Question

Show that a triangle of maximum area that can be inscribed in a circle of radius ‘r’ is an equilateral triangle.

Answer 1

Hint: In this question first of all draw a circle and then insert a triangle in such a way that all vertices are on the circumference of the circle. Mark the centre of the circle as O. Join vertices of the triangle to the centre of the circle, this gives the radius of the circle. Use the formula of area of triangle $A=\dfrac{1}{2}(B)(H)$
Where B and H are the height and base of the triangle. Find the value of ${{A}^{2}}$. Suppose ${{A}^{2}}=Z$Write the expression of area in terms of variable height of the triangle. Use the condition of maxima and minima for the area of the triangle. As for maxima or minima the $\dfrac{dZ}{dH}=0$, and for maxima $\dfrac{{{d}^{2}}Z}{d{{H}^{2}}}<0$.first find $\dfrac{dZ}{dH}$by equating it to zero, find for what values of H it vanishes. Then test for each of these values whether the sign of $\dfrac{{{d}^{2}}Z}{d{{H}^{2}}}.$If the sign is negative then maximum value occurs at that value of H.

Complete step-by-step answer:
As we have to find out the maximum area inscribed in a circle of radius r.










Inside the circle we draw a triangle of side AB, BC and CA. We have to prove that the triangle ABC is an equilateral triangle when its area is maximum.
As R is the radius of the circle. Let us assume that H is the height of the triangle and 2r is the base BC of the triangle.
Let AD be the height, it is perpendicular to BC, so OD is perpendicular to the chord BC. As we know that perpendicular from the centre of circle bisect the chord so we can write
$BD=\dfrac{BC}{2}=\dfrac{2r}{2}=r$
Now let us concentrate in the triangle OBD.
As triangle OBD is right angle triangle so here we can use Pythagoras theorem, so we can write
$O{{B}^{2}}=O{{D}^{2}}+B{{D}^{2}}$,
As from figure we have
$OB=R,OD=AD-OD=H-R\text{ and }BD=r\text{, }$so, we substitute these values we can write
\[\begin{align}
  & {{R}^{2}}={{r}^{2}}+{{(H-R)}^{2}} \\
 & \Rightarrow {{R}^{2}}={{r}^{2}}+{{H}^{2}}+{{R}^{2}}-2HR \\
 & \Rightarrow 0={{r}^{2}}+{{H}^{2}}-2HR \\
 & \Rightarrow {{r}^{2}}=2HR-{{H}^{2}}-----(a) \\
\end{align}\]
Now we have to maximize area of triangle,
$\begin{align}
  & A=\dfrac{1}{2}(BC)(H) \\
 & \Rightarrow A=\dfrac{1}{2}2r(H) \\
 & \Rightarrow A=rH----(b) \\
\end{align}$
As we know the value of ${{r}^{2}}$so in value of r, we will get square root which will be difficult to differentiate so here we will square the value of $A$, So squaring equation $(b)$, we can write
\[{{A}^{2}}={{r}^{2}}{{H}^{2}}\]
Putting the value of ${{r}^{2}}$from equation $(a)$, we can write
${{A}^{2}}=(2HR-{{H}^{2}}){{H}^{2}}$
Let us suppose $Z={{A}^{2}}$so we can write
$Z=2{{H}^{3}}R-{{H}^{4}}$
As we see here H is variable so differentiate with respect to H, we can write
$\dfrac{dZ}{dH}=2R(3{{H}^{2}})-4{{H}^{3}}----(c)$
Here we use the formula $\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}$
For maximum value of Z
$\dfrac{dZ}{dH}=0$
So, we can write
$\begin{align}
  & 2R(3{{H}^{2}})-4{{H}^{3}}=0 \\
 & \Rightarrow {{H}^{2}}(6R-4H)=0 \\
\end{align}$
Hence,
$H=0,\text{ or }H=\dfrac{3R}{2}$
But $H\ne 0,\text{ so }H=\dfrac{3R}{2}$
Now we have to check the sign of $\dfrac{d{{Z}^{2}}}{d{{H}^{2}}}$
Differentiating equation $(c)$we can write
$\dfrac{d{{Z}^{2}}}{d{{H}^{2}}}=6R(2H)-12{{H}^{2}}$
Here we use the formula $\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}$
Now putting the value of $H=\dfrac{3R}{2}$we can write
\[\begin{align}
  & \dfrac{d{{Z}^{2}}}{d{{H}^{2}}}=6R\{(\dfrac{3R}{2})\}-12{{\left( \dfrac{3R}{2} \right)}^{2}} \\
 & \Rightarrow \dfrac{d{{Z}^{2}}}{d{{H}^{2}}}=9{{R}^{2}}-27{{R}^{2}} \\
 & \Rightarrow \dfrac{d{{Z}^{2}}}{d{{H}^{2}}}=-18{{R}^{2}} \\
 & \Rightarrow \dfrac{d{{Z}^{2}}}{d{{H}^{2}}}<0 \\
\end{align}\]
As the value of $\dfrac{d{{Z}^{2}}}{d{{H}^{2}}}$is negative so Z is maximum when $H=\dfrac{3R}{2}$
Now, solving for H and r
${{r}^{2}}=2HR-{{H}^{2}}$ Putting $H=\dfrac{3R}{2}$

