Circular Measure, Worked Examples (KBAT)

The shaded region is an outer sector with an inner sector removed, both share the same centre $O$ and the same angle $1.2$ rad, but have radii $10$ and $4$.

(a) Subtract the small sector from the large one. Because the angle is common, factor it out:

$$A=\tfrac{1}{2}R^{2}\theta-\tfrac{1}{2}r^{2}\theta=\tfrac{1}{2}\theta\,(R^{2}-r^{2})$$

Substitute $R=10$, $r=4$ and $\theta=1.2$:

$$A=\tfrac{1}{2}(1.2)\,(10^{2}-4^{2})=0.6\,(100-16)=0.6\times84=50.4$$

(b) The boundary has four parts: the outer arc $AB$, the inner arc $CD$, and the two straight pieces $CA$ and $DB$ along the radii. Each straight piece is $R-r=10-4=6$ cm.

Find the two arcs with $s=r\theta$:

$$\text{arc } AB=10\times1.2=12\qquad \text{arc } CD=4\times1.2=4.8$$

Add all four edges:

$$P=12+4.8+6+6=28.8$$

(a) The chord and the two radii form an isosceles triangle. Split it down its axis of symmetry into two right-angled triangles; each has hypotenuse $r=10$, opposite side $\tfrac{12}{2}=6$ and angle $\tfrac{\theta}{2}$ at the centre:

$$\sin\!\left(\dfrac{\theta}{2}\right)=\dfrac{6}{10}=0.6$$

Take the inverse sine, then double it to reach $\theta$:

$$\dfrac{\theta}{2}=\sin^{-1}(0.6)=0.6435\;\Rightarrow\;\theta=1.2870$$

(b) The minor segment is the sector minus the triangle: $A=\tfrac{1}{2}r^{2}(\theta-\sin\theta)$. Here $\sin\theta$ is exact by the double-angle identity, since $\cos\!\left(\tfrac{\theta}{2}\right)=\sqrt{1-0.6^{2}}=0.8$:

$$\sin\theta=2\sin\!\left(\dfrac{\theta}{2}\right)\cos\!\left(\dfrac{\theta}{2}\right)=2(0.6)(0.8)=0.96$$

Substitute $r=10$, $\theta=1.2870$ and $\sin\theta=0.96$:

$$A=\tfrac{1}{2}(10^{2})(1.2870-0.96)=50\times0.3270=16.35$$

Write both conditions. Let the radius be $r$ and the angle $\theta$ radians:

$$\text{Area: }\tfrac{1}{2}r^{2}\theta=45\quad\Rightarrow\quad r^{2}\theta=90$$

$$\text{Perimeter: }2r+r\theta=28\quad\Rightarrow\quad r\theta=28-2r$$

Notice $r^{2}\theta=r\,(r\theta)$. Substitute the perimeter result into the area result to remove $\theta$:

$$r\,(r\theta)=90\;\Rightarrow\;r\,(28-2r)=90\;\Rightarrow\;28r-2r^{2}=90$$

Rearrange to a standard quadratic and divide through by $2$:

$$2r^{2}-28r+90=0\;\Rightarrow\;r^{2}-14r+45=0$$

This is the required equation. Factorise it, two numbers with product $45$ and sum $14$ are $5$ and $9$:

$$(r-5)(r-9)=0\;\Rightarrow\;r=5\ \text{or}\ r=9$$

Find each angle from $r\theta=28-2r$. For $r=5$: $5\theta=28-10=18$, so $\theta=3.6$.

For $r=9$: $9\theta=28-18=10$, so $\theta=\dfrac{10}{9}=1.11$ (2 d.p.).

Write both given facts using the same two symbols so you can combine them. Let the radius be $r$ and the angle $\theta$ radians.

$$\text{Area: }\tfrac{1}{2}r^{2}\theta=48\qquad \text{Arc length: }r\theta=12$$

(a) Notice $r^{2}\theta=r\,(r\theta)$, so the area equation is really $\tfrac12 r\,(r\theta)=48$. Substitute the arc length directly, without ever finding $\theta$ first:

$$\tfrac{1}{2}r(12)=48\;\Rightarrow\;6r=48\;\Rightarrow\;r=8$$

(b) With the radius known, the angle follows at once from the arc-length equation.

$$\theta=\dfrac{12}{r}=\dfrac{12}{8}=1.5$$

(c) The chord $AB$ closes off the isosceles triangle formed by the two radii, so use $AB=2r\sin\!\left(\dfrac{\theta}{2}\right)$:

$$AB=2(8)\sin(0.75)=16\times0.6816=10.91$$

Since $O_1O_2=O_1P=O_2P=10$ cm, triangle $O_1O_2P$ has three equal sides, it is equilateral, so every angle inside it is $60^{\circ}=\tfrac{\pi}{3}$ rad.

