# Find the dimensions of the isosceles triangle of largest area that can be inscri

Find the dimensions of the isosceles triangle of largest area that can be inscribed in a circle of radius r.

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Wasither1957

Step 1
From the figure, by Pythagoras theorem we can write
$$\displaystyle{\frac{{{b}}}{{{2}}}}=\sqrt{{{r}^{{{2}}}-{x}^{{{2}}}}}$$
$$\displaystyle{b}={2}\sqrt{{{r}^{{{2}}}-{x}^{{{2}}}}}\rightarrow{\left({1}\right)}$$

Step 2
Area of Triangle $$\displaystyle={\frac{{{1}}}{{{2}}}}$$ (Base)(Height)
$$\displaystyle{A}={\frac{{{1}}}{{{2}}}}{\left({b}\right)}{\left({r}+{x}\right)}$$
Use Eqn(1) to substitute the value of b
$$\displaystyle{A}={\frac{{{1}}}{{{2}}}}{\left({2}\sqrt{{{r}^{{{2}}}-{x}^{{{2}}}}}\right)}{\left({r}+{x}\right)}$$
$$\displaystyle{A}{\left({x}\right)}={\left({r}+{x}\right)}\sqrt{{{r}^{{{2}}}-{x}^{{{2}}}}}$$
Step 3
Differentiate $$\displaystyle{A}{\left({x}\right)}$$
$$\displaystyle{A}'{\left({x}\right)}={\frac{{{d}{\left[{\left({r}+{x}\right)}\sqrt{{{r}^{{{2}}}-{x}^{{{2}}}}}\right]}}}{{{\left.{d}{x}\right.}}}}$$
Use Product Rule
$$\displaystyle{A}'{\left({x}\right)}=\sqrt{{{r}^{{{2}}}-{x}^{{{2}}}}}\cdot{\frac{{{d}{\left[{r}+{x}\right]}}}{{{\left.{d}{x}\right.}}}}+{\left({r}+{x}\right)}\cdot{\frac{{{d}{\left[\sqrt{{{r}^{{{2}}}-{x}^{{{2}}}}}\right]}}}{{{\left.{d}{x}\right.}}}}$$
$$\displaystyle{A}'{\left({x}\right)}=\sqrt{{{r}^{{{2}}}-{x}^{{{2}}}}}\cdot{1}+{\left({r}+{x}\right)}\cdot{\frac{{{d}{\left[\sqrt{{{r}^{{{2}}}-{x}^{{{2}}}}}\right]}}}{{{d}{\left({r}^{{{2}}}-{x}^{{{2}}}\right)}}}}\times{\frac{{{d}{\left({r}^{{{2}}}-{x}^{{{2}}}\right)}}}{{{\left.{d}{x}\right.}}}}$$
$$\displaystyle{A}'{\left({x}\right)}=\sqrt{{{r}^{{{2}}}-{x}^{{{2}}}}}+{\left({r}+{x}\right)}\cdot{\frac{{{1}}}{{{2}\sqrt{{{r}^{{{2}}}-{x}^{{{2}}}}}}}}\times{\left(-{2}{x}\right)}$$
$$\displaystyle{A}'{\left({x}\right)}=\sqrt{{{r}^{{{2}}}-{x}^{{{2}}}}}-{\frac{{{r}{x}+{x}^{{{2}}}}}{{\sqrt{{{r}^{{{2}}}-{x}^{{{2}}}}}}}}$$
Step 4
Solve for $$\displaystyle{A}'{\left({x}\right)}={0}$$
$$\displaystyle\sqrt{{{r}^{{{2}}}-{x}^{{{2}}}}}-{\frac{{{r}{x}+{x}^{{{2}}}}}{{\sqrt{{{r}^{{{2}}}-{x}^{{{2}}}}}}}}={0}$$
$$\displaystyle\sqrt{{{r}^{{{2}}}-{x}^{{{2}}}}}={\frac{{{r}{x}+{x}^{{{2}}}}}{{\sqrt{{{r}^{{{2}}}-{x}^{{{2}}}}}}}}$$
Multiply both sides by $$\displaystyle\sqrt{{{r}^{{{2}}}-{x}^{{{2}}}}}$$
$$\displaystyle{r}^{{{2}}}-{x}^{{{2}}}={r}{x}+{x}^{{{2}}}$$
$$\displaystyle{2}{x}^{{{2}}}+{r}{x}-{r}^{{{2}}}={0}$$
Using the quadratic formula, we can write
$$\displaystyle{x}={\frac{{-{r}\pm\sqrt{{{r}^{{{2}}}-{4}{\left({2}\right)}{\left(-{4}^{{{2}}}\right)}}}}}{{{2}{\left({2}\right)}}}}$$
$$\displaystyle{x}={\frac{{-{r}\pm\sqrt{{{9}{r}^{{{2}}}}}}}{{{4}}}}$$
$$\displaystyle{x}={\frac{{-{r}\pm{3}{r}}}{{{4}}}}$$
$$\displaystyle{x}={\frac{{-{r}-{3}{r}}}{{{4}}}}$$ or $$\displaystyle{x}={\frac{{-{r}+{3}{r}}}{{{4}}}}$$
$$\displaystyle{x}={\frac{{-{4}{r}}}{{{4}}}}$$ or $$\displaystyle{x}={\frac{{{2}{r}}}{{{4}}}}$$
$$\displaystyle{x}=-{r}$$ or $$\displaystyle{x}={\frac{{{r}}}{{{2}}}}$$
Step 5
But x cannot be negative so it must be $$\displaystyle{\frac{{{r}}}{{{2}}}}$$
Substitute $$\displaystyle{x}={\frac{{{r}}}{{{2}}}}$$ in Eqn(1). To get
$$\displaystyle{b}={2}\sqrt{{{r}^{{{2}}}-{\frac{{{r}^{{{2}}}}}{{{4}}}}}}={r}\sqrt{{{3}}}$$
Also the height is $$\displaystyle{x}+{r}={\frac{{{r}}}{{{2}}}}+{r}={\frac{{{3}}}{{{2}}}}{r}$$
For Maximum Area: Base must be $$\displaystyle{r}\sqrt{{{3}}}$$ and the height must be $$\displaystyle{\frac{{{3}}}{{{2}}}}{r}$$
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Elizabeth Witte

