Question

given is the right triangle TUV with altitude UO. Apply the Pythagorean theorem and similar triangles to find the length of line OV


15−−√
square root of 15

15
15

215−−√

Answer 1

I don’t have the side lengths from your diagram — can you confirm the lengths of UT, UV, or TV? (Or upload the picture.)

Meanwhile here is the general method and formula so you can get OV once you know the sides.

Let UT = a, UV = b, and TV (the hypotenuse) = c. Because U is the right angle and UO is the altitude to the hypotenuse, the altitude splits the hypotenuse into two segments: TO and OV. From similar triangles (and the Pythagorean relation c = sqrt(a^2 + b^2)) you get

- TO = a^2 / c
- OV = b^2 / c
- the altitude UO = ab / c

So to find OV compute OV = b^2 / c = b^2 / sqrt(a^2 + b^2).

If you give the numeric values for a and b (or for c and one leg), I’ll compute OV for you.