Question

Find the largest possible area of a right angled triangle whose hypotenuse is 5 cm long.   

Answer 1

$$\text{ Let the base of the right angled triangle be x and its height be y . Then,}$$

$$x^2+y^2=5^2$$

$$\Rightarrow y^2=25-x^2$$

$$\Rightarrow y=\sqrt{25-x^2}$$

$$\text{ As, the area of the triangle, }A=\frac{1}{2}\times x\times y$$

$$\Rightarrow A\left(x\right)=\frac{1}{2}\times x\times \sqrt{25-x^2}$$

$$\Rightarrow A\left(x\right)=\frac{x\sqrt{25-x^2}}{2}$$

$$\Rightarrow A^{\prime }\left(x\right)=\frac{\sqrt{25-x^2}}{2}+\frac{x(-2x)}{4\sqrt{25-x^2}}$$

$$\Rightarrow A^{\prime }\left(x\right)=\frac{\sqrt{25-x^2}}{2}-\frac{x^2}{2\sqrt{25-x^2}}$$

$$\Rightarrow A^{\prime }\left(x\right)=\frac{25-x^2-x^2}{2\sqrt{25-x^2}}$$

$$\Rightarrow A^{\prime }\left(x\right)=\frac{25-2x^2}{2\sqrt{25-x^2}}$$

$$\text{ For maxima or minima, we must have }{f}^{\prime }\left(x\right)=0$$

$$A^{\prime }\left(x\right)=0$$

$$\Rightarrow \frac{25-2x^2}{2\sqrt{25-x^2}}=0$$

$$\Rightarrow 25-2x^2=0$$

$$\Rightarrow 2x^2=25$$

$$\Rightarrow x=\frac{5}{\sqrt{2}}$$

$$\text{ So, y }=\sqrt{25-\frac{25}{2}}$$

$$=\sqrt{\frac{50-25}{2}}$$

$$=\sqrt{\frac{25}{2}}$$

$$=\frac{5}{\sqrt{2}}$$

$$\text{ Also,}{A}^{″}\left(x\right)=\frac{[-4x\sqrt{25-x^2}-\frac{(25-2x^2)(-2x)}{2\sqrt{25-x^2}}]}{25-x^2}$$

$$=\frac{\left[\frac{-4x(25-x^2)+(25x-2x^3)}{\sqrt{25-x^2}}\right]}{25-x^2}$$

$$=\frac{-100x+4x^3+25x-2x^3}{(25-x^2)\sqrt{25-x^2}}$$

$$=\frac{-75x+2x^3}{(25-x^2)\sqrt{25-x^2}}$$

$$\Rightarrow A^{″}\left(\frac{5}{\sqrt{2}}\right)=\frac{-75\left(\frac{5}{\sqrt{2}}\right)+2{\left(\frac{5}{\sqrt{2}}\right)}^3}{{(25-{\left(\frac{5}{\sqrt{2}}\right)}^2)}^{\frac{3}{2}}}<0$$

$$So,x=\left(\frac{5}{\sqrt{2}}\right)\text{ is point of maxima }.$$

$$\therefore \text{ The largest possible area of the triangle }=\frac{1}{2}\times \left(\frac{5}{\sqrt{2}}\right)\times \left(\frac{5}{\sqrt{2}}\right)=\frac{25}{4}\text{ square units }$$