AREAS RELATED TO CIRCLES-Exercise 11.1

Q4. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding:
(i)minor segment
(ii)major sector. (Use \(\pi\) = 3.14)

Solution:

Angle subtended by chord at the center = 90°
Radius of the circle = 10 cm
In \(\triangle AOD\)
Construction: Draw \(OD\perp BC\) (\(OD\) is perpendicular bisector)

$$\begin{aligned}\sin 45^{\circ}&=\dfrac{OD}{OA}\\ OD&=OA\sin 45^{0}\\ &=10\times \dfrac{1}{\sqrt{2}}\\\\ &=\dfrac{10}{\sqrt{2}}\\\\ \cos 45^{0}&=\dfrac{AD}{OA}\\\\ AD&=OA\cos 45^{\circ }\\\\ &=10\times \dfrac{1}{\sqrt2}\\\\ &=\dfrac{10}{\sqrt{2}}\end{aligned}$$

\(AB = 2AD\) (\(OD\) is bisector),
therefore Area \(A_t\) of triangle \(\triangle AOB\)

$$\require{cancel}\begin{aligned}A_{t}&=\dfrac{1}{2}\times AB\times OD\\\\ &=\dfrac{1}{2}\times \dfrac{20}{\sqrt{2}}\times \dfrac{10}{\sqrt{2}}\\\\ &=\dfrac{\cancelto{\cancelto{5}{10}}{20}\times 10}{\cancel{2}\times \cancel{2}}\\\\ &=50\ cm^{2}\end{aligned}$$

Calculate Area of minor sector \(A_{ms}\)

$$\require{cancel}\begin{aligned}A_{ms}&=\dfrac{\theta }{360^{\circ }}\pi r^{2}\\\\ &=\dfrac{\cancel{90^{0}}}{\cancelto{4}{360^{0}}}\times \dfrac{22}{7}\times 10\times 10\\\\ &=\dfrac{1}{\cancelto{\cancel{2}}{4}}\cdot \dfrac{22}{7}\times \cancelto{5}{10}\times \cancelto{5}{10}\\\\ &=3.14\times 25\\\\ &=78.50\end{aligned}$$

Area of segment \(A_{s}\) = Area of minor sector \(A_{ms}\) - Area of \(\triangle\) AOB
therefore,

$$\begin{aligned}A_{s}&=78.50-50\\ &=28.50\ cm^{2}\end{aligned}$$

Area of major sector \(A_{ms}\)

$$\begin{aligned}A_{ms}&=\dfrac{\theta }{360^{\circ }}\times \pi r^{2}\\\\ \theta &=360^{\circ }-90^{\circ }\\\\ &=270^{0}\\\\ \therefore A_{ms}&=\dfrac{270}{360}\times \dfrac{22}{7}\times 10^{2}\\\\ &=\dfrac{3}{4}\times 3.14\times 100\\\\ &=3\times 3.14\times 25\\\\ &=75\times 3.14\\\\ &=235.50\ cm^{2}\end{aligned}$$

Area of Minor Segment = \(28.50\ cm^{2}\)
Area of Major sector = \(235.50\ cm^{2}\)

Q5. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find: (i) the length of the arc
(ii) area of the sector formed by the arc
(iii) area of the segment formed by the corresponding chord

Solution:

Radius of the circle = 21 cm
Angle subtended by arc on the centre of the circle = 60° Length of the arc \((l)\)

$$\require{cancel}\begin{aligned}l&=\dfrac{\theta }{360^{\circ }}\times 2\pi r\\\\ &=\dfrac{\cancel{60^{\circ }}}{\cancelto{6}{360^{\circ}}}\times 2\times\dfrac{22}{\cancel{7}}\times \cancelto{3}{21}\\\\ &=\dfrac{1}{\cancel{6}}\times \cancel{2}\times 22\times \cancel{3}\\\\ &=22\ cm\end{aligned}$$

Area of the sector \((A_s)\)

$$\require{cancel}\begin{aligned}A_{s}&=\dfrac{\theta }{360^{0}}\times \pi r^{2}\\\\ &=\dfrac{60^{0}}{360^{\circ}}\times \dfrac{22}{\cancel{7}}\times 21\times \cancelto{3}{21}\\\\ &=\dfrac{1}{\cancel{6}}\times \cancelto{11}{22}\times \cancel{3}\times 21\\\\ &=11\times 21\\\\ &=231\ cm^{2}\end{aligned}$$

