Class 10 Areas Related to Circles: 14 Practice Questions Solved

Short answer#

Ques­tion 1#

Find the length of an arc of a cir­cle of radius $14$ cm that sub­tends an angle of $45^\circ$ at the cen­tre. Use $\displaystyle \pi = \frac{22}{7}$.

The frac­tion is $\displaystyle \frac{45^\circ}{360^\circ} = \frac{1}{8}$.

$$\displaystyle l = \frac{1}{8} \times 2 \times \frac{22}{7} \times 14 = \frac{1}{8} \times 88 = 11 \text{ cm}$$

Answer: $11$ cm.

Ques­tion 2#

Find the area of a sec­tor of a cir­cle of radius $6$ cm with angle $40^\circ$. Use $\pi = 3.14$.

$$\displaystyle \begin{aligned} \text{Area} &= \frac{40}{360} \times 3.14 \times 36 \\ &= \frac{1}{9} \times 113.04 = 12.56 \text{ cm}^2 \end{aligned}$$

Answer: $12.56$ cm².

Ques­tion 3#

Find the perime­ter of a sec­tor of a cir­cle of radius $10.5$ cm whose angle is $60^\circ$. Use $\displaystyle \pi = \frac{22}{7}$.

First the arc:

$$\displaystyle l = \frac{1}{6} \times 2 \times \frac{22}{7} \times 10.5 = \frac{1}{6} \times 66 = 11 \text{ cm}$$

The perime­ter includes both radii: $11 + 2 \times 10.5 = 11 + 21 = 32$ cm.

Answer: $32$ cm.

Why this works: the bound­ary of a sec­tor is one arc and two straight radii.

Ques­tion 4#

A quad­rant of a cir­cle has area $38.5$ cm². Find the radius. Use $\displaystyle \pi = \frac{22}{7}$.

A quad­rant is a sec­tor of $90^\circ$, so

$$\displaystyle \begin{aligned} \frac{1}{4} \times \frac{22}{7} \times r^2 &= 38.5 \\ r^2 &= 38.5 \times 4 \times \frac{7}{22} = 49 \\ r &= 7 \text{ cm} \end{aligned}$$

Answer: $7$ cm.

Ques­tion 5#

The minute hand of a table clock is $7$ cm long. Find the area swept by it and the dis­tance trav­elled by its tip between 9:00 and 9:35. Use $\displaystyle \pi = \frac{22}{7}$.

In $35$ min­utes the hand turns through $35 \times 6^\circ = 210^\circ$. The frac­tion is $\displaystyle \frac{210}{360} = \frac{7}{12}$.

$$\displaystyle \begin{aligned} \text{Area} &= \frac{7}{12} \times \frac{22}{7} \times 49 = \frac{7}{12} \times 154 = \frac{539}{6} \approx 89.83 \text{ cm}^2 \\ \text{Distance} &= \frac{7}{12} \times 2 \times \frac{22}{7} \times 7 = \frac{7}{12} \times 44 = \frac{77}{3} \approx 25.67 \text{ cm} \end{aligned}$$

Answer: about $89.83$ cm² swept and about $25.67$ cm trav­elled.

Medium#

Ques­tion 6#

An arc of length $22$ cm is cut from a cir­cle of radius $14$ cm. Find the angle at the cen­tre and the area of the sec­tor. Use $\displaystyle \pi = \frac{22}{7}$.

$$\displaystyle \begin{aligned} 22 &= \frac{\theta}{360^\circ} \times 2 \times \frac{22}{7} \times 14 = \frac{\theta}{360^\circ} \times 88 \\ \frac{\theta}{360^\circ} &= \frac{1}{4}, \quad \theta = 90^\circ \end{aligned}$$

Area $\displaystyle = \frac{1}{2} \times l \times r = \frac{1}{2} \times 22 \times 14 = 154$ cm².

Answer: $90^\circ$ and $154$ cm².

Ques­tion 7#

A chord of a cir­cle of radius $14$ cm sub­tends an angle of $60^\circ$ at the cen­tre. Find the area of the minor seg­ment. Use $\displaystyle \pi = \frac{22}{7}$ and $\sqrt{3} = 1.73$.

$$\displaystyle \begin{aligned} \text{Sector} &= \frac{1}{6} \times \frac{22}{7} \times 196 = \frac{616}{6} = \frac{308}{3} \approx 102.67 \text{ cm}^2 \\ \triangle OAB &= \frac{\sqrt{3}}{4} \times 196 = 49 \times 1.73 = 84.77 \text{ cm}^2 \\ \text{Segment} &= \frac{308}{3} - 84.77 \approx 17.90 \text{ cm}^2 \end{aligned}$$

Answer: approx­i­mately $17.90$ cm².

