The legs of a right-angled triangle are 7 dm and 24 dm. Find the segments of the hypotenuse

The legs of a right-angled triangle are 7 dm and 24 dm. Find the segments of the hypotenuse into which the bisector of the right angle divides it.

Let us designate this right-angled triangle ABC, angle B – straight line, leg AB = 7 dm, leg BC = 24 dm.
By the Pythagorean theorem, we find the hypotenuse AC:
AC = √ (AB² + BC²) = √ (49 + 576) = √625 = 25 (dm)
The bisector BE divides the AC hypotenuse into segments AE = x dm, CE = (25 – x) dm.
Let’s use the property of the angle bisector and write down the aspect ratio:
AB / BC = AE / CE
7/24 = x / (25 – x)
24x = 7 * (25 – x)
24x = 175 – 7x
31x = 175
x = 175/31 = 5, 64516 ≈ 5.65 (dm) – segment AE.
25 – 5.65 = 19.35 (dm) – segment CE.
Answer: 5.65 dm, 19.35 dm – hypotenuse segments.