NCERT Solutions for Class 10 Maths Chapter 11 Areas Related to Circles Exercise 11.1#

Using Area of sector $= \dfrac{\theta}{360°} \times \pi r^2$, we find the area:

$$= \frac{60}{360} \times \frac{22}{7} \times 6^2$$

$$= \frac{1}{6} \times \frac{22}{7} \times 36$$

$$= \frac{132}{7}$$

∴ The area of the sector is $\mathbf{\dfrac{132}{7}}$ cm² (about 18.86 cm²).

Let the radius be $r$ cm. The circumference is 22 cm, so:

$$2\pi r = 22$$

$$\Rightarrow 2 \times \frac{22}{7} \times r = 22$$

$$\Rightarrow r = \frac{7}{2}$$

A quadrant is a sector of angle 90°. Using Area of sector $= \dfrac{\theta}{360°} \times \pi r^2$, we find the area:

$$= \frac{90}{360} \times \frac{22}{7} \times \left(\frac{7}{2}\right)^2$$

$$= \frac{1}{4} \times \frac{22}{7} \times \frac{49}{4}$$

$$= \frac{77}{8}$$

∴ The area of the quadrant is $\mathbf{\dfrac{77}{8}}$ cm² (9.625 cm²).

The minute hand turns through 360° in 60 minutes, so in 5 minutes it turns through:

$$\theta = \frac{360°}{60} \times 5 = 30°$$

The area swept is a sector of radius 14 cm and angle 30°. Using Area of sector $= \dfrac{\theta}{360°} \times \pi r^2$, we find the area:

$$= \frac{30}{360} \times \frac{22}{7} \times 14^2$$

$$= \frac{1}{12} \times \frac{22}{7} \times 196$$

$$= \frac{154}{3}$$

∴ The area swept is $\mathbf{\dfrac{154}{3}}$ cm² (about 51.33 cm²).

Using Area of sector $= \dfrac{\theta}{360°} \times \pi r^2$, we find the area of sector OAB:

$$= \frac{90}{360} \times 3.14 \times 10^2$$

$$= \frac{1}{4} \times 314$$

$$= 78.5$$

△OAB is right-angled at O, so we find its area:

$$= \frac{1}{2} \times 10 \times 10$$

$$= 50$$

Area of minor segment = Area of sector OAB − Area of △OAB
$= 78.5 - 50$
$= 28.5$

∴ The area of the minor segment is 28.5 cm².

The angle of the major sector is 360° − 90° = 270°. Using Area of sector $= \dfrac{\theta}{360°} \times \pi r^2$, we find the area:

$$= \frac{270}{360} \times 3.14 \times 10^2$$

$$= \frac{3}{4} \times 314$$

$$= 235.5$$

∴ The area of the major sector is 235.5 cm².

Using Length of arc $= \dfrac{\theta}{360°} \times 2\pi r$, we find the length:

$$= \frac{60}{360} \times 2 \times \frac{22}{7} \times 21$$

$$= \frac{1}{6} \times 132$$

$$= 22$$

∴ The length of the arc is 22 cm.

Using Area of sector $= \dfrac{\theta}{360°} \times \pi r^2$, we find the area:

$$= \frac{60}{360} \times \frac{22}{7} \times 21^2$$

$$= \frac{1}{6} \times 1386$$

$$= 231$$

∴ The area of the sector is 231 cm².

Let the chord be AB and the centre O. OA = OB = 21 cm and ∠AOB = 60°, so the other two angles are each 60° and △OAB is equilateral.

Using Area of equilateral triangle $= \dfrac{\sqrt{3}}{4} \times \text{side}^2$, we find the area of △OAB:

$$= \frac{\sqrt{3}}{4} \times 21^2$$

$$= \frac{441\sqrt{3}}{4}$$

Area of segment = Area of sector − Area of △OAB
$= 231 - \dfrac{441\sqrt{3}}{4}$

∴ The area of the segment is $\mathbf{\left(231 - \dfrac{441\sqrt{3}}{4}\right)}$ cm² (about 40.04 cm²).

Let the chord be AB and the centre O. OA = OB = 15 cm and ∠AOB = 60°, so △OAB is equilateral.

Using Area of sector $= \dfrac{\theta}{360°} \times \pi r^2$, we find the area of sector OAB:

$$= \frac{60}{360} \times 3.14 \times 15^2$$

$$= \frac{1}{6} \times 706.5$$

$$= 117.75$$

Using Area of equilateral triangle $= \dfrac{\sqrt{3}}{4} \times \text{side}^2$, we find the area of △OAB:

$$= \frac{1.73}{4} \times 15^2$$

$$= \frac{389.25}{4}$$

$$= 97.3125$$

Area of minor segment = Area of sector − Area of △OAB
$= 117.75 - 97.3125$
$= 20.4375$

Area of the circle = $3.14 \times 15^2 = 706.5$

Area of major segment = Area of circle − Area of minor segment
$= 706.5 - 20.4375$
$= 686.0625$

∴ The minor segment has area 20.4375 cm² and the major segment has area 686.0625 cm².

Let the chord be AB and the centre O, with ∠AOB = 120°.

