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
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]
There is no reason whatsoever why the triangle should be rightangled.
What you need in order for the given side lengths to form a triangle is that the sum of the lengths of any two of the three sides should be greater than the length of the third side.
(So for example, side lengths 5, 6, 10 could form a triangle but 5, 6, 12 couldn't).
You have three inequalities.
(x + 11) + (2x + 10) > 5x - 9,
(x + 11) + (5x - 9) > 2x + 10,
(2x + 10) + (5x - 9) > x + 11.
Solving the first one,
3x + 21 > 5x - 9,
30 > 2x,
so x < 15.
Now solve the other two, and deduce a possible range of values for x.
-Bertie
