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# Find the Largest Possible Area of a Right Angled Triangle Whose Hypotenuse is 5 Cm Long. - CBSE (Science) Class 12 - Mathematics

#### Question

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

#### Solution

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

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

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

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

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

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

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

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

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

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

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

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

$A'\left( x \right) = 0$

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

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

$\Rightarrow 2 x^2 = 25$

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

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

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

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

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

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

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

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

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

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

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

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

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Solution Find the Largest Possible Area of a Right Angled Triangle Whose Hypotenuse is 5 Cm Long. Concept: Graph of Maxima and Minima.
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