Areas Related to Circles — CBSE Class 10 Maths Textbook Solutions

For a sector of radius $r$r and angle $\theta$: $\text{Area} = \dfrac{\theta}{360} \times \pi r^2$Area = /360 × π r^2.

Here $r = 6$r = 6 cm, $\theta = 60^\circ$= 60^, and $\pi = \dfrac{22}{7}$π = 22/7:

$\text{Area} = \frac{60}{360} \times \frac{22}{7} \times 6^2 = \frac{1}{6} \times \frac{22}{7} \times 36$Area = 60/360 × 22/7 × 6^2 = 1/6 × 22/7 × 36
$= \frac{22 \times 36}{7 \times 6} = \frac{792}{42} = \frac{132}{7}\text{ cm}^2 \approx 18.86\text{ cm}^2$= 22 × 36/7 × 6 = 792/42 = 132/7 cm^2 18.86 cm^2

The area of the sector is $\dfrac{132}{7}\text{ cm}^2$132/7 cm^2.

A quadrant is a sector with angle $90^\circ$90^ (a quarter of the circle).

First find the radius from the circumference $= 2\pi r = 22$= 2π r = 22 cm:

$2 \times \frac{22}{7} \times r = 22 \Rightarrow r = 22 \times \frac{7}{22 \times 2} = \frac{7}{2} = 3.5\text{ cm}$2 × 22/7 × r = 22 r = 22 × 7/22 × 2 = 7/2 = 3.5 cm

Now, area of the quadrant:

$\text{Area} = \frac{90}{360} \times \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times (3.5)^2$Area = 90/360 × π r^2 = 1/4 × 22/7 × (3.5)^2
$= \frac{1}{4} \times \frac{22}{7} \times 12.25 = \frac{22 \times 12.25}{28} = \frac{269.5}{28} = \frac{77}{8}\text{ cm}^2$= 1/4 × 22/7 × 12.25 = 22 × 12.25/28 = 269.5/28 = 77/8 cm^2

The area of the quadrant is $\dfrac{77}{8}\text{ cm}^2 = 9.625\text{ cm}^2$77/8 cm^2 = 9.625 cm^2.

In $60$60 minutes, the minute hand sweeps a full circle, i.e., $360^\circ$360^. So in $1$1 minute it sweeps $\dfrac{360^\circ}{60} = 6^\circ$360^/60 = 6^.

In $5$5 minutes, the angle swept:

$\theta = 5 \times 6^\circ = 30^\circ$= 5 × 6^ = 30^

The minute hand acts as the radius, $r = 14$r = 14 cm, so the area swept is the area of a sector of angle $30^\circ$30^:

$\text{Area} = \frac{30}{360} \times \frac{22}{7} \times 14^2 = \frac{1}{12} \times \frac{22}{7} \times 196$Area = 30/360 × 22/7 × 14^2 = 1/12 × 22/7 × 196
$= \frac{22 \times 196}{7 \times 12} = \frac{22 \times 28}{12} = \frac{616}{12} = \frac{154}{3}\text{ cm}^2$= 22 × 196/7 × 12 = 22 × 28/12 = 616/12 = 154/3 cm^2

The area swept by the minute hand in 5 minutes is $\dfrac{154}{3}\text{ cm}^2 \approx 51.33\text{ cm}^2$154/3 cm^2 51.33 cm^2.

Let $O$O be the centre, radius $r = 10$r = 10 cm, and $AB$AB the chord subtending $\angle AOB = 90^\circ$AOB = 90^ at the centre.

(i) Minor segment

$\text{Area of sector } OAB = \frac{90}{360} \times \pi r^2 = \frac{1}{4} \times 3.14 \times 100 = 78.5\text{ cm}^2$Area of sector OAB = 90/360 × π r^2 = 1/4 × 3.14 × 100 = 78.5 cm^2

Since $\angle AOB = 90^\circ$AOB = 90^ and $OA = OB = r$OA = OB = r, $\triangle OAB$OAB is a right triangle with both legs equal to $r = 10$r = 10 cm:

$\text{Area of } \triangle OAB = \frac{1}{2} \times OA \times OB = \frac{1}{2} \times 10 \times 10 = 50\text{ cm}^2$Area of OAB = 1/2 × OA × OB = 1/2 × 10 × 10 = 50 cm^2

$\text{Area of minor segment} = \text{Area of sector} - \text{Area of } \triangle OAB = 78.5 - 50 = 28.5\text{ cm}^2$Area of minor segment = Area of sector - Area of OAB = 78.5 - 50 = 28.5 cm^2

(ii) Major sector

$\text{Area of major sector} = \pi r^2 - \text{Area of minor sector} = (3.14 \times 100) - 78.5 = 314 - 78.5 = 235.5\text{ cm}^2$Area of major sector = π r^2 - Area of minor sector = (3.14 × 100) - 78.5 = 314 - 78.5 = 235.5 cm^2

So the minor segment is $28.5\text{ cm}^2$28.5 cm^2 and the major sector is $235.5\text{ cm}^2$235.5 cm^2.

