Solved Examples

How to Use This Section

This is your practice bank for Areas Related to Circles. Keep the core formulas in front of you: circle area $\pi r^2$, circumference $2\pi r$, sector area $\tfrac{\theta}{360}\pi r^2$, arc $\tfrac{\theta}{360}\,2\pi r$, and segment $=$ sector $-$ triangle. Use $\pi=\tfrac{22}{7}$ unless a question says $3.14$.

Example 1: Find the area of a circle of diameter 28 cm. Solution: $r=14$; $A=\tfrac{22}{7}\times196=616$ cm$^2$. Answer: 616 cm$^2$.

Example 2: Find the circumference of a circle of area 154 cm$^2$. Solution: $r^2=\tfrac{154\times7}{22}=49\Rightarrow r=7$; $C=2\times\tfrac{22}{7}\times7=44$ cm. Answer: 44 cm.

Example 3: A sector has radius 10.5 cm and angle $60^\circ$. Find its area. $(\pi=\tfrac{22}{7})$ Solution: $\tfrac16\times\tfrac{22}{7}\times10.5^2=\tfrac16\times\tfrac{22}{7}\times110.25=57.75$ cm$^2$. Answer: 57.75 cm$^2$.

Example 4: Find the length of an arc of a circle of radius 14 cm subtending $90^\circ$. $(\pi=\tfrac{22}{7})$ Solution: $\tfrac14\times2\times\tfrac{22}{7}\times14=\tfrac14\times88=22$ cm. Answer: 22 cm.

Example 5: The area of a quadrant of a circle of radius 14 cm is: $(\pi=\tfrac{22}{7})$ Solution: $\tfrac14\times\tfrac{22}{7}\times196=154$ cm$^2$. Answer: 154 cm$^2$.

Example 6: Find the perimeter of a sector of radius 7 cm and angle $60^\circ$. $(\pi=\tfrac{22}{7})$ Solution: arc $=\tfrac16\times2\times\tfrac{22}{7}\times7=\tfrac{44}{6}=7.33$; perimeter $=7.33+14=21.33$ cm. Answer: ~21.33 cm.

Example 7: A chord subtends $90^\circ$ at the centre of a circle of radius 14 cm. Find the minor segment area. $(\pi=\tfrac{22}{7})$ Solution: sector $=\tfrac14\times\tfrac{22}{7}\times196=154$; triangle $=\tfrac12\times14\times14=98$; segment $=154-98=56$ cm$^2$. Answer: 56 cm$^2$.

Example 8: Find the area of the major sector when the minor sector (radius 7, $\pi=\tfrac{22}{7}$) has angle $60^\circ$. Solution: minor $=\tfrac16\times\tfrac{22}{7}\times49=\tfrac{25.67}{1}\approx25.67$; whole $=154$; major $=154-25.67=128.33$ cm$^2$. Answer: ~128.33 cm$^2$.

Example 9: The minute hand of a clock is 12 cm long. Area swept in 35 minutes. $(\pi=\tfrac{22}{7})$ Solution: $35$ min $=35\times6=210^\circ$; area $=\tfrac{210}{360}\times\tfrac{22}{7}\times144=\tfrac{7}{12}\times\tfrac{22}{7}\times144=264$ cm$^2$. Answer: 264 cm$^2$.

Example 10: A horse tied at a square-field corner with a 21 m rope grazes: $(\pi=\tfrac{22}{7})$ Solution: $\tfrac14\times\tfrac{22}{7}\times441=346.5$ m$^2$. Answer: 346.5 m$^2$.

Example 11: A wiper blade 21 cm long sweeps $90^\circ$. Area cleaned. $(\pi=\tfrac{22}{7})$ Solution: $\tfrac14\times\tfrac{22}{7}\times441=346.5$ cm$^2$. Answer: 346.5 cm$^2$.

Example 12: A circular flower bed of radius 7 m is surrounded by a 3.5 m wide path. Find the area of the path. $(\pi=\tfrac{22}{7})$ Solution: $R=10.5$, $r=7$; $\pi(R^2-r^2)=\tfrac{22}{7}(110.25-49)=\tfrac{22}{7}\times61.25=192.5$ m$^2$. Answer: 192.5 m$^2$.

Example 13: Find the area between a circle of radius 7 cm and its largest inscribed square. $(\pi=\tfrac{22}{7})$ Solution: square $=2r^2=98$; circle $=154$; between $=56$ cm$^2$. Answer: 56 cm$^2$.

Example 14: A wheel of diameter 84 cm makes 500 revolutions. Distance covered? $(\pi=\tfrac{22}{7})$ Solution: circumference $=\tfrac{22}{7}\times84=264$ cm; distance $=500\times264=132000$ cm $=1320$ m. Answer: 1320 m.

Example 15: A chord of a circle of radius 12 cm subtends $120^\circ$. Area of the corresponding segment. $(\pi=3.14,\sqrt3=1.73)$ Solution: sector $=\tfrac13\times3.14\times144=150.72$; triangle $=\tfrac{\sqrt3}{4}\times144=62.28$; segment $=150.72-62.28=88.44$ cm$^2$. Answer: 88.44 cm$^2$.

