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Solve for x in the right triangle. i'm stuck on the early units in my final it would mean the world to me if someone could explain this. Thanks in advance.

Solve for x in the right triangle. i'm stuck on the early units in my final it would-example-1
User Noldor
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wel have to use the Pythagorean theorem (a^2) + (b^2) = (c^2)

look at the top triangle inside the big triangle. we know it's 2 sides but not the hypotenuse. one side would be 25 - 9 = 16 and other side is x. So so from theorem we know
16^2 + x^2 = h^2

with bottom triangle we can see the same thing but with 9 and x as the sides so:
9^2 + x ^2 = H^2 (capital H to distinguish them)

we also know that in the main triangle:
h^2 + H^2 = 25^2

now all we have to do is find a way to use these 3 equations to get the answer

we know what h^2 and H^2 both equal and can plug them into equation in orsee to get rid of one and solve for the other. So we use:
9^2 + x ^2 = H^2

and plug it into
h^2 + H^2 = 25^2

and get
h^2 + 9^2 + x ^2 = 25^2

solve for h^2
25^2 - x^2 - 9^2 = h^2

now plug what we got for h^2 into:
16^2 + x^2 = h^2

and get:
16^2 + x^2 = 25^2 - x^2 - 9^2

2x^2 = 25^2 - 16^2 - 9^2
2x^2 = 288
x^2 = 144
x = 12

there is your answer
User Julient
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