i have this problem on my geo homework it is a triangle with x+11, 2x+10, and 5x-9 as the side lengths and it is asking for the possible values for x so help me out here i am confussed on what to do
We are assuming here that it is a right-angle triangle. If so, then you would use Pythagoras's Theorem: Just by looking at the "x's", it is clear that 5x - 9 is the Hypotenuse. Then we have:
[5x - 9]^2=[x +11]^2 + [2x + 10]^2
Solve for x:
(5 x-9)^2 = (x+11)^2+(2 x+10)^2
Expand out terms of the right hand side:
(5 x-9)^2 = 5 x^2+62 x+221
Subtract 5 x^2+62 x+221 from both sides:
-221-62 x-5 x^2+(5 x-9)^2 = 0
Expand out terms of the left hand side:
20 x^2-152 x-140 = 0
Divide both sides by 20:
x^2-(38 x)/5-7 = 0
Add 7 to both sides:
x^2-(38 x)/5 = 7
Add 361/25 to both sides:
x^2-(38 x)/5+361/25 = 536/25
Write the left hand side as a square:
(x-19/5)^2 = 536/25
Take the square root of both sides:
x-19/5 = (2 sqrt(134))/5 or x-19/5 = -(2 sqrt(134))/5
Add 19/5 to both sides:
x = 19/5+(2 sqrt(134))/5 or x-19/5 = -(2 sqrt(134))/5
Add 19/5 to both sides:
Answer: | x = 19/5+(2 sqrt(134))/5 or x = 19/5-(2 sqrt(134))/5=+or- 8.43
If you substitute x=8.43 in the formula above, you get:
[5x - 9]^2=[x +11]^2 + [2x + 10]^2
8.43 X 5 - 9=33.15 This is the Hypotenuse
8.43 X 2 + 10=26.86 One of the other two sides.
8.43 + 11=19.43 This is the third side. So that:
33.15^2=26.86^2 + 19.43^2 [rounded]