Radius $r = 21$r = 21 cm, angle $\theta = 60^\circ$= 60^, $\pi = \dfrac{22}{7}$π = 22/7.

(i) Length of the arc

$\text{Arc length} = \frac{\theta}{360} \times 2\pi r = \frac{60}{360} \times 2 \times \frac{22}{7} \times 21 = \frac{1}{6} \times 2 \times 22 \times 3 = \frac{132}{6} = 22\text{ cm}$Arc length = /360 × 2π r = 60/360 × 2 × 22/7 × 21 = 1/6 × 2 × 22 × 3 = 132/6 = 22 cm

(ii) Area of the sector

$\text{Area of sector} = \frac{60}{360} \times \frac{22}{7} \times 21^2 = \frac{1}{6} \times \frac{22}{7} \times 441 = \frac{1}{6} \times 22 \times 63 = \frac{1386}{6} = 231\text{ cm}^2$Area of sector = 60/360 × 22/7 × 21^2 = 1/6 × 22/7 × 441 = 1/6 × 22 × 63 = 1386/6 = 231 cm^2

(iii) Area of the segment

Since $OA = OB = 21$OA = OB = 21 cm and $\angle AOB = 60^\circ$AOB = 60^, $\triangle OAB$OAB is isosceles with a $60^\circ$60^ angle between the equal sides, so it is in fact equilateral (all angles $60^\circ$60^). Its area:

$\text{Area of } \triangle OAB = \frac{\sqrt{3}}{4} \times 21^2 = \frac{441\sqrt{3}}{4}\text{ cm}^2 \approx 190.97\text{ cm}^2$Area of OAB = √3/4 × 21^2 = 441√3/4 cm^2 190.97 cm^2

$\text{Area of segment} = \text{Area of sector} - \text{Area of } \triangle OAB = 231 - \frac{441\sqrt{3}}{4}\text{ cm}^2 \approx 231 - 190.97 = 40.03\text{ cm}^2$Area of segment = Area of sector - Area of OAB = 231 - 441√3/4 cm^2 231 - 190.97 = 40.03 cm^2

So: arc length $= 22$= 22 cm, sector area $= 231\text{ cm}^2$= 231 cm^2, segment area $= \left(231 - \dfrac{441\sqrt{3}}{4}\right)\text{cm}^2 \approx 40.03\text{ cm}^2$= (231 - 441√3/4)cm^2 40.03 cm^2.

Radius $r = 15$r = 15 cm, $\theta = 60^\circ$= 60^.

$\text{Area of sector} = \frac{60}{360} \times \pi r^2 = \frac{1}{6} \times 3.14 \times 225 = \frac{706.5}{6} = 117.75\text{ cm}^2$Area of sector = 60/360 × π r^2 = 1/6 × 3.14 × 225 = 706.5/6 = 117.75 cm^2

Since $\angle AOB = 60^\circ$AOB = 60^ and $OA = OB = 15$OA = OB = 15 cm, $\triangle OAB$OAB is equilateral with side $15$15 cm:

$\text{Area of } \triangle OAB = \frac{\sqrt{3}}{4} \times 15^2 = \frac{1.73}{4} \times 225 = \frac{389.25}{4} = 97.3125\text{ cm}^2$Area of OAB = √3/4 × 15^2 = 1.73/4 × 225 = 389.25/4 = 97.3125 cm^2

Minor segment:

$\text{Area} = 117.75 - 97.3125 = 20.4375\text{ cm}^2$Area = 117.75 - 97.3125 = 20.4375 cm^2

Major segment:

$\text{Area} = \pi r^2 - \text{minor segment} = (3.14 \times 225) - 20.4375 = 706.5 - 20.4375 = 686.0625\text{ cm}^2$Area = π r^2 - minor segment = (3.14 × 225) - 20.4375 = 706.5 - 20.4375 = 686.0625 cm^2

So the minor segment is $20.4375\text{ cm}^2$20.4375 cm^2 and the major segment is $686.0625\text{ cm}^2$686.0625 cm^2.

Radius $r = 12$r = 12 cm, $\theta = 120^\circ$= 120^.

