Question

If the sum of the lengths of the hypotenuse and a side of a right triangle is constant, the area of the triangle is maximum, when the angle between them is -

• A. (30)^o
• B. (60)^o
• C. (75)^o
• D. None of these

Answer 1

The correct option is

B

(60)^o

From the figure, sum of hypotenuse and a side is constant means b+a=k(constant) or b+c=k

Let’s take the side to be BC = a

$\Rightarrow$ a + b = k where k is a constant.

Using Pythagoras theorem, we get c = $\sqrt{b^2-a^2}$

We want to maximize the area of the triangle.

Area = $\frac{1}{2}$ base $\times$ altitude = $\frac{1}{2}$ a $\times$ c

We found that c = $b^2-a^2$

Substituting it in the expression for area, we get

Area = $\frac{1}{4}$ a $\times$ $\sqrt{b^2-a^2}$

Let A be the area.

So, $A^2$ = $\frac{1}{4}$ $a^2$ $\times$ $(b^2-a^2)$

Also, from a+b = k, we get b = k-a

Substituting it in the expression for area, we get

$A^2$ = $\frac{1}{4}$ $\times$ $(k^2-2Ka)$ $\times$ $a^2$

[b^2 - a^2 = $(k-a{)}^2$ - $a^2$ ]

$\Rightarrow$ $A^2$ = $\frac{1}{4}$ $\times$$(k^2{a}^2-2k{a}^3)$

This is a function which gives the square of area. Since k is a constant, we can say this is a function of a. When the area is maximum, area square will also be maximum. We saw that the derivative will be zero at a point where the function has a maximum value. To find this point, we will differentiate with respect to a and equate to zero.

2A dA / da = 1 /4 $(2a{k}^2–6k{a}^2)$

dA /da = 0 => $(2a{k}^2–6k{a}^2)$ = 0

Or a = k / 3, we neglect other values because side has to be positive

From a+b = k, we get b = k – k /3 = 2k / 3

Let the angle between hypotenuse and base be A.

cos(C) = a\b = 1\2

or C = ${60}^0$