SOLUTION: Hi I need help on how to find the perimeter of a triangle when two angles are congruent and it includes exponents, so for triangle DEF angles D and F are congruent, DE= 10x^2-100 E

Question 673974: Hi I need help on how to find the perimeter of a triangle when two angles are congruent and it includes exponents, so for triangle DEF angles D and F are congruent, DE= 10x^2-100 EF= 5x^2+40x and DF=15x. Thank you!
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
triangle is DEF
Angle D and F are congruent.
this means that sides opposite those angles are also congruent.
that would be sides DE and EF
the length of side DE is 10x^2 - 100
the length of side EF is 5x^2 + 40x
set these sides equal to each other and you get:
10x^2 - 100 = 5x^2 + 40x
subtract 5x^2 and 40x from both sides of the equation and you get:
10x^2 - 100 - 5x^2 - 40x = 0
combine like terms and you get:
5x^2 - 40x - 100 = 0
factor out a 5 and the equation becomes:
x^2 - 8x - 20 = 0
factor this to get:
(x-10) * (x+2) = 0
solve for x to get x = 10 or x = -2.
since x can't be negative, then x = 10 is the solution you are looking for.
now that you know that x = 10, you can solve for the perimeter by replacing x with 10 in each equation and then adding up the result.
your equations are:
10x^2 - 100
5x^2 + 40x
15x
when x = 10, these equations become
10*10^2 - 100 = 1000 - 100 = 900
5*10^2 + 40*10 = 5*100 + 400 = 500 + 400 = 900
15*10 = 150
the sum of all of these is 1800 + 150 which gets you a perimeter of 1950.

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