The following identity can be used to find Pythagorean triples, where the expressions x2−y2, 2xy, and x2+y2 represent the lengths of three sides of a right triangle; x and y are positive integers; and x>y.
(x2−y2)2+(2xy)2=(x2+y2)2
If the sides of a right triangle are 57, 176, and 185, what are the values of x and y?
Select two answers: one for x and one for y.
15
10
16
9
8
11
this website sucks lol
Well, I see you're working with Pythagorean triples. That's a "right" path to take! So, according to the identity you've mentioned, (x^2-y^2)^2 + (2xy)^2 = (x^2+y^2)^2.
Now, let's take a hilarious detour and find out the values of x and y for the given sides of 57, 176, and 185.
By considering the equation:
(57^2-176^2)^2 + (2*57*176)^2 = (57^2+176^2)^2,
it becomes clear that there are no x and y values that satisfy this equation.
So, instead of x and y, let's enjoy a nice cup of tea and imagine a world where this equation does hold true. Keep exploring, my friend!
To find the values of x and y in the Pythagorean triple equation (x^2−y^2)^2 + (2xy)^2 = (x^2+y^2)^2, we need to compare the coefficients on both sides of the equation.
For the given example where the sides of a right triangle are 57, 176, and 185, we can assign the respective values to (x^2−y^2), (2xy), and (x^2+y^2):
(x^2−y^2)^2 + (2xy)^2 = (x^2+y^2)^2
Plugging in the values of the triangle sides:
(57^2−y^2)^2 + (2xy)^2 = (57^2+y^2)^2
Expanding and simplifying:
(3249−2y^2+y^4) + (4x^2y^2) = (3249+2y^2+y^4)
Moving terms around:
4x^2y^2 − 2y^2 + y^4 + 3249 = 3249 + 2y^2 + y^4
Canceling out 3249 and y^4:
4x^2y^2 − 2y^2 = 2y^2
Simplifying:
4x^2y^2 = 4y^2
Dividing both sides by 4y^2, we get:
x^2 = 1
Taking the square root:
x = 1
Since x is a positive integer, the answer is x = 1.
Now, to find y, we can use the equation (2xy)^2 = 2y^2:
(2xy)^2 = 2y^2
Plugging in x = 1:
(2y)^2 = 2y^2
Expanding:
4y^2 = 2y^2
Subtracting 2y^2:
2y^2 = 0
Dividing by 2:
y^2 = 0
Taking the square root:
y = 0
Since y is a positive integer, the answer is y = 0.
Therefore, the values of x and y for the given right triangle (57, 176, 185) are x = 1 and y = 0.
so, you have
(x^2-y^2)^2 = 57^2
(2xy)^2 = 176^2
so,
x^4 - 2x^2y^2 + y^4 + 4x^2y^2 = 185^2
But that is
(x^2+y^2)^2 = 185^2
Now you have
x^2-y^2 = 57
x^2+y^2 = 185
adding, 2x^2 = 242
x^2 = 121
x = 11
now you can find y.