Question

Prove that the area of a right angled triangle of a given hypotenuse

is maximum when the triangle is isosceles. ​

Answer 1

Step-by-step explanation:

area of triangle =1/2 x base x height

give c(hypotenuse)

area= 1/2 * a *b

also

a^2 + b^2 =c^2

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

replace it in the formula

differentiate the formula w.r.t b

we get 1/2 * sqrt(c^2 - b^2) + 1/(4 sqrt(c^2 - b^2)) * (-2b)*b -------(1)

equate this to 0

b^2= c^2 - b^2

b=c/sqrt(2)

again differentiate eq (1) and substitue

the value of b if answer is less than 0 then the area will be maximum

c=sqrt(2) * b

substituting c value in a^2 + b^2 =c^2

we get a=b

hence isosceles triangle

Answer 2

Step-by-step explanation:

Let ABC be the right angled triangle at B and,

let ∠CAB = ∅.

Let l be the length of its hypotenuse, l is fixed.

Then,

AB = l cos∅ and BC = l sin∅.

Let A sq.units be the area of ΔABC, then

A = (1/2) * l cos∅ * l sin∅

  = (l²/4) sin 2∅    ------- (1)

On differentiating w.r.t x, we get

(dA/d∅) = l²/4 cos 2∅ * 2

            = (l²/2) cos 2∅

Again,

differentiating w.r.t x, we get

d²A/d∅ = l²/2 (-sin 2∅) * 2

           = -l² sin 2∅

Now,

dA/d∅ = 0

⇒ l²/2 cos 2∅ = 0

⇒ cos 2∅ = 0

⇒ 2∅ = π/2

⇒ ∅ = π/4

Also,

d²∅/d∅²

⇒ -l² sinπ/2

⇒ -l² * 1

⇒ -l² < 0

A is maximum when ∅ = π/4.

Then,

AB = l cosπ/4

    = l/√2

BC = l sinπ/4

    = l/√2

∴ AB = BC.

Therefore,

Triangle is isosceles. ​

Hope it helps!