$\text{Area of sector} = \frac{120}{360} \times \pi r^2 = \frac{1}{3} \times 3.14 \times 144 = \frac{452.16}{3} = 150.72\text{ cm}^2$Area of sector = 120/360 × π r^2 = 1/3 × 3.14 × 144 = 452.16/3 = 150.72 cm^2

$\triangle OAB$OAB has $OA = OB = 12$OA = OB = 12 cm with included angle $120^\circ$120^, so:

$\text{Area of } \triangle OAB = \frac{1}{2} \times OA \times OB \times \sin 120^\circ = \frac{1}{2} \times 12 \times 12 \times \frac{\sqrt{3}}{2}$Area of OAB = 1/2 × OA × OB × 120^ = 1/2 × 12 × 12 × √3/2
$= 36\sqrt{3} = 36 \times 1.73 = 62.28\text{ cm}^2$= 36√3 = 36 × 1.73 = 62.28 cm^2

$\text{Area of segment} = 150.72 - 62.28 = 88.44\text{ cm}^2$Area of segment = 150.72 - 62.28 = 88.44 cm^2

The area of the segment is $88.44\text{ cm}^2$88.44 cm^2.

The peg is at a corner of the square field, so the interior angle of the square there is $90^\circ$90^. The horse can graze within a sector of angle $90^\circ$90^ and radius equal to the rope length (as long as the rope is shorter than the side of the field, which it is in both parts).

(i) Rope $= 5$= 5 m

$\text{Grazing area} = \frac{90}{360} \times \pi r^2 = \frac{1}{4} \times 3.14 \times 5^2 = \frac{1}{4} \times 3.14 \times 25 = \frac{78.5}{4} = 19.625\text{ m}^2$Grazing area = 90/360 × π r^2 = 1/4 × 3.14 × 5^2 = 1/4 × 3.14 × 25 = 78.5/4 = 19.625 m^2

(ii) Rope $= 10$= 10 m

$\text{Grazing area} = \frac{1}{4} \times 3.14 \times 10^2 = \frac{1}{4} \times 3.14 \times 100 = \frac{314}{4} = 78.5\text{ m}^2$Grazing area = 1/4 × 3.14 × 10^2 = 1/4 × 3.14 × 100 = 314/4 = 78.5 m^2

Increase in grazing area:

$78.5 - 19.625 = 58.875\text{ m}^2$78.5 - 19.625 = 58.875 m^2

So the horse can graze $19.625\text{ m}^2$19.625 m^2 with the 5 m rope, and the grazing area increases by $58.875\text{ m}^2$58.875 m^2 when the rope is lengthened to 10 m.

Diameter $= 35$= 35 mm, so radius $r = 17.5$r = 17.5 mm.

(i) Total length of silver wire

The wire forms the circumference plus the 5 diameters.

$\text{Circumference} = \pi d = \frac{22}{7} \times 35 = 110\text{ mm}$Circumference = π d = 22/7 × 35 = 110 mm

$\text{Length of 5 diameters} = 5 \times 35 = 175\text{ mm}$Length of 5 diameters = 5 × 35 = 175 mm

$\text{Total wire} = 110 + 175 = 285\text{ mm}$Total wire = 110 + 175 = 285 mm

(ii) Area of each sector

The 5 diameters (10 radii) divide the circle into 10 equal sectors, so each sector has angle:

$\theta = \frac{360^\circ}{10} = 36^\circ$= 360^/10 = 36^

$\text{Area of each sector} = \frac{36}{360} \times \pi r^2 = \frac{1}{10} \times \frac{22}{7} \times (17.5)^2$Area of each sector = 36/360 × π r^2 = 1/10 × 22/7 × (17.5)^2
$= \frac{1}{10} \times \frac{22}{7} \times 306.25 = \frac{22 \times 306.25}{70} = \frac{6737.5}{70} = \frac{385}{4}\text{ mm}^2$= 1/10 × 22/7 × 306.25 = 22 × 306.25/70 = 6737.5/70 = 385/4 mm^2

So the total wire needed is $285$285 mm, and each sector has area $\dfrac{385}{4}\text{ mm}^2 = 96.25\text{ mm}^2$385/4 mm^2 = 96.25 mm^2.

There are 8 equally spaced ribs, so they divide the circle into 8 equal sectors, each of angle:

$\theta = \frac{360^\circ}{8} = 45^\circ$= 360^/8 = 45^

With radius $r = 45$r = 45 cm:

$\text{Area between two consecutive ribs} = \frac{45}{360} \times \pi r^2 = \frac{1}{8} \times \frac{22}{7} \times 45^2$Area between two consecutive ribs = 45/360 × π r^2 = 1/8 × 22/7 × 45^2
$= \frac{1}{8} \times \frac{22}{7} \times 2025 = \frac{22 \times 2025}{56} = \frac{44550}{56} = \frac{22275}{28}\text{ cm}^2$= 1/8 × 22/7 × 2025 = 22 × 2025/56 = 44550/56 = 22275/28 cm^2

The area between two consecutive ribs is $\dfrac{22275}{28}\text{ cm}^2 \approx 795.54\text{ cm}^2$22275/28 cm^2 795.54 cm^2.

Each wiper blade acts as the radius of a sector, $r = 25$r = 25 cm, sweeping angle $\theta = 115^\circ$= 115^.

