How do you find the perimeter to a right triangle if the hypotenuse is 20x+1,side a is 18x-5,and side b is 12?

Answer 1

Solve for x in (20x+1)^2 = (18x-5)^2 + 144. Then use this x value in 20x+1 and 18x-5, and add these to the 12.

Answer 2

You would have to use Pythagoras's Theorem about the squares on the short sides equaling the square on the hypotenuse. So your equation would read as follows:

(20x+1)^2 = (18x-5)^2 + 12^2. When you simplify, you get
400x^2 + 40x + 1 = 324x^2 - 180x + 25 + 144. Sort that out and you will get 76x^2 + 220x - 168 = 0. Divide both sides by 4, the common factor of the three terms, you get 19x^2 + 55x - 42 = 0. Now you have to try to factor that. You will have to use the quadratic formula, because it won't factor with whole numbers. That will give you the value of x, from which you can calculate the side lengths and perimeter.

Answer 3

Applying Pythagoras theorem we can find out the value of 'x'. Substituting this value in the sides we can havethe actua values of the sides. Then finding the perimeter is not a problem.
By Pythagoras Thorem:
(18x - 5)² + (12)² = (20x +1)²
324 x² - 180 x + 144 = 400 x² + 40 x + 1
- 76 x² - 220 x + 143 = 0
OR
76 x² + 220 x - 143 = 0
Solve the equation, Substitute the value of x in the expressions of sides, add them up . You are in the destination.

Answer 4

a^2 + b^2 = c^2.

Square each side length

(18x-5)^2 + 12^2 = (20x+1)^2

Solve for x (you'll get two answers), and then plug x back into the original lengths of the sides, and add them up.

Answer 5

Here's my best shot. Please double check the arithmatic.
a^2 + b^2 = c^2
(18x-5)^2 + (12)^2 = (20x+1)^2
324x^2 + 144 = 400x^2 +1
-76x^2 = 1 + 25 +-144
76x^2 =118 and x = ~1.25
Then a = 18(1.25)-5 = 17.5
b = 12
c = 20(1.25)+1 = 26
Add all together and you get 55.5