\[\begin{align}
  & {{r}^{2}}=2\dfrac{3R}{2}(R)-{{\left( \dfrac{3R}{2} \right)}^{2}} \\
 & \Rightarrow {{r}^{2}}=3{{R}^{2}}-\dfrac{9}{4}{{R}^{2}} \\
 & \Rightarrow {{r}^{2}}=\dfrac{3{{R}^{2}}}{4} \\
 & \Rightarrow r=\dfrac{\sqrt{3}}{2}R \\
\end{align}\]
As $BC=2r$so
$BC=\sqrt{3}R-----(1)$
Now, in right angle triangle ABD we use Pythagoras theorem so we can write
\[\begin{align}
  & A{{B}^{2}}=A{{D}^{2}}+B{{D}^{2}} \\
 & \Rightarrow A{{B}^{2}}={{H}^{2}}+{{r}^{2}} \\
 & \Rightarrow A{{B}^{2}}={{\left( \dfrac{3R}{2} \right)}^{2}}+{{\left( \dfrac{\sqrt{3}}{2}R \right)}^{2}} \\
 & \Rightarrow A{{B}^{2}}={{\dfrac{9R}{4}}^{2}}+\dfrac{3{{R}^{2}}}{4} \\
 & \Rightarrow A{{B}^{2}}={{\dfrac{12R}{4}}^{2}} \\
 & \Rightarrow A{{B}^{2}}=3{{R}^{2}} \\
 & \Rightarrow AB=\sqrt{3}R------(2) \\
\end{align}\]
Now, in right angle triangle ACD we use Pythagoras theorem so we can write
\[\begin{align}
  & A{{C}^{2}}=A{{D}^{2}}+C{{D}^{2}} \\
 & \Rightarrow A{{C}^{2}}={{H}^{2}}+{{r}^{2}} \\
 & \Rightarrow A{{C}^{2}}={{\left( \dfrac{3R}{2} \right)}^{2}}+{{\left( \dfrac{\sqrt{3}}{2}R \right)}^{2}} \\
 & \Rightarrow A{{C}^{2}}={{\dfrac{9R}{4}}^{2}}+\dfrac{3{{R}^{2}}}{4} \\
 & \Rightarrow A{{C}^{2}}={{\dfrac{12R}{4}}^{2}} \\
 & \Rightarrow A{{C}^{2}}=3{{R}^{2}} \\
 & \Rightarrow AC=\sqrt{3}R------(3) \\
\end{align}\]
Hence from equation $(1),(2)\text{ and }(3)$
We have $AB=BC=CA$
As all the sides of the triangle are equal so the triangle is equilateral.

Note: When we draw the graph of a function then the maximum or minimum value of function occurs at the point where the tangent drawn is parallel to x-axis. At a point of maxima rate of change of slope of tangent is negative and at the point of minima maxima rate of change of slope of tangent is positive.