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Mathematics · Areas Related to Circles · NCERT Exercises

Exercise 11.1

Complete, independently verified solutions for NCERT Exercise 11.1.

Mathematics · Chapter 11 · NCERT Exercises

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Question

Question 1

area of sector

Find the area of a sector of a circle with radius 6 cm if the angle of the sector is 60°.

Solution

Board-exam working:

Given: r = 6 cm and θ = 60°.

A=θ360∘×πr2A=\frac{\theta}{360^\circ}\times\pi r^2
=60∘360∘×227×62=\frac{60^\circ}{360^\circ}\times\frac{22}{7}\times6^2
=16×227×36=1327 cm2=\frac16\times\frac{22}{7}\times36=\frac{132}{7}\text{ cm}^2

Final answer

1327 cm2\boxed{\frac{132}{7}\text{ cm}^2}

Question

Question 2

quadrant · circumference

Find the area of a quadrant of a circle whose circumference is 22 cm.

Solution

Board-exam working:

Let the radius be r cm.

2πr=222\pi r=22
2×227×r=222\times\frac{22}{7}\times r=22
r=72=3.5 cmr=\frac72=3.5\text{ cm}

A quadrant is one-fourth of a circle.

A=14πr2A=\frac14\pi r^2
=14×227×(72)2=\frac14\times\frac{22}{7}\times\left(\frac72\right)^2
=778 cm2=9.625 cm2=\frac{77}{8}\text{ cm}^2=9.625\text{ cm}^2

Final answer

778 cm2\boxed{\frac{77}{8}\text{ cm}^2}

Question

Question 3

clock · area swept

The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.

Solution

Board-exam working:

The minute hand turns through 360° in 60 minutes.

θ=560×360∘=30∘\theta=\frac{5}{60}\times360^\circ=30^\circ

The swept area is a sector of radius 14 cm.

A=30∘360∘×227×142A=\frac{30^\circ}{360^\circ}\times\frac{22}{7}\times14^2
=112×227×196=\frac1{12}\times\frac{22}{7}\times196
=1543 cm2≈51.33 cm2=\frac{154}{3}\text{ cm}^2\approx51.33\text{ cm}^2

Final answer

1543 cm2≈51.33 cm2\boxed{\frac{154}{3}\text{ cm}^2\approx51.33\text{ cm}^2}

Question

Question 4

minor segment · major sector

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 π = 3.14.

Solution

Board-exam working:

Given: r = 10 cm and θ = 90°.

(i) Area of the 90° sector:

Asector=90∘360∘×3.14×102=78.5 cm2A_{sector}=\frac{90^\circ}{360^\circ}\times3.14\times10^2=78.5\text{ cm}^2

The two radii form a right triangle whose perpendicular sides are 10 cm each.

Atriangle=12×10×10=50 cm2A_{triangle}=\frac12\times10\times10=50\text{ cm}^2
Aminorsegment=78.5−50=28.5 cm2A_{minor segment}=78.5-50=28.5\text{ cm}^2

(ii) The angle of the major sector is:

360∘−90∘=270∘360^\circ-90^\circ=270^\circ
Amajorsector=270∘360∘×3.14×102A_{major sector}=\frac{270^\circ}{360^\circ}\times3.14\times10^2
=235.5 cm2=235.5\text{ cm}^2

Final answer

(i) 28.5 cm2,(ii) 235.5 cm2\text{(i) }28.5\text{ cm}^2,\qquad\text{(ii) }235.5\text{ cm}^2

Question

Question 5

arc length · sector · segment

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) the area of the sector formed by the arc, (iii) the area of the segment formed by the corresponding chord.

Solution

Board-exam working:

Given: r = 21 cm and θ = 60°.

(i) Length of the arc:

l=60∘360∘×2×227×21l=\frac{60^\circ}{360^\circ}\times2\times\frac{22}{7}\times21
=16×132=22 cm=\frac16\times132=22\text{ cm}

(ii) Area of the sector:

Asector=60∘360∘×227×212A_{sector}=\frac{60^\circ}{360^\circ}\times\frac{22}{7}\times21^2
=16×1386=231 cm2=\frac16\times1386=231\text{ cm}^2

(iii) The triangle formed by the two radii and chord has all angles 60°, so it is equilateral with side 21 cm.

Atriangle=34×212=44134 cm2A_{triangle}=\frac{\sqrt3}{4}\times21^2=\frac{441\sqrt3}{4}\text{ cm}^2
Asegment=Asector−AtriangleA_{segment}=A_{sector}-A_{triangle}
=231−44134 cm2=231-\frac{441\sqrt3}{4}\text{ cm}^2

Final answer

(i) 22 cm,(ii) 231 cm2,(iii) 231−44134 cm2\text{(i) }22\text{ cm},\quad\text{(ii) }231\text{ cm}^2,\quad\text{(iii) }231-\frac{441\sqrt3}{4}\text{ cm}^2

Question

Question 6

minor segment · major segment

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. Use π = 3.14 and √3 = 1.73.

