Exercises — Circles — centre, radius, diameter, chord, arc, sector, segment

What each is. $OP$ goes from centre to edge, so $OP$ is a radius. $AB$ is a chord passing through the centre, so $AB$ is a diameter. Why the length. The diameter is a chord through $O$; the two halves $OA$ and $OB$ are each a radius. Since $A,O,B$ lie on one straight line, distances add: $AB=OA+OB=r+r=2r=2(7)=\boxed{14\text{ cm}}.$

Circumference — the distance once around the edge. The number $\pi$ is defined as (distance around) $÷$ (distance across), and distance across is $d=2r$, so: $C=2\pi r=2\pi (10)=20\pi \approx 62.83\text{ cm}.$ Area — the space enclosed: $A=\pi r^2=\pi (10{)}^2=100\pi \approx 314.16{\text{ cm}}^2.$ Note $r$ is squared in area (a 2‑D measure) but not in circumference (a 1‑D length).

Why radians first. The clean formula $s=r\theta$ (arc length equals radius times angle) is only true when $\theta$ is measured in radians — the angle unit built so that "angle $=$ arc$/$radius". So convert: $\theta =40^{\circ }\times \frac{\pi }{180^{\circ }}=\frac{2\pi }{9}\text{ rad}.$ Apply. $s=r\theta =9\times \frac{2\pi }{9}=2\pi \approx 6.28\text{ cm}.$ (Sanity check: $40^{\circ }$ is $\frac{1}{9}$ of $360^{\circ }$, and $\frac{1}{9}$ of the circumference $2\pi (9)=18\pi$ is indeed $2\pi$. ✓)

Fraction of the whole. A sector is a fraction $\frac{\theta }{360^{\circ }}$ of the whole disk. Here that fraction is $\frac{45}{360}=\frac{1}{8}$. $A_{\text{sector}}=\frac{45}{360}\times \pi r^2=\frac{1}{8}\times \pi (14{)}^2=\frac{196\pi }{8}=24.5\pi \approx 76.97{\text{ cm}}^2.$ This is Sectors and Pizza Problem in one line. Equivalently, with $\theta =\frac{\pi }{4}$ rad, $A=\frac{1}{2}{r}^2\theta =\frac{1}{2}(196)(\frac{\pi }{4})=24.5\pi$. ✓

The picture. Drop a perpendicular from centre $O$ to the chord; it meets the chord at its midpoint $M$ (the perpendicular from the centre bisects the chord). This creates a right triangle $OMA$ with hypotenuse the radius $OA=10$, one leg $OM=6$, and the other leg $MA=$ half the chord.

Why Pythagoras. We have a right angle at $M$ and two of three sides, so Pythagorean Theorem finds the third: $MA=\sqrt{O{A}^2-O{M}^2}=\sqrt{10^2-6^2}=\sqrt{100-36}=\sqrt{64}=8\text{ cm}.$ Double it. $MA$ is only half the chord, so $AB=2\times MA=\boxed{16\text{ cm}}.$

Undo $s=r\theta$. Solve for $\theta$ (in radians): $\theta =\frac{s}{r}=\frac{12}{8}=1.5\text{ rad}.$ Back to degrees. Multiply by $\frac{180^{\circ }}{\pi }$: $\theta =1.5\times \frac{180^{\circ }}{\pi }=\frac{270^{\circ }}{\pi }\approx 85.94^{\circ }.$

The picture: segment = sector − triangle. The wedge (sector) is the triangle plus the little cap (segment). So the cap is what's left after removing the triangle.

Radians. $\theta =120^{\circ }\times \frac{\pi }{180}=\frac{2\pi }{3}$ rad.

