Help with pythagorean theorem world problem?

Let a meters be the rise.
Then, the length of the inclined plane, c = (a + 10) meters
and the run b = (a + 8) meters

Apply Pythagorean theorem : c² = a² + b²
(a + 10)² = a² + (a + 8)²
a² + 20a + 100 = a² + a² + 16a + 64
a² - 4a - 36 = 0
a = [4 ± √(4² + 4*36)]/2
a = (4 ± √160)/2
a = (4 ± 4√10)/2
a = 2 + 2√10 or a = 2 - 2√10 (rejected)

Length of the inclined plane
= [2 + 2√10 + 10] meters
= (12 + 2√10) meters
≈ 18.325 meters

The plane (hypotenuse) is 10 meters longer than the rise, and is also 10 - 8= 2 meters longer than the run.

So the plane is 10 + 2 + √(2*10*2) = 12 + 2√10 ≈ 18.3 meters.

We have L = R + 10 and N = R + 8; so L^2 = (R^2 + 20R + 100) = R^2 + (R^2 + 16R + 64) = R^2 + N^2 from the Pythagorean equation. From which we find the quad R^2 -4R - 36 = 0 which has the solution R = 8.33 so that L = 18.33 m ANS.

r = 10 + y; x = 8 + y; x^2 + y^2 = r^2.
Therefore, (8 + y)^2 + y^2 = (10 + y)^2 =>
64 + 16y + 2y^2 = 100 + 20y + y^2 =>
y^2 - 4y - 36 = 0 =>
y = 2 + (1/2)*sqrt(16 + 144)
= 2 + sqrt(40).

Check:
y = 2 + sqrt(40) => x = 10 + sqrt(40) and r = 12 + sqrt(40).
Let's look at x^2 + y^2; it's
4 + 4 sqrt(40) + 40 + 100 + 20 sqrt(40) + 40
= 184 + 24 sqrt(40).
And r^2 = 144 + 24 sqrt(40) + 40 = 184 + 24 sqrt(40).

So that checks out; it's correct.

Let's give variables to your terms.

length of the inclined plane is the hypotenuse of the triangle, so that's "c".
rise and run variables will be "a" and "b" (doesn't matter which is which, so I'll call "a" the rise).

You are told that:

c is 10 meters longer than a:

c = a + 10

and b is 8 meters longer than a:

b = 8 + a

And you want to solve for c. Since we want to solve for c, let's change the first equation to be a in terms of c:

c = a + 10
c - 10 = a

And we want the equation with "b" in it to also have a "c", so let's substitute the expression (c - 10) in for a in that equation to get an equation that is b in terms of c:

b = 8 + a
b = 8 + c - 10
b = c - 2

Now that we have a and b in terms of c, let's go to the Pythagorean Theorem, substitute what we know, then solve for c:

a² + b² = c²
(c - 10)² + (c - 2)² = c²
c² - 20c + 100 + c² - 4c + 4 = c²

Subtract c² from both sides, then simplify:

c² - 20c + 100 - 4c + 4 = 0
c² - 24c + 104 = 0

I'll complete the square, so I'll move the 104 to the other side, then add 144 to both sides:

c² - 24c = -104
c² - 24c + 144 = -104 + 144
c² - 24c + 144 = 40

Now we can factor and get the square root of both sides:

(c - 12)² = 40
c - 12 = ± √40
c = 12 ± √40

Factor out the square factor "4" from the 40 and pull it out of the radical:

c = 12 ± √(4 * 10)
c = 12 ± 2√(10)

So that gives us two possible values for c:

the hypotenuse could be: 12 - 2√10 m (approx 5.675 m) or 12 + 2√10 m (approx 18.325 m)

Now we have to check all of our variables to make sure everything makes sense.

We are told that the hypotenuse was 10 m longer than the rise. So if the hypotenuse is only 5.6 m, the rise will end up being a negative value. So we can then throw this out.

The other one has a rise of 8.325 m and the run is 16.325 m since it's 8 meters longer. Checking the numbers:

a² + b² = c²
8.325² + 16.325² = 18.325²
69.305625 + 266.505325 = 335.805625
335.81095 = 335.805625

While it's not exact due to rounding, it's good to 4 SF (which is the value we did our rounding to) so this would be close enough to be confident in our answer.

So again, the answer to your question is:

12 + 2√10 m (or approx 18.325 m)

L^2 = rise^2 + run^2
(rise + 10)^2 = rise^2 + (rise + 8)^2
r^2 + 20r + 100 = r^2 + r^2 + 16r + 64
r^2 - 4r - 36 = 0
from quadratic formula (rejecting -ve value)
r = 8.325 m

rise = 8.325 m
run = 16.325 m
incline = 18.325 m