Solution

Board-exam working:

Given: r = 15 cm and θ = 60°.

Area of the 60° sector:

Asector=60∘360∘×3.14×152A_{sector}=\frac{60^\circ}{360^\circ}\times3.14\times15^2
=16×3.14×225=117.75 cm2=\frac16\times3.14\times225=117.75\text{ cm}^2

The triangle formed is equilateral with side 15 cm.

Atriangle=34×152A_{triangle}=\frac{\sqrt3}{4}\times15^2
=1.734×225=97.3125 cm2=\frac{1.73}{4}\times225=97.3125\text{ cm}^2

Area of the minor segment:

Aminor=117.75−97.3125=20.4375 cm2A_{minor}=117.75-97.3125=20.4375\text{ cm}^2
Aminor≈20.44 cm2A_{minor}\approx20.44\text{ cm}^2

Area of the circle:

Acircle=3.14×152=706.5 cm2A_{circle}=3.14\times15^2=706.5\text{ cm}^2

Area of the major segment:

Amajor=706.5−20.4375=686.0625 cm2A_{major}=706.5-20.4375=686.0625\text{ cm}^2
Amajor≈686.06 cm2A_{major}\approx686.06\text{ cm}^2

Final answer

Aminor≈20.44 cm2,Amajor≈686.06 cm2A_{minor}\approx20.44\text{ cm}^2,\qquad A_{major}\approx686.06\text{ cm}^2

Question

Question 7

area of segment · 120 degree sector

A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment. Use π = 3.14 and √3 = 1.73.

Solution

Board-exam working:

Given: r = 12 cm and θ = 120°.

Area of the 120° sector:

Asector=120∘360∘×3.14×122A_{sector}=\frac{120^\circ}{360^\circ}\times3.14\times12^2
=13×3.14×144=150.72 cm2=\frac13\times3.14\times144=150.72\text{ cm}^2

Drop a perpendicular from the centre to the chord. The triangle splits into two congruent right triangles.

Atriangle=12r2sin⁡120∘A_{triangle}=\frac12r^2\sin120^\circ
=12×122×32=\frac12\times12^2\times\frac{\sqrt3}{2}
=363=36×1.73=62.28 cm2=36\sqrt3=36\times1.73=62.28\text{ cm}^2

Therefore, area of the segment:

Asegment=150.72−62.28=88.44 cm2A_{segment}=150.72-62.28=88.44\text{ cm}^2

Final answer

88.44 cm2\boxed{88.44\text{ cm}^2}

Question

Question 8

grazing area · quadrant

A horse is tied to a peg at one corner of a square-shaped grass field of side 15 m by a 5 m long rope, as shown in Figure 11.8. Find: (i) the area of the part of the field in which the horse can graze, (ii) the increase in grazing area if the rope were 10 m long instead of 5 m. Use π = 3.14.

Horse tied at a corner of a square grass field, showing the quarter-circular grazing region.
Figure 11-8 from the supplied NCERT chapter.

Solution

Board-exam working:

The peg is at a corner, so the grazing region is a quadrant.

(i) For a 5 m rope:

A5=14πr2A_5=\frac14\pi r^2
=14×3.14×52=19.625 m2=\frac14\times3.14\times5^2=19.625\text{ m}^2

(ii) For a 10 m rope:

A10=14×3.14×102=78.5 m2A_{10}=\frac14\times3.14\times10^2=78.5\text{ m}^2

Increase in grazing area:

A10−A5=78.5−19.625=58.875 m2A_{10}-A_5=78.5-19.625=58.875\text{ m}^2

Final answer

(i) 19.625 m2,(ii) 58.875 m2\text{(i) }19.625\text{ m}^2,\qquad\text{(ii) }58.875\text{ m}^2

Question

Question 9

brooch · circumference · equal sectors

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

Circular brooch divided by five diameters into ten equal sectors.
Figure 11-9 from the supplied NCERT chapter.

Solution

Board-exam working:

Given: diameter d = 35 mm, so radius r = 17.5 mm.

(i) Length of wire used for the circumference:

C=πd=227×35=110 mmC=\pi d=\frac{22}{7}\times35=110\text{ mm}

Length of wire used for 5 diameters:

5d=5×35=175 mm5d=5\times35=175\text{ mm}

Total wire required:

110+175=285 mm110+175=285\text{ mm}

(ii) Area of the circle:

A=πr2=227×17.52=962.5 mm2A=\pi r^2=\frac{22}{7}\times17.5^2=962.5\text{ mm}^2

The 5 diameters form 10 equal sectors.

