You can use the pythagoras’ theorem and just look at one right triangle.
The dimensions for one of the right triangles is (x/2), 8, and square root of 80.
a^2 + b^c = c^2
when c is the square root of 80.
Plug everything in
(x/2)^2 + 8^2 = (square root of 80)^2
That is equivalent to
((x^2)/4) + 64 = 80
Solve for x
((x^2)/4) = 16
Multiple by 4 on each side
x^2 = 64
Take the square root and you have you’re final answer
x = 8
Two consecutive odd integers have a difference of 2. If you let the smaller one of them be x, the other one is x + 2.
Their sum is -36, so add x and x + 2, and set equal to -36. Then solve for x to find the smaller one. Finally, 2 more than x is the greater of the two integers.
x + x + 2 = -36
2x + 2 = -36
2x = -38
x = -19
The smaller one of the two integers is -19.
x + 2 = -19 + 2 = -17
The greater of the two integers is -17.
Answer: The integers are -19 and -17.