Sector area. $A_{\text{sector}}=\frac{1}{2}{r}^2\theta =\frac{1}{2}(36)\left(\frac{2\pi }{3}\right)=12\pi \approx 37.70{\text{ cm}}^2.$

Triangle area. Two radii of length $r$ meet at angle $\theta$; the enclosed triangle has area $\frac{1}{2}{r}^2\sin \theta$ (base $=r$, height $=r\sin \theta$): $A_{△}=\frac{1}{2}{r}^2\sin \theta =\frac{1}{2}(36)\sin 120^{\circ }=18\times \frac{\sqrt{3}}{2}=9\sqrt{3}\approx 15.59{\text{ cm}}^2.$

Subtract. $A_{\text{segment}}=A_{\text{sector}}-A_{△}=12\pi -9\sqrt{3}\approx 37.70-15.59=\boxed{22.11{\text{ cm}}^2}.$

What bounds a sector. Its border is: radius, arc, radius. So the perimeter is $r+s+r=2r+s$, using the arc length, since a sector's curved side is the arc. $s=r\theta =6\times \frac{2\pi }{3}=4\pi \approx 12.57\text{ cm}.$ $P=2r+s=12+4\pi \approx 24.57\text{ cm}.$

(a) Minor segment. $\theta =90^{\circ }=\frac{\pi }{2}$ rad. $A_{\text{sector}}=\frac{1}{2}{r}^2\theta =\frac{1}{2}(100)\left(\frac{\pi }{2}\right)=25\pi \approx 78.54{\text{ cm}}^2.$ $A_{△}=\frac{1}{2}{r}^2\sin 90^{\circ }=\frac{1}{2}(100)(1)=50{\text{ cm}}^2.$ $A_{\text{minor seg}}=25\pi -50\approx 28.54{\text{ cm}}^2.$

(b) Major segment. The chord splits the whole disk into minor + major segment. So: $A_{\text{major seg}}=A_{\text{circle}}-A_{\text{minor seg}}=100\pi -(25\pi -50)=75\pi +50\approx 285.62{\text{ cm}}^2.$ Why this works. The two segments together are the entire disk $\pi r^2=100\pi$; subtracting the minor cap leaves the major cap. Notice the $+50$: the triangle we removed from the minor sector is added back into the major region.

Two equations, two unknowns. Write both facts in radian form: $\frac{1}{2}{r}^2\theta =30\pi \text{and}r\theta =6\pi .$ Eliminate $\theta$. Substitute $r\theta =6\pi$ into the first: $\frac{1}{2}r(r\theta )=30\pi$, i.e. $\frac{1}{2}r(6\pi )=30\pi \Rightarrow 3\pi r=30\pi \Rightarrow r=10\text{ cm}.$ Back-solve $\theta$. $r\theta =6\pi \Rightarrow \theta =\frac{6\pi }{10}=\frac{3\pi }{5}$ rad. $\theta =\frac{3\pi }{5}\times \frac{180^{\circ }}{\pi }=108^{\circ }.$ Check. $\frac{1}{2}(10{)}^2(\frac{3\pi }{5})=50\cdot \frac{3\pi }{5}=30\pi$ ✓.

(a) Chord. Right triangle: radius $5$, distance $3$, half-chord $=\sqrt{5^2-3^2}=\sqrt{16}=4$. Chord $=2\times 4=8$ cm. (b) Angle then arc. Half the central angle $\alpha$ satisfies $\cos \alpha =\frac{3}{5}$, so $\alpha =\arccos (0.6)\approx 53.13^{\circ }$; full central angle $\theta =2\alpha \approx 106.26^{\circ }=1.8546$ rad. $s=r\theta =5(1.8546)\approx 9.27\text{ cm}.$ (c) Segment. $A_{\text{sector}}=\frac{1}{2}{r}^2\theta =\frac{1}{2}(25)(1.8546)\approx 23.18{\text{ cm}}^2,$ $A_{△}=\frac{1}{2}{r}^2\sin \theta =\frac{1}{2}(25)\sin (106.26^{\circ })\approx \frac{1}{2}(25)(0.96)=12.0{\text{ cm}}^2,$ $A_{\text{minor seg}}\approx 23.18-12.0=11.18{\text{ cm}}^2.$