Aonesector=962.510=96.25 mm2A_{one sector}=\frac{962.5}{10}=96.25\text{ mm}^2

Final answer

(i) 285 mm,(ii) 96.25 mm2\text{(i) }285\text{ mm},\qquad\text{(ii) }96.25\text{ mm}^2

Question

Question 10

umbrella · equal sectors

An umbrella has 8 ribs which are equally spaced, as shown in Figure 11.10. Assuming the umbrella to be a flat circle of radius 45 cm, find the area between two consecutive ribs.

Umbrella and its top view showing eight equally spaced ribs.
Figure 11-10 from the supplied NCERT chapter.

Solution

Board-exam working:

Eight equally spaced ribs divide the circle into 8 equal sectors.

θ=360∘8=45∘\theta=\frac{360^\circ}{8}=45^\circ

Area between two consecutive ribs:

A=45∘360∘×227×452A=\frac{45^\circ}{360^\circ}\times\frac{22}{7}\times45^2
=18×227×2025=\frac18\times\frac{22}{7}\times2025
=2227528 cm2=\frac{22275}{28}\text{ cm}^2
≈795.54 cm2\approx795.54\text{ cm}^2

Final answer

2227528 cm2≈795.54 cm2\boxed{\frac{22275}{28}\text{ cm}^2\approx795.54\text{ cm}^2}

Question

Question 11

car wipers · sector area

A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.

Solution

Board-exam working:

Each wiper cleans a sector of radius 25 cm and angle 115°.

Area cleaned by one wiper:

A1=115∘360∘×227×252A_1=\frac{115^\circ}{360^\circ}\times\frac{22}{7}\times25^2

Since the two swept regions do not overlap:

Atotal=2A1A_{total}=2A_1
=2×115360×227×625=2\times\frac{115}{360}\times\frac{22}{7}\times625
=158125126 cm2=\frac{158125}{126}\text{ cm}^2
≈1254.96 cm2\approx1254.96\text{ cm}^2

Final answer

158125126 cm2≈1254.96 cm2\boxed{\frac{158125}{126}\text{ cm}^2\approx1254.96\text{ cm}^2}

Question

Question 12

lighthouse · sector area

To warn ships of underwater rocks, a lighthouse spreads a red-coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned. Use π = 3.14.

Solution

Board-exam working:

The warned region is a sector with radius 16.5 km and angle 80°.

A=80∘360∘×3.14×16.52A=\frac{80^\circ}{360^\circ}\times3.14\times16.5^2
=29×3.14×272.25=\frac29\times3.14\times272.25
=189.97 km2=189.97\text{ km}^2

Final answer

189.97 km2\boxed{189.97\text{ km}^2}

Question

Question 13

table cover · regular hexagon · cost

A round table cover has six equal designs as shown in Figure 11.11. If the radius of the cover is 28 cm, find the cost of making the designs at ₹0.35 per cm². Use √3 = 1.7.

Round table cover with a regular hexagon and six equal decorative circular segments.
Figure 11-11 from the supplied NCERT chapter.

Solution

Board-exam working:

The six radii form a regular hexagon inside the circle. Each of its six triangles is equilateral with side 28 cm.

Area of the circle:

Acircle=πr2=227×282=2464 cm2A_{circle}=\pi r^2=\frac{22}{7}\times28^2=2464\text{ cm}^2

Area of one equilateral triangle:

A1=34a2A_1=\frac{\sqrt3}{4}a^2
=1.74×282=333.2 cm2=\frac{1.7}{4}\times28^2=333.2\text{ cm}^2

Area of the regular hexagon:

Ahexagon=6×333.2=1999.2 cm2A_{hexagon}=6\times333.2=1999.2\text{ cm}^2

Total area of the six designs:

Adesigns=2464−1999.2=464.8 cm2A_{designs}=2464-1999.2=464.8\text{ cm}^2

Cost at ₹0.35 per cm²:

Cost=464.8×0.35=Rs. 162.68Cost=464.8\times0.35=\text{Rs. }162.68

Final answer

₹162.68

Question

Question 14

multiple choice · area of sector

Tick the correct answer. The area of a sector of angle p (in degrees) of a circle with radius R is: (A) p/180 × 2πR, (B) p/180 × πR², (C) p/360 × 2πR, (D) p/720 × 2πR².

Solution

Board-exam working:

The area of a full circle is πR² and its angle is 360°.

Therefore, the area of a sector of angle p is:

A=p360×πR2A=\frac{p}{360}\times\pi R^2

Rewrite this to match the options:

p360×πR2=p720×2πR2\frac{p}{360}\times\pi R^2=\frac{p}{720}\times2\pi R^2

This is option (D).

Final answer

Option (D): p720×2πR2\boxed{\text{Option (D): }\frac{p}{720}\times2\pi R^2}