$\text{Area cleaned by one wiper} = \frac{115}{360} \times \frac{22}{7} \times 25^2 = \frac{115}{360} \times \frac{22}{7} \times 625$Area cleaned by one wiper = 115/360 × 22/7 × 25^2 = 115/360 × 22/7 × 625

$= \frac{115 \times 22 \times 625}{360 \times 7} = \frac{1{,}581{,}250}{2520} = \frac{158125}{252}\text{ cm}^2$= 115 × 22 × 625/360 × 7 = 1,581,2502520 = 158125/252 cm^2

Since the two wipers do not overlap, the total area cleaned by both wipers is twice this:

$\text{Total area} = 2 \times \frac{158125}{252} = \frac{158125}{126}\text{ cm}^2 \approx 1254.96\text{ cm}^2$Total area = 2 × 158125/252 = 158125/126 cm^2 1254.96 cm^2

The total area cleaned at each sweep of the two blades is $\dfrac{158125}{126}\text{ cm}^2 \approx 1254.96\text{ cm}^2$158125/126 cm^2 1254.96 cm^2.

The light spreads over a sector of radius $r = 16.5$r = 16.5 km and angle $\theta = 80^\circ$= 80^.

$\text{Area} = \frac{80}{360} \times \pi r^2 = \frac{2}{9} \times 3.14 \times (16.5)^2$Area = 80/360 × π r^2 = 2/9 × 3.14 × (16.5)^2

$(16.5)^2 = 272.25$(16.5)^2 = 272.25

$\text{Area} = \frac{2}{9} \times 3.14 \times 272.25 = \frac{2 \times 854.865}{9} = \frac{1709.73}{9} = 189.97\text{ km}^2$Area = 2/9 × 3.14 × 272.25 = 2 × 854.865/9 = 1709.73/9 = 189.97 km^2

The area of the sea over which the ships are warned is $189.97\text{ km}^2$189.97 km^2.

Six equal designs mean the circle is divided into 6 equal sectors, each of angle:

$\theta = \frac{360^\circ}{6} = 60^\circ$= 360^/6 = 60^

Radius $r = 28$r = 28 cm, $\pi = \dfrac{22}{7}$π = 22/7.

$\text{Area of one sector} = \frac{60}{360} \times \frac{22}{7} \times 28^2 = \frac{1}{6} \times \frac{22}{7} \times 784 = \frac{17248}{42} = \frac{2464}{6} = 410.6\overline{6}\text{ cm}^2$Area of one sector = 60/360 × 22/7 × 28^2 = 1/6 × 22/7 × 784 = 17248/42 = 2464/6 = 410.66 cm^2

Since $\angle AOB = 60^\circ$AOB = 60^ with $OA = OB = 28$OA = OB = 28 cm, the triangle in each sector is equilateral with side $28$28 cm:

$\text{Area of triangle} = \frac{\sqrt{3}}{4} \times 28^2 = \frac{1.7}{4} \times 784 = \frac{1332.8}{4} = 333.2\text{ cm}^2$Area of triangle = √3/4 × 28^2 = 1.7/4 × 784 = 1332.8/4 = 333.2 cm^2

$\text{Area of one design (segment)} = 410.6\overline{6} - 333.2 = 77.4\overline{6}\text{ cm}^2$Area of one design (segment) = 410.66 - 333.2 = 77.46 cm^2

For all six designs:

$\text{Total design area} = 6 \times 77.4\overline{6} = 464.8\text{ cm}^2$Total design area = 6 × 77.46 = 464.8 cm^2

Cost of making the designs at Rs 0.35 per $\text{cm}^2$cm^2:

$\text{Cost} = 464.8 \times 0.35 = \text{Rs } 162.68$Cost = 464.8 × 0.35 = Rs 162.68

The cost of making the six designs is Rs 162.68.

The standard formula for the area of a sector of angle $\theta$ in a circle of radius $R$R is:

$\text{Area} = \frac{\theta}{360} \times \pi R^2$Area = /360 × π R^2

Here the angle is called $p$p, so the correct expression is $\dfrac{p}{360} \times \pi R^2$p/360 × π R^2.

Check option (D): $\dfrac{p}{720} \times 2\pi R^2 = \dfrac{2p}{720} \times \pi R^2 = \dfrac{p}{360} \times \pi R^2$p/720 × 2π R^2 = 2p/720 × π R^2 = p/360 × π R^2 — this matches the standard formula exactly (the factor of $2$2 in the numerator cancels with the $720$720 in the denominator to give $360$360).

None of options (A), (B), (C) reduce to $\dfrac{p}{360}\pi R^2$p/360π R^2, so they are incorrect.

Answer: (D) $\dfrac{p}{720} \times 2\pi R^2$p/720 × 2π R^2.