Tag: Exercise 11.1 Chapter 11 NCERT Answers and Solutions

Q1

Question. Find the area of a sector of a circle with radius $6$ cm if angle of the sector is ${60}^{\circ }$.
Solution.

$$\text{Area}=\frac{\theta }{360}\pi r^2=\frac{60}{360}\pi (6{)}^2=\frac{1}{6}\pi \cdot 36=6\pi \mathrm{}$$

Using $\pi =\frac{22}{7}$:

$$\text{Area}=6\cdot \frac{22}{7}=\frac{132}{7}\approx \boxed{{\displaystyle 18.86{\text{cm}}^2}}\mathrm{}$$

---

Q2

Question. Find the area of a quadrant of a circle whose circumference is $22$ cm.
Solution. Circumference $=2\pi r=22\Rightarrow r={\displaystyle \frac{11}{\pi }}$ With $\pi =\frac{22}{7}$ we get $r=3.5$.
Quadrant area $={\displaystyle \frac{1}{4}}\pi r^2={\displaystyle \frac{1}{4}}\pi (3.5{)}^2={\displaystyle \frac{1}{4}}\pi (12.25)=3.0625\pi$. With $\pi =\frac{22}{7}$,

$$\text{Area}\approx \boxed{{\displaystyle 9.62{\text{cm}}^2}}\mathrm{.}$$

---

Q3

Question. The length of the minute hand of a clock is $14$ cm. Find the area swept by the minute hand in $5$ minutes.
Solution. In 60 minutes the hand sweeps ${360}^{\circ }$; in 5 minutes it sweeps ${30}^{\circ }$.
Area $={\displaystyle \frac{30}{360}}\pi (14{)}^2={\displaystyle \frac{1}{12}}\pi \cdot 196={\displaystyle \frac{49\pi }{3}}$. With $\pi =\frac{22}{7}$,

$$\text{Area}\approx \boxed{{\displaystyle 51.33{\text{cm}}^2}}\mathrm{.}$$

---

Q4

Question. 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. Here $\theta ={90}^{\circ }$, $r=10$
Sector area $={\displaystyle \frac{90}{360}}\pi r^2=\frac{1}{4}\pi (100)=25\pi =25\times 3.14=78.50{\text{cm}}^2\mathrm{}$
Triangle $\mathrm{△}OAB$ (central angle ${90}^{\circ }$) area $=\frac{1}{2}{r}^2\sin {90}^{\circ }=\frac{1}{2}\cdot 100\cdot 1=50{\text{cm}}^2\mathrm{}$

(i) Minor segment $=$ sector $-$ triangle $=78.50-50=\boxed{{\displaystyle 28.50{\text{cm}}^2}}\mathrm{}$
(ii) Major sector angle $=360-90={270}^{\circ }$. Major sector area $={\displaystyle \frac{270}{360}}\pi r^2=\frac{3}{4}\cdot 100\pi =75\pi =75\times 3.14=\boxed{{\displaystyle 235.50{\text{cm}}^2}}\mathrm{}$

---

Q5

Question. In a circle of radius $21$ cm, an arc subtends an angle of ${60}^{\circ }$ at the centre. Find:
(i) length of the arc (ii) area of the sector formed by the arc (iii) area of the segment formed by the corresponding chord.
Solution. $r=21,\theta ={60}^{\circ },\pi =\frac{22}{7}\mathrm{}$

(i) Arc length $={\displaystyle \frac{\theta }{360}}\cdot 2\pi r={\displaystyle \frac{60}{360}}\cdot 2\pi \cdot 21={\displaystyle \frac{1}{6}}\cdot 42\pi =7\pi =7\frac{22}{7}=\boxed{{\displaystyle 22.00\text{ cm}}}\mathrm{}$

(ii) Sector area $={\displaystyle \frac{60}{360}}\pi r^2={\displaystyle \frac{1}{6}}\pi \cdot 441=73.5\pi$. With $\pi =\frac{22}{7}$ this is $\boxed{{\displaystyle 231.00{\text{cm}}^2}}\mathrm{}$

