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  • If the sum of the lengths of the hypotenuse and a side of a right angled triangle is given, show that the area of the triangle is maximum when the angle between them is .pi/3

If the sum of the lengths of the hypotenuse and a side of a right angled triangle is given, show that the area of the triangle is maximum when the angle between them is \frac{\pi}{3}.

 

Answers (1)

Given: a right-angled triangle with the sum of the lengths of its hypotenuse and side.

To show: at this angle  \frac{\pi}{3}  the area of the triangle is maximum

Explanation:

 

Let ΔABC be the right-angled triangle,

Let hypotenuse, AC = y,

side, BC = x, AB = h

then the calculation of sum of the side and hypotenuse is done using,

⇒ x+y = k, where k is any constant value

⇒ y = k-x………..(i)

Take A as the area of the triangle, as we know

\mathrm{A}=\frac{1}{2} \mathrm{~h} \cdot \mathrm{x} \ldots$(ii)
Then using the Pythagoras theorem, we get
y^{2}=x^{2}+h^{2}$
\Rightarrow \mathrm{h}=\sqrt{\mathrm{y}^{2}-\mathrm{x}^{2}}$
Putting the value from equation (i) in above equation, we get
\Rightarrow \mathrm{h}=\sqrt{(\mathrm{k}-\mathrm{x})^{2}-\mathrm{x}^{2}}$

\\\Rightarrow \mathrm{h}=\sqrt{\mathrm{k}^{2}+\mathrm{x}^{2}-2 \mathrm{kx}-\mathrm{x}^{2}}$\\ $\Rightarrow \mathrm{h}=\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}} \ldots \ldots \ldots$(iii)
Applying the values from equation (iii) into equation (ii), we get
A=\frac{1}{2}\left(\sqrt{k^{2}-2 k x}\right) \cdot x$
The above equation is differentiated with respect to x,
\frac{\mathrm{d} \mathrm{A}}{\mathrm{dx}}=\frac{\mathrm{d}\left(\frac{1}{2}\left(\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}\right) \cdot \mathrm{x}\right)}{\mathrm{dx}}$
Then the constant terms are taken out,
\frac{d A}{d x}=\frac{1}{2} \frac{d\left(\left(\sqrt{k^{2}-2 k x}\right) \cdot x\right)}{d x}$

\begin{aligned} &\text { By using the product rule, }\\ &\frac{\mathrm{d} \mathrm{A}}{\mathrm{dx}}=\frac{1}{2}\left[\left(\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}\right) \frac{\mathrm{d}(\mathrm{x})}{\mathrm{dx}}+(\mathrm{x}) \frac{\mathrm{d}\left(\left(\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}\right)\right)}{\mathrm{dx}}\right]\\ &\Rightarrow \frac{\mathrm{d} \mathrm{A}}{\mathrm{dx}}=\frac{1}{2}\left[\left(\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}\right)(1)+(\mathrm{x}) \frac{\mathrm{d}\left(\left(\mathrm{k}^{2}-2 \mathrm{kx}\right)^{\frac{1}{2}}\right)}{\mathrm{dx}}\right] \end{aligned}

Power rule pf differentiation is applied on the second part of the above equation,

\\ \frac{\mathrm{d} \mathrm{A}}{\mathrm{dx}}=\frac{1}{2}\left[\left(\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}\right)+(\mathrm{x})\left[\frac{1}{2} \cdot\left(\mathrm{k}^{2}-2 \mathrm{kx}\right)^{\frac{1}{2}-1} \cdot \frac{\mathrm{d}\left(\mathrm{k}^{2}-2 \mathrm{kx}\right)}{\mathrm{dx}}\right]\right] \\ \Rightarrow \frac{\mathrm{d} \mathrm{A}}{\mathrm{dx}}=\frac{1}{2}\left[\left(\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}\right)+\frac{1}{2}(\mathrm{x})\left(\mathrm{k}^{2}-2 \mathrm{kx}\right)^{\frac{-1}{2}}\left[\frac{\mathrm{d}\left(\mathrm{k}^{2}\right)}{\mathrm{d} \mathrm{x}}-\frac{\mathrm{d}(2 \mathrm{kx})}{\mathrm{d} \mathrm{x}}\right]\right] \\ \Rightarrow \frac{\mathrm{d} \mathrm{A}}{\mathrm{dx}}=\frac{1}{2}\left[\left(\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}\right)+\frac{(\mathrm{x})}{2 \sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}}[0-2 \mathrm{k}]\right] \\

