NCERT Solutions Class 10 Maths Chapter 11: Areas Related to Circles
Detailed solutions for Exercise 11.1 (Area of Sector and Segment).
#### **Q1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is $60^\circ$.**
#### **A1. Solution:**
Given: Radius ($r$) = 6 cm, Angle ($\theta$) = $60^\circ$.
Area of the sector = $\frac{\theta}{360^\circ} \times \pi r^2$
$= \frac{60^\circ}{360^\circ} \times \frac{22}{7} \times 6 \times 6$
$= \frac{1}{6} \times \frac{22}{7} \times 36$
$= \frac{132}{7}$ cm²
#### **Q2. Find the area of a quadrant of a circle whose circumference is 22 cm.**
#### **A2. Solution:**
Given: Circumference ($2\pi r$) = 22 cm.
$2 \times \frac{22}{7} \times r = 22 \Rightarrow r = \frac{7}{2}$ cm.
For a quadrant, $\theta = 90^\circ$.
Area of quadrant = $\frac{1}{4} \times \pi r^2$
$= \frac{1}{4} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} = \frac{77}{8}$ cm²
#### **Q3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.**
#### **A3. Solution:**
Length of minute hand ($r$) = 14 cm.
Angle swept in 60 min = $360^\circ$.
Angle swept in 5 min ($\theta$) = $\frac{360^\circ}{60} \times 5 = 30^\circ$.
Area swept = $\frac{30^\circ}{360^\circ} \times \frac{22}{7} \times 14 \times 14$
$= \frac{1}{12} \times 22 \times 2 \times 14 = \frac{154}{3}$ cm²
#### **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 $\pi = 3.14$)**
#### **A4. Solution:**
Given: $r = 10$ cm, $\theta = 90^\circ$.
(i) Area of minor segment = Area of sector - Area of $\Delta OAB$
$= (\frac{90^\circ}{360^\circ} \times 3.14 \times 10^2) - (\frac{1}{2} \times 10 \times 10)$
$= 78.5 - 50 = 28.5$ cm².
(ii) Area of major sector = Area of circle - Area of minor sector
$= (3.14 \times 10^2) - 78.5 = 314 - 78.5 = 235.5$ cm².
#### **Q5. In a circle of radius 21 cm, an arc subtends an angle of $60^\circ$ at the centre. Find: (i) arc length (ii) area of sector (iii) area of segment.**
#### **A5. Solution:**
Given: $r = 21$ cm, $\theta = 60^\circ$.
(i) Arc length = $\frac{60^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 21 = 22$ cm.
(ii) Area of sector = $\frac{60^\circ}{360^\circ} \times \frac{22}{7} \times 21^2 = 231$ cm².
(iii) Area of segment = Sector area - Area of equilateral $\Delta OAB$
$= 231 - (\frac{\sqrt{3}}{4} \times 21^2) = (231 - \frac{441\sqrt{3}}{4})$ cm².
#### **Q6. A chord of a circle of radius 15 cm subtends $60^\circ$ at the centre. Find areas of minor and major segments. (Use $\pi=3.14, \sqrt{3}=1.73$)**
#### **A6. Solution:**
Area of minor sector = $\frac{60^\circ}{360^\circ} \times 3.14 \times 15^2 = 117.75$ cm².
Area of $\Delta OAB = \frac{\sqrt{3}}{4} \times 15^2 = \frac{1.73 \times 225}{4} = 97.3125$ cm².
Minor segment area = $117.75 - 97.3125 = 20.4375$ cm².
Major segment area = Area of circle - Minor segment
$= (3.14 \times 15^2) - 20.4375 = 706.5 - 20.4375 = 686.0625$ cm².
#### **Q7. A chord of a circle of radius 12 cm subtends $120^\circ$ at the centre. Find the area of the corresponding segment. (Use $\pi=3.14, \sqrt{3}=1.73$)**
#### **A7. Solution:**
Area of sector = $\frac{120^\circ}{360^\circ} \times 3.14 \times 12^2 = 150.72$ cm².
