Question

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

Answer 1

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

\[ x^2 + y^2 = 5^2 \]

\[ \Rightarrow y^2 = 25 - x^2 \]

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

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

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

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

\[ \Rightarrow A'\left( x \right) = \frac{\sqrt{25 - x^2}}{2} + \frac{x\left( - 2x \right)}{4\sqrt{25 - x^2}}\]

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

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

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

\[\text { For maxima or minima, we must have } f'\left( x \right) = 0\]

\[A'\left( x \right) = 0\]

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

\[ \Rightarrow 25 - 2 x^2 = 0\]

\[ \Rightarrow 2 x^2 = 25\]

\[ \Rightarrow x = \frac{5}{\sqrt{2}}\]

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

\[ = \sqrt{\frac{50 - 25}{2}}\]

\[ = \sqrt{\frac{25}{2}}\]

\[ = \frac{5}{\sqrt{2}}\]

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

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

\[ = \frac{- 100x + 4 x^3 + 25x - 2 x^3}{\left( 25 - x^2 \right)\sqrt{25 - x^2}}\]

\[ = \frac{- 75x + 2 x^3}{\left( 25 - x^2 \right)\sqrt{25 - x^2}}\]

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

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

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