\Rightarrow \frac{\mathrm{d} \mathrm{A}}{\mathrm{dx}}=\frac{1}{2}\left[\left(\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}\right)-\frac{2 \mathrm{k}(\mathrm{x})}{2 \sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}}\right]

\begin{aligned} &\Rightarrow \frac{\mathrm{d} \mathrm{A}}{\mathrm{dx}}=\frac{1}{2}\left[\left(\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}\right)-\frac{\mathrm{kx}}{\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}}\right] \ldots . \text { (iv) }\\ &\text { When the first derivative is equated with } 0, \text { it gives critical point, }\\ &\frac{\mathrm{d} \mathrm{A}}{\mathrm{dx}}=0 \end{aligned}

\\ \Rightarrow \frac{1}{2}\left[\left(\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}\right)-\frac{\mathrm{kx}}{\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}}\right]=0 \\ \Rightarrow\left(\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}\right)-\frac{\mathrm{kx}}{\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}}=0 \\ \Rightarrow\left(\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}\right)=\frac{\mathrm{kx}}{\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}} \\ \Rightarrow\left(\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}\right)\left(\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}\right)=\mathrm{kx} \\ \Rightarrow \mathrm{k}^{2}-2 \mathrm{kx}=\mathrm{kx} \\ \Rightarrow \mathrm{k}^{2}=2 \mathrm{kx}+\mathrm{kx} \\ \Rightarrow \mathrm{k}^{2}=3 \mathrm{kx} \\ \Rightarrow \mathrm{k}=3 \mathrm{x} \\ \Rightarrow \mathrm{x}=\frac{\mathrm{k}}{3}

Once again, differentiating equation (iv) with respect to x, we get

\begin{aligned} &\Rightarrow \frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{dx}^{2}}=\frac{\mathrm{d}\left(\frac{1}{2}\left[\left(\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}\right)-\frac{\mathrm{kx}}{\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}}\right]\right)}{\mathrm{dx}}\\ &\text { Taking out the constant term and taking the LCM, we get }\\ &\Rightarrow \frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{dx}^{2}}=\frac{1}{2} \frac{\mathrm{d}\left(\left[\frac{\left(\mathrm{k}^{2}-2 \mathrm{kx}\right)-\mathrm{kx}}{\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}}\right]\right)}{\mathrm{dx}}\\ &\Rightarrow \frac{d^{2} A}{d x^{2}}=\frac{1}{2} \frac{d\left(\left[\frac{\left(k^{2}-3 k x\right)}{\sqrt{k^{2}-2 k x}}\right]\right)}{d x}\\ &\Rightarrow \frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{dx}^{2}}=\frac{1}{2} \frac{\mathrm{d}\left(\left(\mathrm{k}^{2}-3 \mathrm{kx}\right)\left(\mathrm{k}^{2}-2 \mathrm{kx}\right)^{\frac{-1}{2}}\right)}{\mathrm{dx}} \end{aligned}

Using the product rule of differentiation,

\Rightarrow \frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{dx}^{2}}=\frac{1}{2}\left[\left(\left(\mathrm{k}^{2}-2 \mathrm{kx}\right)^{\frac{-1}{2}} \frac{\mathrm{d}\left(\mathrm{k}^{2}-3 \mathrm{kx}\right)}{\mathrm{dx}}\right)-\left(\mathrm{k}^{2}-3 \mathrm{kx}\right) \frac{\mathrm{d}\left(\left[\left(\mathrm{k}^{2}-2 \mathrm{kx}\right)^{\frac{-1}{2}}\right]\right)}{\mathrm{dx}}\right]$
Then using the power rule of differentiation,

\begin{aligned} \Rightarrow \frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{dx}^{2}}=\frac{1}{2} &\left[\left(\left(\mathrm{k}^{2}-2 \mathrm{kx}\right)^{\frac{-1}{2}}\left(\frac{\mathrm{d}\left(\mathrm{k}^{2}\right)}{\mathrm{d} \mathrm{x}}-\frac{\mathrm{d}(3 \mathrm{kx})}{\mathrm{d} \mathrm{x}}\right)\right)\right.\\ &\left.-\left(\mathrm{k}^{2}-3 \mathrm{kx}\right)\left(-\frac{1}{2} \cdot\left(\mathrm{k}^{2}-2 \mathrm{kx}\right) \frac{-1}{2}-1 \cdot \frac{\mathrm{d}\left(\left[\mathrm{k}^{2}-2 \mathrm{kx}\right]\right)}{\mathrm{dx}}\right)\right] \\ \Rightarrow \frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{dx}^{2}}=\frac{1}{2} &\left[\left(\mathrm{k}^{2}-2 \mathrm{kx}\right)^{\frac{-1}{2}}(-3 \mathrm{k})\right) \\ &\left.-\left(\mathrm{k}^{2}-3 \mathrm{kx}\right)\left(-\frac{1}{2} \cdot\left(\mathrm{k}^{2}-2 \mathrm{kx}\right)^{\frac{-1}{2}} \cdot\left(\mathrm{k}^{2}-2 \mathrm{kx}\right)^{-1} \cdot(-2 \mathrm{k})\right)\right] \end{aligned}