Step 1
The equation of circle will be
$$\displaystyle{x}^{{{2}}}+{y}^{{{2}}}={r}^{{{2}}}$$

Step 2
Now, area of triangle
$$\displaystyle{A}={\frac{{{1}}}{{{2}}}}{\left({2}{x}\right)}{\left({r}+{y}\right)}$$
$$\displaystyle={x}{\left({r}+{y}\right)}$$
$$\displaystyle{A}{\left({y}\right)}=\sqrt{{{r}^{{{2}}}-{y}^{{{2}}}}}{\left({r}+{y}\right)}$$
triangle derivative
$$\displaystyle{A}'{\left({y}\right)}=\sqrt{{{r}^{{{2}}}-{y}^{{{2}}}}}+{\left({r}+{y}\right)}\cdot{\frac{{-{2}{y}}}{{{2}\sqrt{{{r}^{{{2}}}-{y}^{{{2}}}}}}}}$$
$$\displaystyle={\frac{{{r}^{{{2}}}-{r}{y}-{2}{y}^{{{2}}}}}{{\sqrt{{{r}^{{{2}}}-{y}^{{{2}}}}}}}}$$
equating to zero
$$\displaystyle{\frac{{{r}^{{{2}}}-{r}{y}-{2}{y}^{{{2}}}}}{{\sqrt{{{r}^{{{2}}}-{y}^{{{2}}}}}}}}={0}$$
$$\displaystyle{r}^{{{2}}}-{r}{y}-{2}{y}^{{{2}}}={0}$$
$$\displaystyle{y}={\frac{{{r}\pm\sqrt{{{9}{r}^{{{2}}}}}}}{{-{4}}}}\Rightarrow{y}=-{r}\ {\quad\text{or}\quad}\ {y}={\frac{{{r}}}{{{2}}}}$$
rejecting, $$\displaystyle{y}=-{r}$$
Therefore, we check to see $$\displaystyle{y}={\frac{{{r}}}{{{2}}}}$$ is maximum
$$\displaystyle{A}'{\left({y}\right)}{>}{0}$$ for $$\displaystyle{y}{<}{\frac{{{r}}}{{{2}}}}$$ and $$\displaystyle{A}'{\left({y}\right)}{<}{0}$$ for $$\displaystyle{y}{>}{\frac{{{r}}}{{{2}}}}$$
$$\displaystyle\therefore{y}={\frac{{{r}}}{{{2}}}}$$ is maximum
$$\displaystyle{x}^{{{2}}}+{\left({\frac{{{r}}}{{{2}}}}\right)}^{{{2}}}={r}^{{{2}}}\Rightarrow{x}={\frac{{\sqrt{{{3}}}}}{{{2}}}}{r}$$
Dimension of triangle with maximum area is base
$$\displaystyle={2}{x}={2}\cdot{\frac{{\sqrt{{{3}}}}}{{{2}}}}{r}=\sqrt{{{3}}}{r}$$
$$\displaystyle{h}={r}+{y}={r}+{\frac{{{r}}}{{{2}}}}={\frac{{{3}}}{{{2}}}}{r}$$