Why this works: with a $60^\circ$ angle and two equal radii, tri­an­gle $OAB$ is equi­lat­eral.

Ques­tion 8#

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 $\pi = 3.14$.

$$\displaystyle \begin{aligned} \text{Quadrant} &= \frac{1}{4} \times 3.14 \times 400 = 314 \text{ cm}^2 \\ \triangle OAB &= \frac{1}{2} \times 20 \times 20 = 200 \text{ cm}^2 \\ \text{Minor segment} &= 314 - 200 = 114 \text{ cm}^2 \\ \text{Circle} &= 3.14 \times 400 = 1256 \text{ cm}^2 \\ \text{Major segment} &= 1256 - 114 = 1142 \text{ cm}^2 \end{aligned}$$

Answer: minor seg­ment $114$ cm², major seg­ment $1142$ cm².

Ques­tion 9#

A chord of a cir­cle of radius $21$ cm sub­tends an angle of $120^\circ$ at the cen­tre. Find the area of the minor seg­ment. Use $\displaystyle \pi = \frac{22}{7}$ and $\sqrt{3} = 1.73$.

$$\displaystyle \begin{aligned} \text{Sector} &= \frac{1}{3} \times \frac{22}{7} \times 441 = \frac{1386}{3} = 462 \text{ cm}^2 \end{aligned}$$

For $\triangle OAB$, drop the per­pen­dic­u­lar $BM$ from $B$ to $AO$ extended beyond $O$. Since $\angle AOB$ and $\angle BOM$ form a lin­ear pair, $\angle BOM = 180^\circ - 120^\circ = 60^\circ$, so in right tri­an­gle $OMB$ the height is $BM = 21 \sin 60^\circ$ and

$$\displaystyle \begin{aligned} \triangle OAB &= \frac{1}{2} \times 21 \times 21 \times \frac{\sqrt{3}}{2} = \frac{441 \times 1.73}{4} = 190.7325 \text{ cm}^2 \\ \text{Segment} &= 462 - 190.7325 = 271.2675 \text{ cm}^2 \end{aligned}$$

Answer: approx­i­mately $271.27$ cm².

Long answer#

Ques­tion 10#

A cir­cu­lar gar­den of radius $42$ m is divided into $8$ equal flower beds by straight paths run­ning from the cen­tre to the edge. Ignor­ing the width of the paths, find the area of one flower bed. Use $\displaystyle \pi = \frac{22}{7}$.

The $8$ paths divide the full turn into $8$ equal angles of $\displaystyle \frac{360^\circ}{8} = 45^\circ$.

$$\displaystyle \begin{aligned} \text{Circle} &= \frac{22}{7} \times 42 \times 42 = 5544 \text{ m}^2 \\ \text{One flower bed} &= \frac{45^\circ}{360^\circ} \times 5544 = \frac{5544}{8} = 693 \text{ m}^2 \end{aligned}$$

Answer: $693$ m².

Ques­tion 11#

A goat is tied with a $7$ m rope to the post at one cor­ner of a rec­tan­gu­lar field mea­sur­ing $25$ m by $20$ m. (a) Find the area of the field it can graze. (b) Find the increase in the graz­ing area if the rope is replaced by one $14$ m long. Use $\displaystyle \pi = \frac{22}{7}$.

Each cor­ner of a rec­tan­gle is a right angle, and the rope is shorter than both sides, so the goat can graze a quad­rant.

$$\displaystyle \begin{aligned} \text{(a)} \quad & \frac{1}{4} \times \frac{22}{7} \times 7 \times 7 = \frac{154}{4} = 38.5 \text{ m}^2 \\ \text{(b)} \quad & \frac{1}{4} \times \frac{22}{7} \times 14 \times 14 = \frac{616}{4} = 154 \text{ m}^2 \\ & \text{Increase} = 154 - 38.5 = 115.5 \text{ m}^2 \end{aligned}$$

Answer: (a) $38.5$ m², (b) an increase of $115.5$ m².

Why this works: $14$ m is still less than both sides ($20$ m and $25$ m), so the graz­ing region is still a full quad­rant.

Ques­tion 12#

A dec­o­ra­tive tile is made from a square $ABCD$ of side $14$ cm with a semi­cir­cle drawn out­wards on each side. Find the area and the perime­ter of the tile. Use $\displaystyle \pi = \frac{22}{7}$.