In \(\triangle AOD\)
OD is drawn perpendicular bisector of AB \((OB\perp AB)\)

$$\require{cancel}\begin{aligned}\sin 30^{\circ }&=\dfrac{AD}{OA}\\\\ AD&=OA\sin 30^{\circ }\\\\ &=21\times \dfrac{1}{2}\\\\ &=\dfrac{21}{2}\\\\ AB&=2AD\\\\ &=\dfrac{21}{\cancel{2}}\times \cancel{2}\\\\ &=21\ cm\\\\ \cos 30^{0}&=\dfrac{OD}{OA}\\\\ OD&=OA\cos 30^{\circ }\\\\ OD&=\dfrac{21\times \sqrt{3}}{2}\\\\ OD&=\dfrac{21\sqrt{3}}{2}\ cm\end{aligned}$$

Area of \(\triangle AOB (A_t)\)

$$\begin{aligned}A_{t}&=\dfrac{1}{2}\times AB\times OD\\\\ &=\dfrac{1}{2}\times \dfrac{21\times 21\sqrt{3}}{2}\\\\ &=\dfrac{441\sqrt{3}}{4}\end{aligned}$$

Area of segment \(A_{s}\) = Area of sector \(A_s\) - Area of \(\triangle AOB\)

$$\begin{aligned}A_{s}&=231-\dfrac{441\sqrt{3}}{4}\ cm^{2}\\\\ &=231-\left( \dfrac{441\sqrt{3}}{4}\right)\ cm^{2}\end{aligned}$$

1. Length of Arc is 22 cm


2. Area of the sector formed by the arc = \(231\ cm^{2}\)


3. Area of the segment formed by the corresponding chord = \(231-\left( \dfrac{441\sqrt{3}}{4}\right)\ cm^{2}\)

Q6. A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use \(\pi\) = 3.14 and 3 = 1.73)

Solution:

Radius of the circle = 15 cm
Angle subtended by the chord at the centre = 60°
Area of Sector \((A_s)\)

$$\require{cancel}\begin{aligned}A_{s}&=\dfrac{\theta }{360}\times \pi r^{2}\\\\ &=\dfrac{60}{360}\times \dfrac{22}{7}\times 15^{2}\\\\ &=\dfrac{1}{\cancel{6}}\times \dfrac{\cancelto{11}{22}}{7}\times \cancelto{5}{15}\times 15\\\\ &=\dfrac{11}{7}\times 5\times 15\\\\ &=\dfrac{11\times 75}{7}\\\\ &=\dfrac{825}{7}\\\\ &\approx 117.857\end{aligned}$$

Lets find base and height of \(\triangle AOB\) where OD is drawn perpendicular bisector of AB \(OD\perp AB\)

$$\begin{aligned}\sin 30^{\circ }&=\dfrac{AC}{OA}\\\\ AC&=15\sin 30^{\circ}\\\\ &=15\times \dfrac{1}{2}\\\\ &=\dfrac{15}{2}\\\\ \cos 30^{\circ }&=\dfrac{OC}{15}\\\\ OC&=15\cos 30^{\circ }\\\\ &=\dfrac{15\times \sqrt{3}}{2}\\\\ &=\dfrac{15\sqrt{3}}{2} cm\\\\ AB&=2AC\\\\ &=\cancel{2}\times \dfrac{15}{\cancel{2}}\\\\ &=15 cm\end{aligned}$$

Area of ΔAOB

$$\begin{aligned}A_{\Delta }&=\dfrac{1}{2}\times 15\times \dfrac{15\sqrt{3}}{2}\\\\ &=\dfrac{225\sqrt{3}}{4}\\\\ &\approx 97.427\end{aligned}$$

Area of minor segment \(A_{sg}\) = Area of minor Arc \(A_{s}\) - Area of Δ AOB \(A_{\Delta }\)

$$\begin{aligned}A_{sg}&=117.857-97.427\\ &=20.43\end{aligned}$$

Area of circle \(A_{\circ}\)

$$\begin{aligned}A_{\circ}&=\pi r^{2}\\\\ &=\dfrac{22}{7}\times 15\times 15\\\\ &=\dfrac{22}{7}\times 225\\\\ &\approx 707.142\end{aligned}$$

