Question

Find the area of the largest trapezoid that can be inscribed in a circle of radius 1 and whose base is a diameter of the circle.

Short answer

The largest area is \(\frac{{8\sqrt 2 }}{9}\).

Step-by-step solution

Given Data

The largest trapezoid that can be inscribed in a circle of radius 1.

Solution

Let\(\left( {0,0} \right)\)be the centre of the circle and the trapezoid be drawn on the diameter of the circle as shown in the figure below,

\(\left( {x,y} \right)\)and\(\left( { - x,y} \right)\)be the coordinates of the vertices of the shorter side of the trapezoid consider the radius of the circle is 1

Then, height of the trapezoid=y

Shorter side=2x and

Longer side=2

Area of the trapezoid is given by:

\(A = \frac{1}{2}y\left( {2 + 2x} \right)\)

To express y in term of x, using the equation of a circle,

\(\begin{aligned}{c}{x^2} + {y^2} &= {r^2}\\{x^2} + {y^2} &= 1\\y &= \sqrt {1 - {x^2}} \end{aligned}\)

 Step 3: Find area

Now, the area can be written as:

\(\begin{aligned}{c}A &= \frac{1}{2}\sqrt {1 - {x^2}} \left( {2 + 2x} \right)\\ &= \sqrt {1 - {x^2}} \left( {1 + x} \right)\end{aligned}\)

Find the derivative of Aas:

\(A' = - \frac{{2x\left( {1 + x} \right)}}{{\sqrt {1 - {x^2}} }} + \sqrt {1 - {x^2}} \)

The area is maximum or minimum when the derivative is zero.

\(\begin{aligned}{c} - \frac{{2x\left( {1 + x} \right)}}{{\sqrt {1 - {x^2}} }} + \sqrt {1 - {x^2}} &= 0\\\frac{{2x\left( {1 + x} \right)}}{{\sqrt {1 - {x^2}} }} &= \sqrt {1 - {x^2}} \\\left( {1 - {x^2}} \right) &= 2x + 2{x^2}\\3{x^2} + 2x - 1 &= 0\\\left( {3x - 1} \right)\left( {x + 1} \right) &= 0\\x &= - 1\;\;\;{\rm{or}}\;\;\;x &= \frac{1}{3}\end{aligned}\)

The area is maximum when\(x = \frac{1}{3}\)therefore the maximum area:

\(\begin{aligned}{c}A &= \sqrt {1 - {x^2}} \left( {1 + x} \right)\\ &= \sqrt {1 - \frac{1}{9}} \left( {1 + \frac{1}{3}} \right)\\ &= \frac{{\sqrt 8 }}{3}.\frac{4}{3}\\ &= \frac{{8\sqrt 2 }}{9}\end{aligned}\)

Hence the largest area is \(\frac{{8\sqrt 2 }}{9}\).