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  • AB is a diameter of a circle and C is any point on the circle. Show that the area of Δ ABC is maximum, when it is isosceles.

AB is a diameter of a circle and C is any point on the circle. Show that the area of Δ ABC is maximum when it is isosceles.

Answers (1)

Given: a circle with AB as diameter and C is any point on the circle

To show: area of Δ ABC is maximum, when it is isosceles

Explanation: If we take r as the radius of the circle, the diameter will become 2r= AB

This indicates that any angle in a semicircle is 90°.

Hence ∠ACB = 90°

Now let AC = x and BC = y

Using the Pythagoras theorem in this right-angled triangle ABC,

\\{(2r)\textsuperscript{2} = x\textsuperscript{2}+y\textsuperscript{2}}\\ { \Rightarrow y\textsuperscript{2} = 4r\textsuperscript{2}-x\textsuperscript{2}}\\

\begin{aligned} &\Rightarrow \mathrm{y}=\sqrt{4 \mathrm{r}^{2}-\mathrm{x}^{2}} \ldots \ldots\\ &\text { Now we know area of } \triangle A B C \text { is }\\ &A=\frac{1}{2} x y \end{aligned}

Then putting the values from the equation (i), we get
\mathrm{A}=\frac{1}{2} \mathrm{x} \sqrt{4 \mathrm{r}^{2}-\mathrm{x}^{2}} \ldots \ldots$(ii)
By finding the first derivative of the area,
$$ \mathrm{A}^{\prime}=\frac{\mathrm{d}\left(\frac{1}{2} \mathrm{x} \sqrt{4 \mathrm{r}^{2}-\mathrm{x}^{2}}\right)}{\mathrm{dx}} $$
Simultaneously using the product rule of differentiation and also taking out the constant terms,
$$ \mathrm{A}^{\prime}=\frac{1}{2}\left[\mathrm{x} \frac{\mathrm{d}\left(\left(4 \mathrm{r}^{2}-\mathrm{x}^{2}\right)^{\frac{1}{2}}\right)}{\mathrm{dx}}+\left(4 \mathrm{r}^{2}-\mathrm{x}^{2}\right)^{\frac{1}{2}} \cdot \frac{\mathrm{dx}}{\mathrm{dx}}\right] $$

Using the power rule of differentiation,

\\ A^{\prime}=\frac{1}{2}\left[x \cdot \frac{1}{2} \cdot\left(4 r^{2}-x^{2}\right)^{\frac{1}{2}-1} \frac{d\left(4 r^{2}-x^{2}\right)}{d x}+\left(4 r^{2}-x^{2}\right)^{\frac{1}{2}}\right] \\ A^{\prime}=\frac{1}{2}\left[x \cdot \frac{1}{2} \cdot\left(4 r^{2}-x^{2}\right)^{\frac{-1}{2}}(-2 x)+\left(4 r^{2}-x^{2}\right)^{\frac{1}{2}}\right] \\ A^{\prime}=\frac{1}{2}\left[\frac{-x^{2}}{\left(4 r^{2}-x^{2}\right)^{\frac{1}{2}}}+\left(4 r^{2}-x^{2}\right)^{\frac{1}{2}}\right] \\ A^{\prime}=\frac{1}{2}\left[\frac{-x^{2}+\left(4 r^{2}-x^{2}\right)}{\left(4 r^{2}-x^{2}\right)^{\frac{1}{2}}}\right] \\ A^{\prime}=\left[\frac{\left(2 r^{2}-x^{2}\right)}{\sqrt{4 r^{2}-x^{2}}}\right] \ldots \ldots \text { (iii) }

Critical point can be calculated by putting the first derivative equal to 0

\begin{aligned} &\mathrm{A}^{\prime}=0\\ &\Rightarrow \frac{\left(2 r^{2}-x^{2}\right)}{\sqrt{4 r^{2}-x^{2}}}=0\\ &\Rightarrow 2 r^{2}-x^{2}=0\\ &\Rightarrow 2 r^{2}=x^{2}\\ &\Rightarrow \mathrm{r}^{2}=\frac{\mathrm{x}^{2}}{2}\\ &\Rightarrow \mathrm{r}=\pm \sqrt{\frac{\mathrm{x}^{2}}{2}}\\ &\Rightarrow \mathrm{r}=\pm \frac{\mathrm{x}}{\sqrt{2}}\\ &0 \Rightarrow x=\pm r \sqrt{2}\\ &\text { Now we know, } r \neq 0 \end{aligned}
Hence r=\frac{x}{\sqrt{2}}$ and $x=r \sqrt{2}$
Differentiating the equation (ii) will give us the second derivative of the equation
\\\mathrm{A}^{\prime \prime}=\frac{\mathrm{d}\left(\left[\frac{\left(2 \mathrm{r}^{2}-\mathrm{x}^{2}\right)}{\sqrt{4 \mathrm{r}^{2}-\mathrm{x}^{2}}}\right]\right)}{\mathrm{dx}}$ \\$\mathrm{A}^{\prime \prime}=\frac{\mathrm{d}\left(\left(2 \mathrm{r}^{2}-\mathrm{x}^{2}\right)\left(4 \mathrm{r}^{2}-\mathrm{x}^{2}\right)^{-\frac{1}{2}}\right)}{\mathrm{dx}}$
Removing all the constant terms and then using the product rule of differentiation,
\\\mathrm{A}^{\prime \prime}=\left(2 \mathrm{r}^{2}-\mathrm{x}^{2}\right) \frac{\mathrm{d}\left(\left(4 \mathrm{r}^{2}-\mathrm{x}^{2}\right)^{-\frac{1}{2}}\right)}{\mathrm{dx}}+\left(4 \mathrm{r}^{2}-\mathrm{x}^{2}\right)^{-\frac{1}{2}} \frac{\mathrm{d}\left(2 \mathrm{r}^{2}-\mathrm{x}^{2}\right)}{\mathrm{dx}}$

