Question

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

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{base},\mathrm{height}\mathrm{and}\mathrm{area}\mathrm{of}\mathrm{a}\mathrm{right}\mathrm{angled}\mathrm{triangle}\mathrm{respectively}\mathrm{be}b,h\mathrm{and}A.\mathrm{Then}, \\ {b}^2=25-h^2...\left(1\right) \\ \mathrm{Area},A=\frac{1}{2}bh \\ \Rightarrow A=\frac{1}{2}h\left(\sqrt{25-h^2}\right)\left[\mathrm{Using}\left(1\right)\right] \\ \mathrm{On}\mathrm{squaring}\mathrm{both}\mathrm{side},\mathrm{we}\mathrm{get} \\ \Rightarrow A^2=\frac{1}{4}\left(25-h^2\right)h^2 \\ \Rightarrow A^2=\frac{25h^2}{4}-\frac{h^4}{4} \\ \mathrm{Let}Z=A^2=\frac{25h^2}{4}-\frac{h^4}{4} \\ \Rightarrow \frac{dZ}{dh}=\frac{50h}{4}-h^3 \\ \mathrm{For}\mathrm{maximum}\mathrm{or}\mathrm{minimum}\mathrm{values}\mathrm{of}Z,\mathrm{we}\mathrm{must}\mathrm{have} \\ \frac{dZ}{dh}=0 \\ \Rightarrow \frac{50h}{4}-h^3=0 \\ \Rightarrow \frac{50h}{4}=h^3 \\ \Rightarrow 4h^2=50 \\ \Rightarrow h=\frac{5}{\sqrt{2}} \\ \mathrm{So},b^2=25-{\left(\frac{5}{\sqrt{2}}\right)}^2=\frac{25}{2} \\ \Rightarrow b=\frac{5}{\sqrt{2}} \\ \mathrm{Now}, \\ \frac{d^{2}Z}{d{h}^2}=\frac{25}{2}-3h^2 \\ \Rightarrow \frac{d^{2}Z}{d{h}^2}=\frac{25}{2}-\frac{75}{2}=-25<0 \\ \mathrm{So},\mathrm{area}\mathrm{is}\mathrm{maximum}\mathrm{for}h=b=\frac{5}{\sqrt{2}} \\ \mathrm{Maximum}\mathrm{area}=\frac{1}{2}bh \\ =\frac{1}{2}\times \frac{5}{\sqrt{2}}\times \frac{5}{\sqrt{2}} \\ =\frac{25}{4}{\mathrm{cm}}^2 \end{gathered}$$