# Find the largest area of an isosceles triangle inscribed in a circle of radius 3.

The aim of the question is to find the largest area of the triangle enclosed by the circle of radius of 3.

The basic concept is the Equation of the Circle, which is defined as:

$x^2+y^2=p^2$

To solve this question, first we have to find the equations for x or y and then put them in the equation of a circle to get the other variable and find the triangle’s area.

We know that the area of a triangle can be written as:

$Area$ $of$ $Triangle$ $= \dfrac {1}{2} \times base \times height$

Here, Base $=b$

Height $=p+x$

Where $p =$ radius of circle enclosing the triangle

$x =$ Center of the circle to the base of triangle

Figure 1

$Area\ of\ Triangle = \frac {1}{2} \times b \times (p+x)$

To find base $b$, by applying the Pythagoras theorem we get:

$\frac{b}{2} = \sqrt {p^2-x^2}$

$b = 2 \times \sqrt {p^2-x^2}$

Putting value of $b$ in area of triangle:

$Area = \frac {1}{2} (2 \times \sqrt {p^2-x^2}) \times (p+x)$

$Area = \sqrt {p^2-x^2} \times (p+x)$

Taking derivative with respect to $x$ on both sides:

$\frac{d}{dx}Area =\frac{d}{dx}\ \left[\sqrt{p^2-x^2}\ \times\left(p+x\right)\ \ \right]$

$\frac{d}{dx}Area =\frac{d}{dx}\ \left[\sqrt{p^2-x^2}\ \right]\left(p+x\right)+\sqrt{p^2-x^2}\frac{d}{dx}\left[p+x\right]$

$\frac{d}{dx}Area =\frac{d}{dx}\ \left[\sqrt{p^2-x^2}\ \right]\ \left(p+x\right)+\sqrt{p^2-x^2}\ \ [0+1]$

$\frac{d}{dx}Area =\frac{d}{dx}\ \left[\sqrt{p^2-x^2}\ \right]\ \left(p+x\right)+\sqrt{p^2-x^2}\ [1]$

$\frac{d}{dx}Area =\frac{1}{2\ \sqrt {p^2-x^2}\ }(-2x)\ \times \left(p+x\right)+\sqrt{p^2-x^2}$

$\frac{d}{dx}Area=\frac{\left(-x\right)\left(p+x\right)}{\sqrt{p^2-x^2}}+\sqrt{p^2-x^2}$

$\frac{d}{dx}Area=\frac{-x\ -\ x^2}{\sqrt{p^2-x^2}}+\sqrt{p^2-x^2}$

$\frac{d}{dx}Area=\frac{(-x\ -\ x^2)(\sqrt{p^2-x^2})}{\sqrt{p^2-x^2}}$

$\frac{d}{dx}Area=\frac{p^2-px\ -2x^2}{\sqrt{p^2-x^2}}$

Putting the equation equal to zero, we get:

$\frac{p^2-px\ -2x^2}{\sqrt{p^2-x^2}}\ =\ 0$

$p^2-px\ -2x^2\ =\ 0$

Now to get the value of $x$ we will apply the Quadratic Formula which is given by:

$x=\ \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\ \frac{p\pm\sqrt{{9p}^2}}{-4}$

Solving the above equation:

$x = -p\ and\ x = \frac{p}{2}$

As value of $x$ can not be negative so ignoring the negative value and confirming the positive value to be maximum we have:

$Area^\prime\left(x\right)>0\ when\ x<\frac{p}{2}$

$Area^\prime\left(x\right)<0\ when\ \ x>\frac{p}{2}$

So we can say that:

$x=\ \frac{p}{2}$

And this value is maximum.

Now to find the value of $y$ we know that the equation of a circle is:

$x^2+y^2=p^2$

Putting value of $x$ in the above equation:

$(\frac{p}{2}\ )^2+y^2=p^2$

$y^2=p^2\ -\ (\frac{p}{2}\ )^2$

$y^2=\frac{4p^2-\ p^2}{4}\$

Taking under root both the sides, we get:

$y=\frac{\sqrt 3}{2}\ p\$

## Numerical Result

Base of the triangle:

$b = 2 \times \sqrt {p^2-x^2}$

Putting value of $x$ here:

$b = 2 \times \sqrt {p^2-(\frac{p}{2})^2}$

$b = \sqrt {3} p$

given $p = 3$

$b = \sqrt {3} (3)$

$b =5.2$

Height of triangle:

$Height = p+x$

Putting value of $x$:

$Height = p+ {\frac {p}{2}}$

$Height =\frac {3p}{2}$

Given $p=3$

$Height =\frac {3(3)}{2}$

$Height =4.5$

$Area\ of\ Triangle = \dfrac {1}{2} \times base \times height$

$Area = 5.2 \times 4.5$

$Area = 23.4$

## Example

Find area of triangle with base $2$ and height $3$.

$Area\ of\ Triangle =\dfrac {1}{2} \times base \times height$

$Area = \dfrac {1}{2} \times 2 \times 3$

$Area =3$

Image/Mathematical drawings are created in Geogebra.