Question

Find the largest possible area of a right angled triangle whose hypotenuse is 5 cm long. [CBSE 2000]

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{base}\mathrm{of}\mathrm{the}\mathrm{right}\mathrm{angled}\mathrm{triangle}\mathrm{be}x\mathrm{and}\mathrm{its}\mathrm{height}\mathrm{be}y.\mathrm{Then}, \\ {x}^2+y^2=5^2 \\ \Rightarrow y^2=25-x^2 \\ \Rightarrow y=\sqrt{25-x^2} \\ \mathrm{As},\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{triangle},A=\frac{1}{2}\times x\times y \\ \Rightarrow A\left(\mathit{x}\right)=\frac{1}{2}\times \mathrm{x}\times \sqrt{25-{\mathrm{x}}^2} \\ \Rightarrow A\left(x\right)=\frac{x\sqrt{25-x^2}}{2} \\ \Rightarrow A\text{'}\left(x\right)=\frac{\sqrt{25-x^2}}{2}+\frac{x\left(-2x\right)}{4\sqrt{25-x^2}} \\ \Rightarrow A\text{'}\left(x\right)=\frac{\sqrt{25-x^2}}{2}-\frac{x^2}{2\sqrt{25-x^2}} \\ \Rightarrow A\text{'}\left(x\right)=\frac{25-x^2-x^2}{2\sqrt{25-x^2}} \\ \Rightarrow A\text{'}\left(x\right)=\frac{25-2x^2}{2\sqrt{25-x^2}} \\ \mathrm{For}\mathrm{maxima}\mathrm{or}\mathrm{minima},\mathrm{we}\mathrm{must}\mathrm{have}f\text{'}\left(x\right)=0 \\ A\text{'}\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}} \\ \mathrm{So},y=\sqrt{25-\frac{25}{2}} \\ =\sqrt{\frac{50-25}{2}} \\ =\sqrt{\frac{25}{2}} \\ =\frac{5}{\sqrt{2}} \\ \mathrm{Also},A\text{'}\text{'}\left(x\right)=\frac{\left[-4x\sqrt{25-x^2}-{\displaystyle \frac{\left(25-2x^2\right)\left(-2x\right)}{2\sqrt{25-x^2}}}\right]}{25-x^2} \\ =\frac{\left[{\displaystyle \frac{-4x\left(25-x^2\right)+\left(25x-2x^3\right)}{\sqrt{25-x^2}}}\right]}{25-x^2} \\ =\frac{-100x+4x^3+25x-2x^3}{\left(25-x^2\right)\sqrt{25-x^2}} \\ =\frac{-75x+2x^3}{\left(25-x^2\right)\sqrt{25-x^2}} \\ \Rightarrow A\text{'}\text{'}\left(\frac{5}{\sqrt{2}}\right)=\frac{-75\left({\displaystyle \frac{5}{\sqrt{2}}}\right)+2{\left({\displaystyle \frac{5}{\sqrt{2}}}\right)}^3}{{\left(25-{\left({\displaystyle \frac{5}{\sqrt{2}}}\right)}^2\right)}^{{\displaystyle \frac{3}{2}}}}<0 \\ \mathrm{So},x=\left(\frac{5}{\sqrt{2}}\right)\mathrm{is}\mathrm{point}\mathrm{of}\mathrm{maxima}. \\ \therefore \mathrm{The}\mathrm{largest}\mathrm{possible}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{triangle}=\frac{1}{2}\times \left(\frac{5}{\sqrt{2}}\right)\times \left(\frac{5}{\sqrt{2}}\right)=\frac{25}{4}\mathrm{square}\mathrm{units} \end{gathered}$$