Each semi­cir­cle has diam­e­ter $14$ cm, so radius $7$ cm.

$$\displaystyle \begin{aligned} \text{Square} &= 14 \times 14 = 196 \text{ cm}^2 \\ \text{One semicircle} &= \frac{1}{2} \times \frac{22}{7} \times 49 = 77 \text{ cm}^2 \\ \text{Tile} &= 196 + 4 \times 77 = 504 \text{ cm}^2 \end{aligned}$$

The outer bound­ary is made only of the four semi­cir­cu­lar arcs; the sides of the square are inside the tile. Four semi­cir­cles make two full cir­cles of radius $7$ cm:

$$\displaystyle \text{Perimeter} = 2 \times 2 \times \frac{22}{7} \times 7 = 88 \text{ cm}$$

Answer: area $504$ cm², perime­ter $88$ cm.

Ques­tion 13#

A square $PQRS$ is drawn inside a cir­cle of radius $10$ cm with all four ver­tices on the cir­cle. Find the area of the cir­cle out­side the square. Use $\pi = 3.14$.

The diag­o­nals of a square are equal and bisect each other, so their meet­ing point is the same dis­tance from all four ver­tices. The only point at the same dis­tance from four points of a cir­cle is its cen­tre, so each diag­o­nal passes through $O$ and is a diam­e­ter: $PR = 20$ cm. For a square, area $\displaystyle = \frac{1}{2} \times d^2$.

$$\displaystyle \begin{aligned} \text{Square} &= \frac{1}{2} \times 20 \times 20 = 200 \text{ cm}^2 \\ \text{Circle} &= 3.14 \times 100 = 314 \text{ cm}^2 \\ \text{Shaded} &= 314 - 200 = 114 \text{ cm}^2 \end{aligned}$$

Check: the shaded part is four equal $90^\circ$ seg­ments, each $78.5 - 50 = 28.5$ cm², and $4 \times 28.5 = 114$ cm².

Answer: $114$ cm².

Ques­tion 14#

Two con­cen­tric cir­cles have radii $14$ cm and $7$ cm. Two radii of the larger cir­cle make an angle of $60^\circ$ at the com­mon cen­tre. Find the area of the part of the ring between the cir­cles that lies between these radii. Use $\displaystyle \pi = \frac{22}{7}$.

The region is the sec­tor of the larger cir­cle minus the sec­tor of the smaller cir­cle, both of angle $60^\circ$.

$$\displaystyle \begin{aligned} \text{Area} &= \frac{60}{360} \times \frac{22}{7} \times 14^2 - \frac{60}{360} \times \frac{22}{7} \times 7^2 \\ &= \frac{1}{6} \times \frac{22}{7} \times (196 - 49) \\ &= \frac{1}{6} \times \frac{22}{7} \times 147 = \frac{1}{6} \times 462 = 77 \text{ cm}^2 \end{aligned}$$

Answer: $77$ cm².

Why this works: both sec­tors share the same angle, so the frac­tion $\displaystyle \frac{1}{6}$ can be taken out as a com­mon fac­tor.

Key terms#

Sec­tor

The region between two radii and the arc join­ing their ends.

Seg­ment

The region between a chord and an arc.

Quad­rant

A sec­tor of angle $90^\circ$.

Perime­ter of a sec­tor

Arc length plus twice the radius.

Con­cen­tric cir­cles

Cir­cles with the same cen­tre and dif­fer­ent radii.

Com­bined fig­ure

A shape made of cir­cles, sec­tors or seg­ments together with straight-edged fig­ures.

Com­mon ques­tions#

How do I know whether to add or sub­tract areas?#

Shade the required region men­tally. If it is made of pieces placed side by side, add them. If it is what is left after remov­ing a piece, sub­tract.

What if the ques­tion gives a diam­e­ter?#

Halve it first. Almost every for­mula in this chap­ter uses the radius.

How many dec­i­mal places should I give?#

Two dec­i­mal places is the usual con­ven­tion, unless the answer is exact or the ques­tion asks for some­thing else.

Can I leave the answer in terms of π?#

Only if the ques­tion does not give a value of $\pi$. When it does, sub­sti­tute that value.

Ref­er­ences#

1. National Coun­cil of Edu­ca­tional Research and Train­ing. Math­e­mat­ics: Text­book for Class X. NCERT, New Delhi.
2. Aggar­wal, R. S. Sec­ondary School Math­e­mat­ics for Class 10. Bharati Bhawan.
3. Sharma, R. D. Math­e­mat­ics for Class 10. Dhan­pat Rai Pub­li­ca­tions.