Area of $\Delta OAB = r^2 \sin(\frac{\theta}{2}) \cos(\frac{\theta}{2}) = 12^2 \sin 60^\circ \cos 60^\circ = 144 \times \frac{\sqrt{3}}{2} \times \frac{1}{2} = 36\sqrt{3}$.
Area of $\Delta OAB = 36 \times 1.73 = 62.28$ cm².
Area of segment = $150.72 - 62.28 = 88.44$ cm².
#### **Q8. A horse is tied to a corner of a square field (side 15m) with a 5m rope. Find: (i) grazing area (ii) increase if rope is 10m. (Use $\pi=3.14$)**
#### **A8. Solution:**
(i) Grazing area ($r=5, \theta=90^\circ$) = $\frac{1}{4} \times 3.14 \times 5^2 = 19.625$ m².
(ii) Area with 10m rope = $\frac{1}{4} \times 3.14 \times 10^2 = 78.5$ m².
Increase in area = $78.5 - 19.625 = 58.875$ m².
#### **Q9. A brooch (diameter 35mm) has 5 diameters dividing it into 10 sectors. Find: (i) total silver wire (ii) area of each sector.**
#### **A9. Solution:**
(i) Total wire = Circumference + 5 × Diameter
$= (\frac{22}{7} \times 35) + (5 \times 35) = 110 + 175 = 285$ mm.
(ii) Area of each sector = $\frac{1}{10} \times \text{Area of circle}$
$= \frac{1}{10} \times \frac{22}{7} \times \frac{35}{2} \times \frac{35}{2} = \frac{385}{4} = 96.25$ mm².
#### **Q10. An umbrella has 8 ribs (radius 45cm). Find area between two consecutive ribs.**
#### **A10. Solution:**
Angle between two ribs ($\theta$) = $\frac{360^\circ}{8} = 45^\circ$.
Area = $\frac{45^\circ}{360^\circ} \times \frac{22}{7} \times 45^2 = \frac{1}{8} \times \frac{22}{7} \times 2025 = \frac{22275}{28}$ cm².
#### **Q11. A car has two wipers (blade 25cm, angle $115^\circ$). Find total area cleaned.**
#### **A11. Solution:**
Total area = $2 \times (\text{Area of one sector})$
$= 2 \times \frac{115^\circ}{360^\circ} \times \frac{22}{7} \times 25^2 = \frac{23}{18} \times \frac{11}{7} \times 625 = \frac{158125}{126}$ cm².
#### **Q12. A lighthouse spreads light over $80^\circ$ to 16.5 km. Find the area. (Use $\pi=3.14$)**
#### **A12. Solution:**
Area = $\frac{80^\circ}{360^\circ} \times 3.14 \times 16.5^2$
$= \frac{2}{9} \times 3.14 \times 272.25 = 189.97$ km².
#### **Q13. A round table cover (radius 28cm) has 6 designs. Find cost at ₹0.35 per cm². (Use $\sqrt{3}=1.7$)**
#### **A13. Solution:**
For 6 designs, $\theta = 60^\circ$.
Area of 1 design = Area of sector - Area of $\Delta$
$= (\frac{60^\circ}{360^\circ} \times \frac{22}{7} \times 28^2) - (\frac{\sqrt{3}}{4} \times 28^2) = 410.67 - 333.2 = 77.47$ cm².
Total Area = $6 \times 77.47 = 464.82$ cm².
Cost = $464.82 \times 0.35 = ₹ 162.68$.
#### **Q14. Area of a sector of angle $p$ (in degrees) of a circle with radius $R$ is:**
**(A) $\frac{p}{180} \times 2\pi R$ (B) $\frac{p}{180} \times \pi R^2$ (C) $\frac{p}{360} \times 2\pi R$ (D) $\frac{p}{720} \times 2\pi R^2$**
#### **A14. Solution:**
Area = $\frac{p}{360} \times \pi R^2$. Multiplying numerator and denominator by 2:
Area = $\frac{p}{720} \times 2\pi R^2$.
**Correct Option: (D)**