 - 94x - 2209 = x^2 = 4x + 4#
#x62 - 98x - 2216 = 0#

The ABC Formula tells us that in order to calculate #x# we'll use the following method:

#x = (-b ± sqrt(b^2 - 4ac))/(2a)#
with #a = 1#, #b = -98# and #c = -2216#

when we fill that in we get:
#x = (98 + sqrt(-98^2 - (4*-2216)))/(2) = 49 + 9sqrt(57)#
or
#x = (98 - sqrt(-98^2 - (4*-2216)))/(2) = 49 - 9sqrt(57)#

#49 - 9sqrt(57)# equals to a negative number, and since we know that x is the length of side b we know that #x# cannot be negative.
therefore we know that #x = 49 + 9sqrt(57)#

we fill that in for the sides:

#a = x - 47# #-># #a = 2 + 9sqrt(57)#
#b = x -> b = 49 + 9sqrt(57)#
#c = x + 2 -> c = 51 + 9sqrt(57)#

and there's your answer!