Get the solution to "Find the largest possible area of a right..."

meedipvf

Answered question

2023-02-02

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

Natalia Arroyo

Beginner2023-02-03Added 4 answers

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 = 25 - x^{2} As, the area of the triangle, A = \frac{1}{2} \times x \times y \Rightarrow A (x) = \frac{1}{2} \times x \times 25 - x^{2} \Rightarrow A (x) = \frac{x 25 - x^{2}}{2} \Rightarrow A' (x) = \frac{25}{-} + \frac{x (- 2 x)}{425 - x^{2}} \Rightarrow A' (x) = \frac{25}{-} - \frac{x^{2}}{225 - x^{2}} \Rightarrow A' (x) = \frac{25 - x^{2} - x^{2}}{225 - x^{2}} \Rightarrow A' (x) = \frac{25 - 2 x^{2}}{225 - x^{2}} For maxima or minima, we must have f' (x) = 0 A' (x) = 0 \Rightarrow \frac{25 - 2 x^{2}}{225 - x^{2}} = 0 \Rightarrow 25 - 2 x^{2} = 0 \Rightarrow 2 x^{2} = 25 \Rightarrow x = \frac{5}{2} So, y = 25 - \frac{25}{2} = \frac{50 - 25}{2} = \frac{25}{2} = \frac{5}{2} Also, A'' (x) = \frac{(- 4 x 25 - x^{2} - \frac{(25 - 2 x^{2}) (- 2 x)}{225 - x^{2}})}{25 - x^{2}} = \frac{(\frac{- 4 x (25 - x^{2}) + (25 x - 2 x^{3})}{25})}{25 - x^{2}} = \frac{- 100 x + 4 x^{3} + 25 x - 2 x^{3}}{(25 - x^{2}) 25 - x^{2}} = \frac{- 75 x + 2 x^{3}}{(25 - x^{2}) 25 - x^{2}} \Rightarrow A'' (\frac{5}{2}) = \frac{- 75 (\frac{5}{2}) + 2 {(\frac{5}{2})}^{3}}{{(25 - {(\frac{5}{2})}^{2})}^{\frac{3}{2}}} < 0 So, x = (\frac{5}{2}) is point of maxima . ∴ The largest possible area of the triangle = \frac{1}{2} \times (\frac{5}{2}) \times (\frac{5}{2}) = \frac{25}{4} square units

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