What's the area of this square if it sits in this sector?

What's the area of this square if it sits in this sector?

Question

$ABCD$ is a square lying symmetrically inside a sector of a circle whose radius is $1\,\text{in.}$, the angle of the sector being $60^\circ$. $A$ and $B$ lie on the straight boundaries, $C$ and $D$ on the curved boundary. Calculate:

1. the length of the arc $CD$,

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1. the area $ABCD$,

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1. the area of the minor segment of the circle cut off by the chord $CD$.

[Answers to (i) and (iii) may be left in terms of $\pi$.]

Solution

We have $\angle OAB = \angle OBA$, and $\angle AOB = 60^\circ$, so the triangle $ABO$ is equilateral. Now consider the triangle $ADO$. It is isosceles, with sides $a, a$ and $1$.

We have $\angle OAD$ is $150^\circ$, so we can say $2a \sin 75 = 1$, and so $a = \dfrac{1}{2 \sin 75} (\approx 0.5176...).$

Now

$\angle DOA = 15^\circ = \angle COB$

, since triangles

$DOA$

and

$COB$

are isosceles. Thus

$\angle DOC = 30^\circ$

, and so

$$\begin{align*} \text{length of the arc }CD\text{} &= \frac{\angle DOC}{360^\circ} \times \text{circumference of the circle} \\ &= \frac{30^\circ}{360^\circ} \times 2\pi \\ &= \frac{\pi}{6}\,\text{in.} \end{align*}$$

This is simply $a^2$, which is $\left(\dfrac{1}{2 \sin 75}\right)^2 = 0.268(3sf)\,\text{sq.in.}$

We are not asked to find this exactly, but this is not hard to do. We have $\sin 75 = \sin30\cos45+\sin45\cos30 = \dfrac{\sqrt{2}+\sqrt{6}}{4}.$

So $\left(\dfrac{1}{2 \sin 75}\right)^2 = \left(\dfrac{2}{\sqrt{2}+\sqrt{6}}\right)^2 = \dfrac{1}{2+\sqrt{3}} = 2-\sqrt{3} = 0.268(3sf)\,\text{sq.in.}$

We need the area of the sector $DOC$, and then subtract from this the area of the triangle $DOC$.

Thus we have $\pi \times 1^2 \times \dfrac{30}{360} - 0.5 \times 1 \times 1 \times \sin 30 = \dfrac{\pi}{12} - \dfrac{1}{4}\,\text{sq.in.}$

Can you find any other ways of doing this question?