A right triangle with a perimeter of 60 units has an altitude (to the hypotenuse) of length 10 units. Find the sum of the lengths of the two (non-hypotenuse) legs of this triangle.

https://web2.0calc.com/questions/geometry_84069#r1

Using this equation, we can calculate the length of a longer leg: b ≈ 23.204

Let's see if this equation works:

Angle A = 25.528 degrees

c = sqrt(b^{2} - 10^{2}) + tan(A) * 10

c ≈ 25.714

a = sqrt(c^{2} - b^{2})

a ≈ 11.082

a + b + c = 11.082 + 23.204 + 25.714 = 60

according to the pythagorean theorem, a^2 + b^2(the two legs) = c^2

we also know that a + b + c = 60

a + b = 50

a^2 + b^2 = 100

(a+b)^2 = 50^2: a^2+b^2+2ab = 2500

a*b = 2400

you can solve from there to get the two legs.

What makes you think that a+b = 50 ? Would you explain, please?

a + b + c = 60