Question

If the hypotenuse of a right triangle is
20 cm, then what is the maximum area of
the triangle ?

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that ABC be right-angled triangle right-angled at B

Let assume that BC = x cm and AB = y cm

It is given that Hypotenuse, AC = 20 cm

Now, In right triangle ABC

$\rm :\longmapsto\: {x}^{2} + {y}^{2} = {20}^{2}$

$\rm :\longmapsto\: {y}^{2} = 400 - {x}^{2}$

$\rm\implies \:y = \sqrt{400 - {x}^{2} } - - - (1)$

Now,

$\rm :\longmapsto\:Area_{(triangle)}, \: A = \dfrac{1}{2}xy$

$\rm :\longmapsto\:A = \dfrac{1}{2}x \sqrt{400 - {x}^{2} }$

can be rewritten as

$\rm :\longmapsto\:2A = x \sqrt{400 - {x}^{2} }$

On squaring both sides, we get

$\rm :\longmapsto\: {4A}^{2} = {x}^{2}(400 - {x}^{2})$

Let assume that

$\rm :\longmapsto\: f(x) = {4A}^{2} = {x}^{2}(400 - {x}^{2})$

$\rm :\longmapsto\: f(x) = 400 {x}^{2} - {x}^{4}$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx} f(x) = \dfrac{d}{dx}(400 {x}^{2} - {x}^{4} )$

$\rm :\longmapsto\:f'(x) = 800x - {4x}^{3}$

For maxima or minima, we have

$\rm :\longmapsto\:f'(x) = 0$

$\rm :\longmapsto\:800x - {4x}^{3} = 0$

$\rm :\longmapsto\: {4x}^{3} = 800x$

$\rm :\longmapsto\: {x}^{2} = 200$

$\bf\implies \:x = 10 \sqrt{2} \: cm$

Now,

$\rm :\longmapsto\:f'(x) = 800x - {4x}^{3}$

So,

$\rm :\longmapsto\:f''(x) = 800 - {12x}^{2}$

$\rm :\longmapsto\:f''(10\sqrt{2} ) = 800 - 2400 = - 1600 < 0$

$\bf\implies \:f(x) \: is \: maximum$

$\bf\implies \:Area \: of \: triangle \: is \: maximum$

On substituting the value of x in equation (1), we get

$\rm :\longmapsto\:y = \sqrt{400 - 200}$

$\rm :\longmapsto\:y = \sqrt{200}$

$\bf\implies \:y = 10 \sqrt{2} \: cm$

So, Maximum area of triangle ABC is

$\rm :\longmapsto\:Area_{(triangle)}, \: A = \dfrac{1}{2} \times 10 \sqrt{2} \times 10 \sqrt{2}$

$\bf\implies \:\:Area_{(triangle)}, \: A = 100 \: {cm}^{2}$