Area of major segment \(A_{msg}\) = Area of circle \(A_{\circ}\) - Area of minor segment \(A_{sg}\)

$$\begin{aligned}A_{msg}&=707.142-20.43\\ &\approx 686.71\end{aligned}$$

Hence,
Area of major segment = 686.71 cm²
Area of minor segment = 20. 43 cm²

Q7. A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle. (Use \(\pi\) = 3.14 and 3 = 1.73)

Solution:

Radius of the circle r = 12 cm
Angle chord subtended by chord on the centre = 120°
Area of the sector \((A_s)\)

$$\require{cancel}\begin{aligned}A_{s}&=\dfrac{\theta }{360}\pi r^{2}\\\\ &=\dfrac{\cancel{120}}{\cancelto{3}{360}}\times \dfrac{22}{7}\times 12\times 12\\\\ &=\dfrac{1}{\cancel{3}}\times \dfrac{22}{7}\times \cancelto{4}{12}\times 12\\\\ &=\dfrac{22}{7}\times 4\times 12\\\\ &=\dfrac{22\times 48}{7}\\\\ &=\dfrac{1056}{7}\\\\ &=150.857\end{aligned}$$

In \(\triangle AOB,\ OC\ \) is perpendicular bisector on chord \(AB\; \angle AOC = \frac{1}{2} 120° = 60°\)

$$\require{cancel}\begin{aligned}\sin 60^{\circ}&=\dfrac{AC}{OA}\\\\ AC&=OA\sin 60^{\circ}\\\\ &=\cancelto{6}{12}\times \dfrac{\sqrt{3}}{\cancel{2}}\\\\ &=6\sqrt{3}\\\\ AB&=2AC\\\\ AB&=6\sqrt{3}\times 2\\\\ &=12\sqrt{3} cm\\\\ \cos 60^{\circ }&=\dfrac{OC}{AC}\\\\ OC&=AC\times \cos 60^{\circ }\\\\ &=\cancelto{6}{12}\times \dfrac{1}{\cancel{2}}\\\\ &=6\ cm\end{aligned}$$

Area of \(\triangle AOB\ (A_{\Delta })\)

$$\require{cancel}\begin{aligned}A_{\Delta }&=\dfrac{1}{2}\times AB\times OC\\\\ &=\dfrac{1}{\cancel{2}}\times 12\sqrt{3}\times \cancelto{3}{6}\\\\ &=36\sqrt{3}\ cm\end{aligned}$$

Area of segment \((A_{sg})\) = Area of minor Sector \((A_{s})\) - Area of \(\triangle AOB\ (A_{\Delta })\)

$$\begin{aligned}A_{sg}&=150.857-36\sqrt{3}\\ &=150.857-36\times 1.732\\ &=150.857-62.352\\ &\approx 88.505\ cm^2\end{aligned}$$

Area of segment = 88. 50 cm²

Q8. A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope (see Fig. 11.8). Find
(i) the area of that part of the field in which the horse can graze.
(ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use \(\pi\) = 3.14)

Solution:

Side of square = 15 m
Length of rope = radius of circle = 5 m
The horse is tied at one corner of the square, so it can graze only in one quadrant of the circle, corresponding to a central angle of \(90^\circ\)

$$\require{cancel}\begin{aligned}A&=\dfrac{\theta }{360}\cdot \pi r^{2}\\\\ &=\dfrac{\cancel{90}}{\cancelto{4}{360}}\times \dfrac{22}{7}\times 5\times 5\\\\ &=\dfrac{1}{4}\times \dfrac{22}{7}\times 25\\\\ &=\dfrac{3.14\times 25}{4}\\\\ &=19.625\ cm^2\end{aligned}$$

Horse will graze 19.625 m²

Q9. A brooch is made with silver wire in the form of a circle with a diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in Fig. 11.9. Find :
(i) the total length of the silver wire required.
(ii) the area of each sector of the brooch.