$$\angle PO_1O_2=\dfrac{\pi}{3}$$

By symmetry, $Q$ is the mirror image of $P$ across the line $O_1O_2$, so $\angle QO_1O_2$ is also $\tfrac{\pi}{3}$. The full angle $\angle PO_1Q$ at the centre of circle 1 is therefore double this, and the same reasoning gives an identical angle at $O_2$:

$$\theta=\angle PO_1Q=\angle PO_2Q=2\times\dfrac{\pi}{3}=\dfrac{2\pi}{3}$$

The overlapping region (the "lens") is made of two identical segments, one cut from each circle by the common chord $PQ$, each with radius $10$ and angle $\tfrac{2\pi}{3}$. Find one segment, using $\sin\theta=\sin120^{\circ}=\tfrac{\sqrt3}{2}$:

$$A_{\text{segment}}=\tfrac12r^{2}(\theta-\sin\theta)=\tfrac12(10^{2})\left(\dfrac{2\pi}{3}-\dfrac{\sqrt3}{2}\right)=50(2.0944-0.8660)=61.42$$

Double it, since both circles contribute an equal segment to the lens:

$$A_{\text{lens}}=2\times61.42=122.84$$

Away from the shed, the goat sweeps a full circle of radius $7$ m except for the $90^{\circ}$ wedge blocked by the shed itself at $A$, so the main region is a sector of angle $360^{\circ}-90^{\circ}=270^{\circ}=\tfrac{3\pi}{2}$ rad:

$$A_1=\tfrac12(7^{2})\left(\dfrac{3\pi}{2}\right)=\tfrac12(49)\left(\dfrac{3\pi}{2}\right)=36.75\pi$$

When the rope is pulled taut around the $4$ m side to the next corner $B$, only $7-4=3$ m of rope remains beyond $B$, sweeping a further quarter-turn (the shed turns another $90^{\circ}=\tfrac{\pi}{2}$ rad there):

$$A_2=\tfrac12(3^{2})\left(\dfrac{\pi}{2}\right)=\tfrac12(9)\left(\dfrac{\pi}{2}\right)=2.25\pi$$

Wrapping the other way, around the $3$ m side to corner $D$, leaves $7-3=4$ m of rope beyond $D$, sweeping another quarter-turn there:

$$A_3=\tfrac12(4^{2})\left(\dfrac{\pi}{2}\right)=\tfrac12(16)\left(\dfrac{\pi}{2}\right)=4\pi$$

In each case the leftover rope exactly equals the shed's remaining side ($3$ m and $4$ m), so it reaches the far corner $C$ with nothing left over, no third sector is needed on either side. Add all three areas:

$$A=36.75\pi+2.25\pi+4\pi=43\pi\approx135.09$$

Let the radius be $r$ and the angle $\theta$ radians. The perimeter is two radii plus the arc:

$$2r+r\theta=24\;\Rightarrow\;r\theta=24-2r\;\Rightarrow\;\theta=\dfrac{24-2r}{r}$$

Substitute this $\theta$ into the area formula so that area depends on $r$ alone:

$$A=\tfrac12r^{2}\theta=\tfrac12r^{2}\left(\dfrac{24-2r}{r}\right)=\tfrac12r(24-2r)=12r-r^{2}$$

This is the required expression. Complete the square to locate its maximum, since the $r^{2}$ coefficient is negative:

$$A=-(r^{2}-12r)=-\left[(r-6)^{2}-36\right]=36-(r-6)^{2}$$

The term $(r-6)^{2}$ can never be negative, so $A$ is largest, equal to $36$, exactly when $(r-6)^{2}=0$, that is, when $r=6$.

The two semicircular ends of a lane always join up to make one full circle, so a lap's distance is twice the straight length plus that circle's circumference.

$$\text{Inner lane: } D_{\text{in}}=2(40)+2\pi r=80+2\pi r$$

$$\text{Outer lane: } D_{\text{out}}=2(40)+2\pi(r+1)=80+2\pi r+2\pi$$

(a) Subtract the two distances. The straight sections and the $2\pi r$ term are identical in both, so they cancel completely:

$$D_{\text{out}}-D_{\text{in}}=(80+2\pi r+2\pi)-(80+2\pi r)=2\pi$$

This difference has no $r$ left in it at all, so it holds for every radius, the extra distance is always $2\pi\approx6.28$ m, regardless of how wide the bend is.

(b) Substitute $r=30$ into the outer-lane formula:

$$D_{\text{out}}=80+2\pi(31)=80+62\pi=80+194.78=274.78$$