MATHEMATICS: AREAS RELATED TO CIRCLES
Class X · Unit Test (Sectors, Segments, Arc Length)
- All questions are compulsory. The paper has five sections: A (10 × 1 = 10), B (5 × 2 = 10), C (5 × 4 = 20), D (4 × 5 = 20), E (2 case studies × 10 = 20).
- Use $\pi=\frac{22}{7}$ unless stated otherwise. Diagrams are not to scale.
- Show all steps. Marks are awarded for method.
SECTION A: Multiple choice (1 mark each)
1.
The length of the arc of a sector of angle $90^\circ$ in a circle of radius $14$ cm is
[1](A) 11 cm(B) 22 cm(C) 44 cm(D) 154 cm
2.
The area of a quadrant of a circle of radius $7$ cm is
[1](A) 38.5 cm²(B) 77 cm²(C) 154 cm²(D) 19.25 cm²
3.
The angle swept by a minute hand in $20$ minutes is
[1](A) 60°(B) 90°(C) 120°(D) 180°
4.
If the angle of a minor sector is $70^\circ$, the angle of the corresponding major sector is
[1](A) 110°(B) 290°(C) 250°(D) 70°
5.
A sector has area equal to $\frac16$ of the circle. Its central angle is
[1](A) 30°(B) 45°(C) 60°(D) 90°
6.
The perimeter of a quadrant of a circle of radius $14$ cm is
[1](A) 22 cm(B) 36 cm(C) 50 cm(D) 64 cm
7.
A chord of a circle of radius $4$ cm subtends $60^\circ$ at the centre. The area of $\triangle OAB$ is
[1](A) $4\sqrt3$ cm²(B) $8\sqrt3$ cm²(C) $16\sqrt3$ cm²(D) $2\sqrt3$ cm²
8.
In the figure, the shaded region (minor segment) has area equal to
[1](A) sector − $\triangle OAB$(B) sector + $\triangle OAB$(C) $\triangle OAB$ − sector(D) $\pi r^2$ − sector
9.
A sector of a circle of radius $7$ cm has an arc of length $22$ cm. Its area is
[1](A) 38.5 cm²(B) 77 cm²(C) 154 cm²(D) 11 cm²
10.
Assertion (A): A semicircular region of radius $r$ has area $\frac12\pi r^2$.
Reason (R): Area of a sector of angle $\theta$ is $\frac{\theta}{360}\pi r^2$.
[1]Reason (R): Area of a sector of angle $\theta$ is $\frac{\theta}{360}\pi r^2$.
(A) Both A and R are true and R explains A.(B) Both are true but R does not explain A.(C) A is true, R is false.(D) A is false, R is true.
SECTION B: Very short answer (2 marks each)
11.
Find the area of a sector of radius $12$ cm and angle $30^\circ$ (use $\pi=3.14$).
[2]12.
Find the length of the arc of a sector of radius $21$ cm and angle $30^\circ$.
[2]13.
The angle of a minor sector of a circle of radius $7$ cm is $90^\circ$. Find the area of the major sector.
[2]14.
A chord of a circle of radius $14$ cm subtends a right angle at the centre. Find the area of the minor segment.
[2]15.
The hour hand of a clock is $7$ cm long. Find the area swept by it in $4$ hours.
[2]SECTION C: Short answer (4 marks each)
16.
The perimeter of a sector of a circle of radius $14$ cm is $50$ cm. Find the angle of the sector and its area.
[4]17.
A goat is tied at one corner of a square field of side $21$ m by a rope of length $7$ m. Find (i) the area of the field the goat can graze, (ii) the increase in grazing area if the rope is $14$ m long.
[4]18.
A car has two wipers which do not overlap. Each blade is $21$ cm long and sweeps through $120^\circ$. Find the total area cleaned in one sweep.
[4]19.
A chord of a circle of radius $8$ cm subtends $60^\circ$ at the centre. Find the area of the minor segment (use $\pi=3.14$, $\sqrt3=1.73$).
[4]20.
A round table cover of radius $14$ cm has six equal designs formed between a regular hexagon and the rim. Find the cost of making the designs at ₹0.50 per cm² (use $\sqrt3=1.7$).
[4]SECTION D: Long answer (5 marks each)
21.
A chord of a circle of radius $7$ cm subtends $120^\circ$ at the centre. Find the area of the corresponding minor segment (use $\sqrt3=1.73$).
[5]22.
A chord of a circle of radius $20$ cm subtends a right angle at the centre. Find the areas of the minor and major segments (use $\pi=3.14$).
[5]23.
A brooch is made of silver wire in the form of a circle of diameter $28$ mm. The same wire makes $4$ diameters dividing the circle into $8$ equal sectors. Find (i) the total length of wire, (ii) the area of each sector.
[5]24.
An umbrella has $8$ equally spaced ribs. Taking it as a flat circle of radius $35$ cm, find (i) the area between two consecutive ribs, (ii) the length of the cloth edge (arc) between them.