Solution:

Diameter of brooch = 35 mm
Hence Radius of brooch = 35/2 = 17.5 mm
Total length of wire = circumference of the brooch + 5 diameters

$$\begin{aligned}\text{Circumference} &=2\pi r\\\\ &=2\times \dfrac{22}{\cancel{7}}\times \cancelto{2.5}{17.5}\\\\ &=110\end{aligned}$$

Total length of diameters

$$35\times 5=175\ mm$$

Sum of length

$$110+175=285\ mm$$

Angle subtended by each arc at the centre of the circle = 360/10 = 36°
Area of each sector

$$\require{cancel}\begin{aligned}A&=\dfrac{\theta }{360}\pi r^{2}\\\\ &=\dfrac{36}{360}\cdot \dfrac{22}{\cancel{7}}\times 17.5\times \cancelto{2.5}{17.5}\\\\ &=\dfrac{1}{\cancelto{\cancel{5}}{10}}\times \cancelto{11}{22}\times 17.5\times \cancelto{0.5}{2.5}\\\\ &=11\times 0.5\times 17.5\\\\ &=5.5\times 17.5\\\\ &=96.25\ mm^{2}\end{aligned}$$ Total length of silver wire is 285 mm and area of each sector is 96.25 mm²

Q10. An umbrella has 8 ribs which are equally spaced (see Fig. 11.10). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

Solution:

Radius of the umbrella = 45 cm
Total Ribs = 8
Angle subtended at the centre by the arc between two ribs = 360/8 = 45°
Area between two consecutive Rib =

$$\require{cancel}\begin{aligned}A&=\dfrac{\theta }{360}\times \pi r^{2}\\\\ &=\dfrac{\cancelto{5}{45}}{\cancelto{40}{360}}\times \dfrac{22}{7}\times 45\times 45\\\\ &=\dfrac{\cancel{5}}{\cancelto{8}{40}}\times \dfrac{22}{7}\times 45\times 45\\\\ &=\dfrac{1}{8}\times \dfrac{22}{7}\times 2025\\\\ &=\dfrac{11\times 2025}{28}\\\\ &=\dfrac{22275}{28}\\ &\approx 795.54 cm^{2}\end{aligned}$$

Q13. A round table cover has six equal designs as shown in Fig. 11.11. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of ₹ 0.35 per cm2. (Use \(\sqrt{3}= 1.7\))

Solution:

Radius of the cover = 28 cm
Angle subtended by each sector = 360/6 = 60°
Area of design of each part = Area of minor sector - Area of Δ made by the sector
Area of one sector

$$\require{cancel}\begin{aligned}A&=\dfrac{\theta }{360^{\circ }}\pi r^{2}\\\\ &=\dfrac{\cancel{60}}{\cancelto{6}{360}}\times\dfrac{22}{7}\times 28\times 28\\\\ &=\dfrac{1}{6}\times \dfrac{22}{\cancel{7}}\times \cancelto{4}{28}\times 28\\\\ &=\dfrac{1}{6}\times 22\times 4\times 28\\\\ &\approx 410.67\end{aligned}$$

In Δ AOB, OC is perpendicular bisector
\(\angle AOC = 30°\)

$$\require{cancel}\begin{aligned}\sin 30^{\circ}&=\dfrac{AC}{OA}\\\\ AC&=OA\sin 30^{\circ }\\\\ &=\cancelto{14}{28}\times \dfrac{1}{\cancel{2}}\\\\ &=14\\\\ AB&=2AC\\\\ &=14\times 2\\\\ &=28\\\\ \cos 30^{0}&=\dfrac{OC}{AC}\\\\ OC&=AC\cos 30^{0}\\\\ &=\dfrac{AC\sqrt{3}}{2}\\\\ &=\dfrac{\cancelto{14}28\sqrt{3}}{\cancel{2}}\\\\ &=14\sqrt{3}\end{aligned}$$

Area of \(\triangle AOB\)

$$\require{cancel}\begin{aligned}A&=\dfrac{1}{2}\times AB\times OC\\ &=\dfrac{1}{\cancel{2}}\times \cancelto{14}{28}\times 14\sqrt{3}\\\\ &=14\times 14\sqrt{3}\\\\ &=196\sqrt{3}\end{aligned}$$

Area of one design Part = Area of Sector - Area of \(\triangle AOB\)

$$\begin{aligned}410.67-339.48\\ =71.19\ cm^{2}\end{aligned}$$

Area of 6 equal parts

$$71.19\times 6=427.14\ cm^{2}$$ Cost of Design = Area x Rate $$427.14\times 0.35=150$$