Question

If the sum of lengths of the hypotenuse and a side of a right angled triangle is given,

show that the area of the triangle is maximum, when the angle between them is π

3

.

Answer 1

Let a triangle ABC is given in such a what that hypotenuse AC = x , base BC = y and altitude AB = √{ x² - y²}
A/C to question,
x + y = constant {let R}

Area of ∆, A = 1/2 × base × height
A = 1/2 × y × √{x² - y²}
= 1/2y √{(R - y)² - y²}
= 1/2y √{R² - 2Ry}
= 1/2√{R²y² - 2Ry³}
Now, differenciate with respect to R
dA/dR = (2Ry - 6Ry²)/2√{R²y² - 2Ry³}
0 = (2R²y - 6Ry²)
Hence, y = R/3 ∴ x = 2R/3 [ because x + y = R ]
Then, √{x² - y²} = √3R/3

Now, angle between BC and AC , sinC = AB/AC =√3R/3/2R/3 = √3/2.
So, C = 60°
Hence, angle between them = 60°