(iii) Triangle $OAB$ area $={\displaystyle \frac{1}{2}}{r}^2\sin {60}^{\circ }={\displaystyle \frac{1}{2}}\cdot 441\cdot \frac{\sqrt{3}}{2}={\displaystyle \frac{441\sqrt{3}}{4}}\approx 110.25\sqrt{3}$. Using $\sqrt{3}\approx 1.732$ gives triangle area $\approx 190.96{\text{cm}}^2$
Segment area $=$ sector $-$ triangle $\approx 231.00-190.96=\boxed{{\displaystyle 40.04{\text{cm}}^2}}\mathrm{}$

---

Q6

Question. A chord of a circle of radius $15$ cm subtends an angle of ${60}^{\circ }$ at the centre. Find the areas of the corresponding minor and major segments.
(Use $\pi =3.14$ and $\sqrt{3}=1.73$)
Solution. $r=15,\theta ={60}^{\circ }\mathrm{}$

Sector area $={\displaystyle \frac{1}{6}}\pi r^2={\displaystyle \frac{1}{6}}\cdot 3.14\cdot 225=117.75{\text{cm}}^2\mathrm{}$
Triangle area $=r^2\cdot \frac{\sqrt{3}}{4}=225\cdot \frac{1.73}{4}=97.31{\text{cm}}^2\mathrm{}$

Minor segment $=117.75-97.31=\boxed{{\displaystyle 20.44{\text{cm}}^2}}\mathrm{}$
Major segment $=$ total circle area $-$ minor segment

$=\pi r^2-$ minor segment$=3.14\cdot 225-20.44=706.50-20.44=\boxed{{\displaystyle 686.06{\text{cm}}^2}}\mathrm{}$

---

Q7

Question. A chord of a circle of radius $12$ cm subtends an angle of ${120}^{\circ }$at the centre. Find the area of the corresponding segment.
(Use $\pi =3.14$ and $\sqrt{3}=1.73$).
Solution. $r=12,\theta ={120}^{\circ }\mathrm{}$

Sector area $={\displaystyle \frac{120}{360}}\pi r^2=\frac{1}{3}\cdot 3.14\cdot 144=150.72{\text{cm}}^2\mathrm{}$
Triangle area $=r^2\cdot \frac{\sqrt{3}}{4}=144\cdot \frac{1.73}{4}=62.28{\text{cm}}^2\mathrm{}$

Segment area $=150.72-62.28=\boxed{{\displaystyle 88.44{\text{cm}}^2}}\mathrm{}$

---

Q8

Question. A horse is tied to a peg at one corner of a square grass field of side $15$ m by a rope of length $5$ m. (i) area of the part of field the horse can graze; (ii) increase in grazing area if rope were $10$ m long instead of $5$ m. (Use $\pi =3.14$).
Solution. The peg at a corner restricts the grazing inside the square to a quarter circle of radius equal to rope length (rope $<$ side).

(i) For rope $=5$ m: area $={\displaystyle \frac{1}{4}}\pi (5{)}^2={\displaystyle \frac{1}{4}}\cdot 3.14\cdot 25=19.625\approx \boxed{{\displaystyle 19.62{\text{m}}^2}}\mathrm{}$
(ii) For rope $=10$ m: area $={\displaystyle \frac{1}{4}}\pi (10{)}^2={\displaystyle \frac{1}{4}}\cdot 3.14\cdot 100=78.50{\text{m}}^2\mathrm{}$
Increase $=78.50-19.62=\boxed{{\displaystyle 58.88{\text{m}}^2}}\mathrm{}$

---

Q9

Question. A brooch is made with silver wire in the form of a circle with diameter $35$ mm. The wire is also used in making $5$ diameters which divide the circle into $10$ equal sectors. Find: (i) total length of silver wire required. (ii) area of each sector of the brooch.
Solution. Diameter $d=35$, radius $r=17.5$ mm, $\pi =\frac{22}{7}$

(i) Total wire $=$ circumference $+$ $5$ diameters $=\pi d+5d=d(\pi +5)$. Numerically,

$$\text{circumference}=\pi d=\frac{22}{7}\cdot 35=110\text{ mm},5d=175\text{ mm}\mathrm{.}$$