Example 16: Two circles have radii 8 cm and 6 cm. Find the radius of the circle whose area equals the sum of their areas. Solution: $\pi R^2=\pi(64+36)=100\pi\Rightarrow R=10$ cm. Answer: 10 cm.

Example 17: Two circles have radii 24 cm and 7 cm. Find the radius of the circle whose circumference equals the sum of their circumferences. Solution: $2\pi R=2\pi(24+7)\Rightarrow R=31$ cm. Answer: 31 cm.

Example 18: The area of a sector is $\tfrac{77}{2}$ cm$^2$ and its radius is 7 cm. Find its angle. $(\pi=\tfrac{22}{7})$ Solution: $\tfrac{\theta}{360}\times\tfrac{22}{7}\times49=38.5\Rightarrow\tfrac{\theta}{360}\times154=38.5\Rightarrow\tfrac{\theta}{360}=\tfrac14\Rightarrow\theta=90^\circ$. Answer: $90^\circ$.

Example 19: Find the area of a sector of angle $45^\circ$ in a circle of radius 28 cm. $(\pi=\tfrac{22}{7})$ Solution: $\tfrac{45}{360}\times\tfrac{22}{7}\times784=\tfrac18\times2464=308$ cm$^2$. Answer: 308 cm$^2$.

Example 20: A square of side 4 cm has a quarter circle of radius 4 cm drawn with centre at one corner. Find the area of the region inside the square but outside the quarter circle. $(\pi=3.14)$ Solution: square $=16$; quarter $=\tfrac14\times3.14\times16=12.56$; region $=16-12.56=3.44$ cm$^2$. Answer: 3.44 cm$^2$.

Example 21: The length of the minute hand of a clock is 14 cm. Find the area swept between 9:00 a.m. and 9:35 a.m. $(\pi=\tfrac{22}{7})$ Solution: 35 min $=210^\circ$; area $=\tfrac{210}{360}\times\tfrac{22}{7}\times196=\tfrac{7}{12}\times616=359.33$ cm$^2$. Answer: ~359.33 cm$^2$.

Example 22: A circle is inscribed in an equilateral triangle… (skip trig) Instead: find area of a semicircle of diameter 14 cm. $(\pi=\tfrac{22}{7})$ Solution: $r=7$; $\tfrac12\times\tfrac{22}{7}\times49=77$ cm$^2$. Answer: 77 cm$^2$.

Example 23: The perimeter of a semicircular protractor of radius 7 cm (including the diameter). $(\pi=\tfrac{22}{7})$ Solution: curved $=\pi r=22$; plus diameter $14$; total $=36$ cm. Answer: 36 cm.

Example 24: From a square of side 14 cm, four quarter circles (radius 7) at the corners and one circle (radius 7) at the centre… simpler: a circle of radius 7 cm is removed from a square of side 14 cm. Area left? $(\pi=\tfrac{22}{7})$ Solution: $196-154=42$ cm$^2$. Answer: 42 cm$^2$.

Example 25: Find the diameter of a circle whose area equals the sum of the areas of two circles of diameters 10 cm and 24 cm. Solution: radii 5, 12; $R^2=25+144=169\Rightarrow R=13$, so diameter $=26$ cm. Answer: 26 cm.

Example 26: A sector of a circle of radius 21 cm has arc length 22 cm. Find its area. Solution: area $=\tfrac12\ell r=\tfrac12\times22\times21=231$ cm$^2$. Answer: 231 cm$^2$.

Example 27: An arc subtends $30^\circ$ at the centre of a circle of radius 42 cm. Find the arc length. $(\pi=\tfrac{22}{7})$ Solution: $\tfrac{30}{360}\times2\times\tfrac{22}{7}\times42=\tfrac{1}{12}\times264=22$ cm. Answer: 22 cm.

Example 28: The wheels of a bicycle have diameter 70 cm. How many revolutions to travel 11 km? $(\pi=\tfrac{22}{7})$ Solution: circumference $=\tfrac{22}{7}\times70=220$ cm $=2.2$ m; revolutions $=\tfrac{11000}{2.2}=5000$. Answer: 5000.

Example 29: A chord of a circle of radius 10 cm subtends $90^\circ$. Find the area of the major segment. $(\pi=3.14)$ Solution: minor segment $=28.5$; whole $=3.14\times100=314$; major $=314-28.5=285.5$ cm$^2$. Answer: 285.5 cm$^2$.

Example 30: A round table cover of radius 28 cm has six equal segment-shaped designs (each subtending $60^\circ$). Find the total area of the designs. $(\pi=\tfrac{22}{7},\ \sqrt3=1.73)$ Solution:

1. Each design is a segment with $\theta=60^\circ$: sector $=\tfrac16\times\tfrac{22}{7}\times784=410.67$; triangle (equilateral) $=\tfrac{\sqrt3}{4}\times784=339.08$.
2. One design $=410.67-339.08=71.59$ cm$^2$.
3. Six designs $=6\times71.59\approx429.5$ cm$^2$.

Final Answer: $\approx 429.5$ cm$^2$.

Takeaway: $n$ equal designs around a circle $\Rightarrow$ each subtends $\tfrac{360^\circ}{n}$; total $=n\times(\text{one segment})$.