ABC is a right-angled triangle of given area k. Find the sides of the…

The correct answer is Option (1) → The triangle is isosceles (two sides are equal).

Let ABC be a right-angled triangle at B.

Let hypotenuse $AC = l$ and side $AB = x$, then

$BC^2=l^2-x^2⇒ BC= \sqrt{l^2 - x^2}$   ...(i)

Area of $ΔABC =\frac{1}{2} AB × BC = k$ (given)

$⇒\frac{1}{2}x \sqrt{l^2 - x^2} = k⇒ l^2 − x^2 =\frac{4k^2}{x^2}$

$⇒l^2=x^2+\frac{4k^2}{x^2}$   ...(ii)

Since $∠ABC =\frac{\pi}{2}$, therefore, AC is a diameter of the circumcircle of ΔABC.

∴ Radius of circumcircle = $\frac{1}{2}AC = \frac{l}{2}$.

Let A be the area of the circumcircle of ΔABC, then $A = π (\frac{l}{2})^2$

$⇒ A= \frac{1}{4}πl^2=\frac{1}{4}π\left(x^2 +\frac{4k^2}{x^2}\right)$   (using (ii))

$⇒\frac{dA}{dx}=\frac{π}{4}\left(2x-\frac{8k^2}{x^3}\right)$ and $\frac{d^2A}{dx^2}=\frac{π}{4}\left(2+\frac{24k^2}{x^4}\right)$

Now $\frac{dA}{dx}=0⇒2x-\frac{8k^2}{x^3}=0 ⇒x^4 = 4k^2⇒x^2 = 2k⇒x=\sqrt{2k}$   $(∵x>0)$

$\left(\frac{d^2A}{dx^2}\right)_{x=\sqrt{2k}}=\frac{π}{4}\left(2+\frac{24k^2}{x^4}\right)=2\pi>0$ ⇒ A is least when $x=\sqrt{2k}$

From (ii), when $x = \sqrt{2k}, l^2 = 2k +\frac{4k^2}{2k}=4k⇒l=2\sqrt{k}$.

From (i), $BC =\sqrt{l^2 - x^2} = \sqrt{4k - 2k} =\sqrt{2k}$.

Hence, the sides of the triangle are $\sqrt{2k}$, \sqrt{2k}$, 2\sqrt{k}$.