After using the power rule of differentiation,

\\ A^{\prime \prime}=\left(2 r^{2}-x^{2}\right) \cdot\left(-\frac{1}{2}\right) \cdot\left(4 r^{2}-x^{2}\right)^{-\frac{1}{2}-1} \frac{d\left(4 r^{2}-x^{2}\right)}{d x}+\left(4 r^{2}-x^{2}\right)^{-\frac{1}{2}}(-2 x) \\ A^{\prime \prime}=\left(2 r^{2}-x^{2}\right) \cdot\left(-\frac{1}{2}\right) \cdot\left(4 r^{2}-x^{2}\right)^{-\frac{3}{2}}(-2 x)+\left(4 r^{2}-x^{2}\right)^{-\frac{1}{2}}(-2 x) \\ A^{\prime \prime}=-\left(2 r^{2}-x^{2}\right) \cdot\left(4 r^{2}-x^{2}\right)^{-\frac{3}{2}}-2 x\left(4 r^{2}-x^{2}\right)^{-\frac{1}{2}} \\ A^{\prime \prime}=-\frac{\left(2 r^{2}-x^{2}\right)}{\left(4 r^{2}-x^{2}\right) \sqrt{4 r^{2}-x^{2}}}-\frac{2 x}{\sqrt{4 r^{2}-x^{2}}} \\ A^{\prime \prime}=\frac{-\left(2 r^{2}-x^{2}\right)-2 x\left(\left(4 r^{2}-x^{2}\right)\right)}{\left(4 r^{2}-x^{2}\right) \sqrt{4 r^{2}-x^{2}}} \\ A^{\prime \prime}=\frac{-\left(2 r^{2}-x^{2}\right)-8 x r^{2}+2 x^{3}}{\left(4 r^{2}-x^{2}\right) \sqrt{4 r^{2}-x^{2}}}

For x=r\sqrt 2 in above equation, we get

\\\mathrm{A}_{\mathrm{x}=\mathrm{r} \sqrt{2}}^{\prime \prime}=\frac{\left.-\left(2 \mathrm{r}^{2}-(\mathrm{r} \sqrt{2})^{2}\right)-8(\mathrm{r} \sqrt{2}) \mathrm{r}^{2}+2(\mathrm{r} \sqrt{2})^{3}\right)}{\left(4 \mathrm{r}^{2}-(\mathrm{r} \sqrt{2})^{2}\right) \sqrt{4 \mathrm{r}^{2}-(\mathrm{r} \sqrt{2})^{2}}}$ \\$A_{x=r \sqrt{2}}^{\prime \prime}=\frac{-\left(2 r^{2}-2 r^{2}\right)-8 \sqrt{2} r^{3}+4 \sqrt{2} r^{3}}{\left(4 r^{2}-2 r^{2}\right) \sqrt{4 r^{2}-2 r^{2}}}$ \\$\mathrm{A}_{\mathrm{x}=\mathrm{r} \sqrt{2}}^{\prime \prime}=\frac{-(0)-4 \sqrt{2} \mathrm{r}^{3}}{\left(2 \mathrm{r}^{2}\right) \sqrt{2 \mathrm{r}^{2}}}$ \\$A_{x=r \sqrt{2}}^{\prime \prime}=\frac{-4 \sqrt{2} r^{3}}{\left(2 r^{3}\right) \sqrt{2}}<0$
Thus, for $\mathrm{x}=\mathrm{r} \sqrt{2}$, the area of $\triangle \mathrm{ABC}$ is maximum
The maximum value can be calculated by substituting $\mathrm{x}=\mathrm{r} \sqrt{2}$  in equation (i),
\Rightarrow \mathrm{y}=\sqrt{4 \mathrm{r}^{2}-(\mathrm{r} \sqrt{2})^{2}}$

\\ \Rightarrow y=\sqrt{4 r^{2}-2 r^{2}} \\ \Rightarrow y=\sqrt{2 r^{2}} \\ \Rightarrow y=r \sqrt 2=x

i.e., the two sides of the $\triangle \mathrm{ABC}$ are equal

Hence, the area of $\triangle \mathrm{ABC}$ is maximum, when it is isosceles

Hence, proved.

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infoexpert22

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