Using Area of sector $= \dfrac{\theta}{360°} \times \pi r^2$, we find the area of sector OAB:

$$= \frac{120}{360} \times 3.14 \times 12^2$$

$$= 3.14 \times 48$$

$$= 150.72$$

Area of △OAB: Draw OM ⊥ AB. △OAB is isosceles (OA = OB), so OM bisects ∠AOB and AB:
$\angle AOM = 60°$
$AB = 2AM$

In right △OMA:

$$OM = OA \cos 60° = 12 \times \frac{1}{2} = 6$$

$$AM = OA \sin 60° = 12 \times \frac{\sqrt{3}}{2} = 6\sqrt{3}$$

So $AB = 12\sqrt{3}$, and we find the area of △OAB:

$$= \frac{1}{2} \times AB \times OM$$

$$= \frac{1}{2} \times 12\sqrt{3} \times 6$$

$$= 36\sqrt{3}$$

$$= 36 \times 1.73$$

$$= 62.28$$

Area of segment = Area of sector − Area of △OAB
$= 150.72 - 62.28$
$= 88.44$

∴ The area of the segment is 88.44 cm².

The corner of the square is 90°, so the horse can graze a quadrant of radius 5 m. Using Area of sector $= \dfrac{\theta}{360°} \times \pi r^2$, we find the area:

$$= \frac{90}{360} \times 3.14 \times 5^2$$

$$= \frac{1}{4} \times 78.5$$

$$= 19.625$$

∴ The horse can graze 19.625 m².

With a 10 m rope, the horse can graze a quadrant of radius 10 m. We find its area:

$$= \frac{90}{360} \times 3.14 \times 10^2$$

$$= \frac{1}{4} \times 314$$

$$= 78.5$$

Increase in grazing area = 78.5 − 19.625
Increase in grazing area = 58.875

∴ The grazing area increases by 58.875 m².

Circumference of the circle = $\pi d$
Circumference of the circle = $\dfrac{22}{7} \times 35 = 110$ mm

Length of 5 diameters = $5 \times 35 = 175$ mm

Total length of wire = 110 + 175
Total length of wire = 285 mm

∴ The total length of silver wire required is 285 mm.

The 10 sectors are equal, so each has angle $\dfrac{360°}{10} = 36°$. The radius is $\dfrac{35}{2}$ mm.

Using Area of sector $= \dfrac{\theta}{360°} \times \pi r^2$, we find the area of each sector:

$$= \frac{36}{360} \times \frac{22}{7} \times \left(\frac{35}{2}\right)^2$$

$$= \frac{1}{10} \times \frac{22}{7} \times \frac{1225}{4}$$

$$= \frac{385}{4}$$

∴ The area of each sector is $\mathbf{\dfrac{385}{4}}$ mm² (96.25 mm²).

The 8 ribs are equally spaced, so the angle between two consecutive ribs is $\dfrac{360°}{8} = 45°$.

Using Area of sector $= \dfrac{\theta}{360°} \times \pi r^2$, we find the area:

$$= \frac{45}{360} \times \frac{22}{7} \times 45^2$$

$$= \frac{1}{8} \times \frac{22}{7} \times 2025$$

$$= \frac{22275}{28}$$

∴ The area between two consecutive ribs is $\mathbf{\dfrac{22275}{28}}$ cm² (about 795.54 cm²).

Each wiper sweeps a sector of radius 25 cm and angle 115°. The wipers do not overlap, so the total area is twice the area of one sector.

We find the total area:

$$= 2 \times \frac{115}{360} \times \frac{22}{7} \times 25^2$$

$$= \frac{115 \times 22 \times 625}{180 \times 7}$$

$$= \frac{158125}{126}$$

∴ The total area cleaned at each sweep is $\mathbf{\dfrac{158125}{126}}$ cm² (about 1254.96 cm²).

The warned area is a sector of radius 16.5 km and angle 80°. Using Area of sector $= \dfrac{\theta}{360°} \times \pi r^2$, we find the area:

$$= \frac{80}{360} \times 3.14 \times 16.5^2$$

$$= \frac{2}{9} \times 3.14 \times 272.25$$

$$= \frac{1709.73}{9}$$

$$= 189.97$$

∴ The ships are warned over an area of about 189.97 km².

Each design is a segment whose chord subtends $\dfrac{360°}{6} = 60°$ at the centre. The radii are 28 cm, so each triangle formed with the centre is equilateral.

Using Area of sector $= \dfrac{\theta}{360°} \times \pi r^2$, we find the area of one sector:

$$= \frac{60}{360} \times \frac{22}{7} \times 28^2$$

$$= \frac{1232}{3}$$

$$= 410.67$$

Using Area of equilateral triangle $= \dfrac{\sqrt{3}}{4} \times \text{side}^2$, we find the area of one triangle:

$$= \frac{1.7}{4} \times 28^2$$

$$= 333.2$$

Area of one design = 410.67 − 333.2
Area of one design = 77.47

Area of six designs = $6 \times 77.47 = 464.8$ (about)

Cost of making the designs = $464.8 \times 0.35$
Cost of making the designs = ₹162.68

∴ The cost of making the designs is ₹162.68.

The area of a sector of angle $p$ is $\dfrac{p}{360} \times \pi R^2$.

We simplify option (D), $\dfrac{p}{720} \times 2\pi R^2$:

$$= \frac{2p\pi R^2}{720}$$

$$= \frac{p}{360} \times \pi R^2$$$$\Rightarrow \frac{p}{720} \times 2\pi R^2 = \frac{p}{360} \times \pi R^2$$

∴ Answer: (D) $\dfrac{p}{720} \times 2\pi R^2$