So total $=110+175=\boxed{{\displaystyle 285.00\text{ mm}}}\mathrm{}$

(ii) Area of whole circle $=\pi r^2=\frac{22}{7}\cdot (17.5{)}^2=962.50\pi \mathrm{/}??$ (numeric) — but each sector is $1\mathrm{/}10$ of circle:

$$\text{area each}=\frac{1}{10}\pi r^2=\frac{1}{10}\cdot \frac{22}{7}\cdot (17.5{)}^2\approx \boxed{{\displaystyle 96.25{\text{mm}}^2}}\mathrm{.}$$

---

Q10

Question. An umbrella has $8$ ribs equally spaced. Assuming umbrella is a flat circle of radius $45$ cm, find the area between two consecutive ribs.
Solution. Angle between consecutive ribs $={360}^{\circ }\mathrm{/}8={45}^{\circ }$

Area of sector

$$=\frac{45}{360}\pi (45{)}^2=\frac{1}{8}\pi \cdot 2025=\frac{2025\pi }{8}\mathrm{.}$$

With $\pi =\frac{22}{7}$,

$$\text{Area}=\frac{2025}{8}\cdot \frac{22}{7}\approx \boxed{{\displaystyle 795.54{\text{cm}}^2}}\mathrm{}$$

---

Q11

Question. A car has two wipers which do not overlap. Each wiper has a blade of length $25$ cm sweeping through an angle of ${115}^{\circ }$. Find the total area cleaned at each sweep of the blades.
Solution. Area cleaned by one wiper $={\displaystyle \frac{115}{360}}\pi (25{)}^2$. Two wipers: double that.

$$\text{Total}=2\cdot \frac{115}{360}\pi \cdot 625.$$

With $\pi =\frac{22}{7}$ this evaluates to $\boxed{{\displaystyle 1254.96{\text{cm}}^2}}$

---

Q12

Question. A lighthouse spreads red light over a sector of angle ${80}^{\circ }$ to a distance $16.5$ km. Find the area of sea warned. (Use $\pi =\frac{22}{7}$).
Solution. $r=16.5$ km, $\theta ={80}^{\circ }\mathrm{<span\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;">.</span>}$Area $={\displaystyle \frac{80}{360}}\pi (16.5{)}^2\approx \boxed{{\displaystyle 190.14{\text{km}}^2}}$

---

Q13

Question. A round table cover has six equal designs as shown in figure. If radius $=28$, find the cost of making the designs at rate ₹0.35 per cm${}^2$. (Use $\sqrt{3}=1.7$).
Solution. Each design is the segment corresponding to $\theta ={60}^{\circ }$ (since $360\mathrm{/}6={60}^{\circ }$). For one segment:
Sector area $={\displaystyle \frac{60}{360}}\pi r^2={\displaystyle \frac{1}{6}}\cdot \frac{22}{7}\cdot {28}^2={\displaystyle \frac{1232}{3}}{\text{cm}}^2\approx 410.67$
Triangle $\mathrm{△}OAB$ (with central angle ${60}^{\circ }$) is equilateral of side $28$; area $={\displaystyle \frac{\sqrt{3}}{4}}\cdot {28}^2=196\sqrt{3}$. Using $\sqrt{3}=1.7$,

$$\text{triangle area}=196\times 1.7=333.20{\text{cm}}^2\mathrm{.}$$

Segment area $=$ sector $-$ triangle $\approx 410.67-333.20=77.47{\text{cm}}^2\mathrm{.}$ For six designs total area $=6\times 77.47=464.80{\text{cm}}^2\mathrm{.}$
Cost $=464.80\times 0.35=\boxed{{\displaystyle \text{₹}162.68}}\mathrm{}$

---

Q14

Question. Tick the correct answer: Area of a sector of angle $p$ (degrees) of circle radius $R$ is — (options include the standard forms).
Solution. Formula is ${\displaystyle \frac{p}{360}}\pi R^2$. So the correct option is (C).