One-Third of a Circle's Area

circarea.html

Exploration

Show that the angle, θ, subtended by a segment of a circle whose area is one-third of the full circle is the root of the equation

.

Also, show that θ is within 1/2 percent of 5π/6 radians.

Approach

Create an expression for the area, , of a segment in terms of a generic angular span, . Create an expression for the area of a full circle, . Solve the equation for .

Initialize

To initialize Descarta2D, select the input cell bracket and press SHIFT-Enter.

This initialization assumes that the Descarta2D software has been copied into one of the standard directories for AddOns which are on the Mathematica search path, $Path.

<<Descarta2D`

Solution

Find the area of a circle's segment.

Clear[r,t1];
A1=SegmentArea2D[c1=Circle2D[{0,0},r],{0,t1}]

Find the area of a full circle.

A2=Area2D[c1]

Form the equation.

eq1=A1-A2/3==0

Divide both sides by .

eq2=eq1 /. r^2->1

Discussion

Solve the equation for .

rt=FindRoot[Pi/3==(t1-Sin[t1])/2,
            {t1,Pi}]

Show that is close to 5π/6.

Rationalize[N[t1/Pi] /. rt,.005]

Perform a numerical check.

{SegmentArea2D[Circle2D[{0,0},1],{0,t1 /. rt}],
Area2D[Circle2D[{0,0},1]]/3} //N