[5]SECTION E: Case-based questions (10 marks each)
25.
The Park Lawn. A circular park has radius $21$ m. A sector-shaped lawn $OAB$ with $\angle AOB=60^\circ$ is turfed, and the chord $AB$ is lined with a path (use $\sqrt3=1.73$).
[10](i) Find the area of the sector-shaped lawn. [2]
(ii) Find the length of the arc $AB$. [2]
(iii) Find the length of fencing needed round the sector. [2]
(iv) Name the type of $\triangle OAB$ and find its area. [2]
(v) Find the area of the segment between chord $AB$ and the arc. [2]
26.
The Pizza Party. A pizza of radius $14$ cm is cut into $8$ equal slices.
[10](i) Find the angle of each slice. [2]
(ii) Find the area of one slice. [2]
(iii) Find the length of the crust on one slice. [2]
(iv) Find the perimeter of one slice. [2]
(v) Two slices are eaten. Find the area of pizza left. [2]
ANSWER KEY AND MARKING SCHEME
Marks are shown in brackets. Alternative correct methods earn full marks.
Section A
1. (B) 22 cm. Arc $=\frac{90}{360}\times2\times\frac{22}{7}\times14=\frac14\times88=22$.
2. (A) 38.5 cm². Quadrant $=\frac14\times\frac{22}{7}\times49=\frac{154}{4}=38.5$.
3. (C) 120°. $\frac{20}{60}\times360^\circ=120^\circ$.
4. (B) 290°. $360^\circ-70^\circ=290^\circ$.
5. (C) 60°. $\frac{\theta}{360}=\frac16\Rightarrow\theta=60^\circ$.
6. (C) 50 cm. Arc $=22$, plus two radii $28$: $22+28=50$.
7. (A) $4\sqrt3$ cm². $OA=OB$ and $\angle O=60^\circ$, so the triangle is equilateral: $\frac{\sqrt3}{4}\times16=4\sqrt3$.
8. (A). Segment = sector − triangle $OAB$.
9. (B) 77 cm². $22=\frac{\theta}{360}\times44\Rightarrow\theta=180^\circ$. Area $=\frac12\times\frac{22}{7}\times49=77$. (Or $\frac12\times r\times\text{arc}=\frac12\times7\times22=77$.)
10. (A). Putting $\theta=180^\circ$ in the sector formula gives $\frac12\pi r^2$, so R explains A.
Section B
11. Area $=\frac{30}{360}\times3.14\times144$ [1] $=\frac{1}{12}\times452.16=\mathbf{37.68\text{ cm}^2}$ [1]
12. Arc $=\frac{30}{360}\times2\times\frac{22}{7}\times21$ [1] $=\frac1{12}\times132=\mathbf{11\text{ cm}}$ [1]
13. Whole circle $=\frac{22}{7}\times49=154$. Minor sector $=\frac14\times154=38.5$ [1]. Major sector $=154-38.5=\mathbf{115.5\text{ cm}^2}$ [1]
14. Sector $=\frac14\times\frac{22}{7}\times196=154$. Triangle $=\frac12\times14\times14=98$ [1]. Segment $=154-98=\mathbf{56\text{ cm}^2}$ [1]
15. Hour hand turns $30^\circ$ per hour, so in $4$ h: $120^\circ$ [1]. Area $=\frac{120}{360}\times\frac{22}{7}\times49=\frac{154}{3}=\mathbf{51.33\text{ cm}^2}$ [1]
Section C
16. Perimeter $=2r+\text{arc}$, so arc $=50-28=22$ cm [1]. $\frac{\theta}{360}\times2\times\frac{22}{7}\times14=22\Rightarrow\frac{\theta}{360}\times88=22\Rightarrow\theta=\mathbf{90^\circ}$ [1]. Area $=\frac{90}{360}\times\frac{22}{7}\times196$ [1] $=\mathbf{154\text{ cm}^2}$ [1]
17. The grazed region is a quadrant ($90^\circ$). (i) $\frac14\times\frac{22}{7}\times49=\mathbf{38.5\text{ m}^2}$ [2]. (ii) With $14$ m: $\frac14\times\frac{22}{7}\times196=154$ [1]. Increase $=154-38.5=\mathbf{115.5\text{ m}^2}$ [1]. (A $14$ m rope still fits inside the $21$ m field.)