\\ \Rightarrow \frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{dx}^{2}}=\frac{1}{2}\left[-\frac{3 \mathrm{k}}{\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}}-\left(\mathrm{k}^{2}-3 \mathrm{kx}\right)\left(\frac{\mathrm{k}}{\left(\mathrm{k}^{2}-2 \mathrm{kx}\right) \sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}}\right)\right] \\ \Rightarrow \frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{dx}^{2}}=\frac{1}{2}\left[-\frac{3 \mathrm{k}}{\sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}}-\frac{\mathrm{k}\left(\mathrm{k}^{2}-3 \mathrm{kx}\right)}{\left(\mathrm{k}^{2}-2 \mathrm{kx}\right) \sqrt{\mathrm{k}^{2}-2 \mathrm{kx}}}\right]

Putting \mathrm{x}=\frac{\mathrm{k}}{3}$, in above equation, we get
\\\Rightarrow\left(\frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{dx}^{2}}\right)_{\mathrm{x}=\frac{\mathrm{k}}{3}}=\frac{1}{2}\left[-\frac{3 \mathrm{k}}{\sqrt{\mathrm{k}^{2}-2 \mathrm{k}\left(\frac{\mathrm{k}}{3}\right)}}-\frac{\mathrm{k}\left(\mathrm{k}^{2}-3 \mathrm{k}\left(\frac{\mathrm{k}}{3}\right)\right)}{\left(\mathrm{k}^{2}-2 \mathrm{k}\left(\frac{\mathrm{k}}{3}\right)\right) \sqrt{\mathrm{k}^{2}-2 \mathrm{k}\left(\frac{\mathrm{k}}{3}\right)}}\right]$ \\$\Rightarrow\left(\frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{dx}^{2}}\right)_{\mathrm{x}=\frac{\mathrm{k}}{3}}=\frac{1}{2}\left[-\frac{3 \mathrm{k}}{\sqrt{\mathrm{k}^{2}\left(1-\frac{2}{3}\right)}}-\frac{\mathrm{k}\left(\mathrm{k}^{2}-\mathrm{k}^{2}\right)}{\left(\mathrm{k}^{2}\left(1-\frac{2}{3}\right)\right) \sqrt{\mathrm{k}^{2}\left(1-\frac{2}{3}\right)}}\right]$

\\ \Rightarrow\left(\frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{dx}^{2}}\right)_{\mathrm{x}=\frac{\mathrm{k}}{3}}=\frac{1}{2}\left[-\frac{3 \mathrm{k}}{\mathrm{k} \sqrt{\left(\frac{1}{3}\right)}}-0\right] \\ \Rightarrow\left(\frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{dx}^{2}}\right)_{\mathrm{x}=\frac{\mathrm{k}}{3}}=\frac{1}{2}\left[-\frac{3}{\left.\sqrt{\left(\frac{1}{3}\right)}\right]}\right. \\ \Rightarrow\left(\frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{dx}^{2}}\right)_{\mathrm{x}=\frac{\mathrm{k}}{3}}=\frac{1}{2}[-3 \sqrt{3}] \\ \Rightarrow\left(\frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{dx}^{2}}\right)_{\mathrm{x}=\frac{\mathrm{k}}{3}}=\frac{-3 \sqrt{3}}{2}<0

Hence the maximum value of A is at X=\frac{k}{3}

We know,

\cos \theta=\frac{\text { adjacent side }}{\text { hypotenuse }}$
Then from figure,
\Rightarrow \cos \theta=\frac{x}{y}$
Applying the value of y=k -x from equation (i), we get

\Rightarrow \cos \theta=\frac{x}{k-x}$
Putting the value of x=\frac{k}{3},$ we get
\\\Rightarrow \cos \theta=\frac{\frac{\mathrm{k}}{3}}{\mathrm{k}-\frac{\mathrm{k}}{3}}$ \\$\Rightarrow \cos \theta=\frac{1}{2}$
This possibility is present when \theta=\frac{\pi}{3}$
Therefore, the area of the triangle is maximum only when the angle between them is \frac{\pi}{3}$

Posted by

infoexpert22

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