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(b2 - 102) + tan(A) * 10
c ≈ 25.714
a = sqrt(c2 - b2)
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