18. One blade: $\frac{120}{360}\times\frac{22}{7}\times441=\frac13\times1386=462$ [2]. Two blades: $2\times462=\mathbf{924\text{ cm}^2}$ [2]
19. Sector $=\frac16\times3.14\times64=33.49$ [1]. Triangle is equilateral (standard equilateral triangle formula): $\frac{\sqrt3}{4}\times64=16\sqrt3=27.68$ [2]. Segment $=33.49-27.68=\mathbf{5.81\text{ cm}^2}$ (approx.) [1]
20. Each hexagon side subtends $60^\circ$, so the triangles are equilateral. One sector $=\frac16\times\frac{22}{7}\times196=\frac{308}{3}$ [1]. One triangle $=\frac{\sqrt3}{4}\times196=49\sqrt3=83.3$ [1]. Six segments $=6\times\frac{308}{3}-6\times83.3=616-499.8=116.2\text{ cm}^2$ [1]. Cost $=116.2\times0.50=\mathbf{₹58.10}$ [1]
Section D
21. Sector $=\frac13\times\frac{22}{7}\times49=\frac{154}{3}=51.33$ [1]. Draw $OM\perp AB$; by RHS congruence $M$ is the midpoint and $\angle AOM=60^\circ$ [1]. $OM=7\cos60^\circ=3.5$ and $AM=7\sin60^\circ=\frac{7\sqrt3}{2}$, so $AB=7\sqrt3$ [1]. Triangle $=\frac12\times7\sqrt3\times3.5=12.25\sqrt3=21.19$ [1]. Segment $=51.33-21.19=\mathbf{30.14\text{ cm}^2}$ (approx.) [1]
22. Sector $=\frac14\times3.14\times400=314$ [1]. Triangle $=\frac12\times20\times20=200$ [1]. Minor segment $=314-200=\mathbf{114\text{ cm}^2}$ [1]. Circle $=3.14\times400=1256$ [1]. Major segment $=1256-114=\mathbf{1142\text{ cm}^2}$ [1]. Check: $114+1142=1256$ ✓
23. (i) Circumference $=\pi d=\frac{22}{7}\times28=88$ mm [1]. Four diameters $=4\times28=112$ mm [1]. Total wire $=88+112=\mathbf{200\text{ mm}}$ [1]. (ii) $\theta=\frac{360}{8}=45^\circ$, $r=14$ [1]. Area $=\frac{45}{360}\times\frac{22}{7}\times196=\frac18\times616=\mathbf{77\text{ mm}^2}$ [1]
24. Angle between ribs $=\frac{360}{8}=45^\circ$ [1]. (i) Area $=\frac18\times\frac{22}{7}\times1225=\frac{3850}{8}=\mathbf{481.25\text{ cm}^2}$ [2]. (ii) Arc $=\frac18\times2\times\frac{22}{7}\times35=\frac{220}{8}=\mathbf{27.5\text{ cm}}$ [2]
Section E
25. (i) $\frac{60}{360}\times\frac{22}{7}\times441=\frac{1386}{6}=\mathbf{231\text{ m}^2}$ [2]
(ii) $\frac16\times2\times\frac{22}{7}\times21=\frac{132}{6}=\mathbf{22\text{ m}}$ [2]
(iii) Fence $=2r+\text{arc}=42+22=\mathbf{64\text{ m}}$ [2]
(iv) Equilateral (since $OA=OB$ and $\angle O=60^\circ$) [1]. Area $=\frac{\sqrt3}{4}\times441=\frac{762.93}{4}=\mathbf{190.73\text{ m}^2}$ [1]
(v) Segment $=231-190.73=\mathbf{40.27\text{ m}^2}$ (approx.) [2]
(ii) $\frac16\times2\times\frac{22}{7}\times21=\frac{132}{6}=\mathbf{22\text{ m}}$ [2]
(iii) Fence $=2r+\text{arc}=42+22=\mathbf{64\text{ m}}$ [2]
(iv) Equilateral (since $OA=OB$ and $\angle O=60^\circ$) [1]. Area $=\frac{\sqrt3}{4}\times441=\frac{762.93}{4}=\mathbf{190.73\text{ m}^2}$ [1]
(v) Segment $=231-190.73=\mathbf{40.27\text{ m}^2}$ (approx.) [2]
26. (i) $\frac{360}{8}=\mathbf{45^\circ}$ [2]
(ii) Whole pizza $=\frac{22}{7}\times196=616$. One slice $=\frac{616}{8}=\mathbf{77\text{ cm}^2}$ [2]
(iii) Circumference $=2\times\frac{22}{7}\times14=88$. Crust $=\frac{88}{8}=\mathbf{11\text{ cm}}$ [2]
(iv) Perimeter $=2r+\text{arc}=28+11=\mathbf{39\text{ cm}}$ [2]
(v) Two slices $=2\times77=154$. Left $=616-154=\mathbf{462\text{ cm}^2}$ [2]
(ii) Whole pizza $=\frac{22}{7}\times196=616$. One slice $=\frac{616}{8}=\mathbf{77\text{ cm}^2}$ [2]
(iii) Circumference $=2\times\frac{22}{7}\times14=88$. Crust $=\frac{88}{8}=\mathbf{11\text{ cm}}$ [2]
(iv) Perimeter $=2r+\text{arc}=28+11=\mathbf{39\text{ cm}}$ [2]
(v) Two slices $=2\times77=154$. Left $=616-154=\mathbf{462\text{ cm}^2}$ [2]
Marks summary
Section A: 10 · Section B: 10 · Section C: 20 · Section D: 20 · Section E: 20 · Total: 80
