Class 10 Mathematics Chapter 12 Question Answer | Areas Related to Circles | English Medium

Class 10 Mathematics Chapter 12 Question Answer | Areas Related to Circles | English Medium | ASSEB

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Hello dear student! Welcome to HSLC Guru. In this lesson you will find the complete English-medium solutions for ASSEB (Assam State School Education Board) Class 10 Mathematics, Chapter 12 — Areas Related to Circles. The circle is the simplest closed curve in plane geometry, and yet its perimeter and area are governed by the most famous irrational number in mathematics — $\pi$. In earlier classes you learnt the two fundamental formulas: the circumference $C = 2\pi r$ and the area $A = \pi r^2$. In this chapter we extend those ideas to study the parts of a circle — arcs, sectors and segments — and to compute the area of combinations of plane figures, where a circle (or part of a circle) is glued together with a square, rectangle, triangle or another circle. The full textbook exercises 12.1, 12.2 and 12.3 are solved here step-by-step, followed by an additional-questions section that frequently appears in HSLC examinations. Unless the question states otherwise we use $\pi = \dfrac{22}{7}$ throughout.

Summary

A circle of radius $r$ has circumference $C = 2\pi r$ and area $A = \pi r^2$. If we cut the circle with two radii enclosing a central angle $\theta$, we get a sector; the curved part of its boundary is an arc. The arc length and sector area are simply the corresponding fractions $\theta/360°$ of the full circumference and full area: $\ell = \frac{\theta}{360°} \cdot 2\pi r$ and $A_{\text{sec}} = \frac{\theta}{360°} \cdot \pi r^2$. A chord of the circle divides it into two segments; the area of a segment is found by subtracting the triangle formed by the two radii from the corresponding sector. Many real-world figures — a race-track, an annular ring, a window with a semicircular top, a tile pattern — are combinations of these basic shapes, and their areas are computed by adding or subtracting the parts. Mastery of this chapter prepares you for surface-area and volume problems in Chapter 13 (Surface Areas and Volumes), where the same circle-formulas reappear inside cylinders, cones and spheres.

Historical Note: Where Does $\pi$ Come From?

The number $\pi$ (Greek letter pi) is one of the oldest numbers in mathematics. As early as 2000 BCE the Babylonians used the value $3\dfrac{1}{8} = 3.125$, and the Egyptian Rhind Papyrus (c. 1650 BCE) used $\left(\dfrac{16}{9}\right)^2 \approx 3.1605$. The ancient Indian mathematician Aryabhata (5th century CE) gave $\pi \approx \dfrac{62832}{20000} = 3.1416$ — accurate to four decimal places — almost a millennium before the same accuracy was achieved in Europe. The symbol $\pi$ itself was popularised by William Jones (1706) and Leonhard Euler (1737). It is now known that $\pi$ is an irrational number (Lambert, 1768) and even transcendental (Lindemann, 1882) — i.e. it is not the root of any polynomial equation with integer coefficients. For the ASSEB syllabus we use the rational approximation $\pi = \dfrac{22}{7}$, which gives an error of less than $0.04\%$ — perfectly adequate for textbook calculations.

The formal statement of the relationship is: in any circle, the ratio of the circumference to the diameter is constant. We call this constant $\pi$. Hence:

$$\pi \;=\; \frac{C}{d} \;=\; \frac{C}{2r} \quad\Longrightarrow\quad C \;=\; 2\pi r$$

For the area, divide a circle into many thin sectors and rearrange them as an approximate parallelogram of base $\pi r$ (half the circumference) and height $r$, giving area $\pi r\cdot r = \pi r^2$. The two great formulas of circle geometry follow.

Key Formulas at a Glance

Quantity	Formula	Units
Circumference of a circle	$C = 2\pi r = \pi d$	cm, m, …
Area of a circle	$A = \pi r^2$	cm$^2$, m$^2$, …
Length of arc (angle $\theta$)	$\ell = \dfrac{\theta}{360°}\cdot 2\pi r$	cm, m, …
Area of sector (angle $\theta$)	$A_{\text{sec}} = \dfrac{\theta}{360°}\cdot \pi r^2$	cm$^2$, m$^2$, …
Area of sector in terms of arc	$A_{\text{sec}} = \dfrac{1}{2}\, r\, \ell$	cm$^2$, m$^2$, …
Area of minor segment	$A_{\text{seg}} = A_{\text{sec}} – A_{\triangle}$	cm$^2$, m$^2$, …
Area of semicircle	$\dfrac{1}{2}\pi r^2$	cm$^2$, m$^2$, …
Area of quadrant	$\dfrac{1}{4}\pi r^2$	cm$^2$, m$^2$, …
Area of annular ring	$\pi(R^2 – r^2)$	cm$^2$, m$^2$, …

Useful values to remember (with $\pi = 22/7$):

$\pi \approx 3.14159$, and the rational approximation $\pi \approx 22/7$ is exact enough for textbook problems.
For a triangle in a sector with central angle $60°$, the triangle is equilateral; for $90°$ it is right-isosceles; for $120°$ each base angle is $30°$.
Area of an equilateral triangle of side $a$ is $\dfrac{\sqrt{3}}{4}a^2$.

Basic Figures Used in This Chapter

(i) Sector of a circle — the region OAPB enclosed between two radii OA, OB and the arc APB.

The minor sector is the smaller region (with angle $\theta < 180°$); the rest of the disc is the major sector.

(ii) Segment of a circle — the region cut off by a chord. The minor segment is between the chord and the minor arc; the major segment lies on the other side.

Area of segment $= $ Area of sector $-$ Area of triangle $OAB$.

$$A_{\text{seg}} \;=\; \frac{\theta}{360°}\pi r^2 \;-\; \frac{1}{2}r^2 \sin\theta$$

Exercise 12.1 — Perimeter and Area of a Circle

(Throughout, take $\pi = 22/7$ unless stated otherwise.)

Q1. The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.

Answer: Let the required radius be $R$. Sum of two circumferences $= 2\pi(19) + 2\pi(9) = 2\pi(19+9) = 2\pi(28)$.

Equating, $2\pi R = 2\pi(28) \implies R = 28$ cm.

Q2. The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.

Answer: Let the required radius be $R$. Sum of areas $= \pi(8)^2 + \pi(6)^2 = \pi(64+36) = 100\pi$.

$\pi R^2 = 100\pi \implies R^2 = 100 \implies R = 10$ cm.

Q3. The figure depicts an archery target marked with its five scoring regions from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is 21 cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions.

Answer: Radii of the five concentric circles (innermost to outermost):

$r_1 = 10.5$ cm (Gold)
$r_2 = 10.5 + 10.5 = 21$ cm (up to Red)
$r_3 = 21 + 10.5 = 31.5$ cm (up to Blue)
$r_4 = 31.5 + 10.5 = 42$ cm (up to Black)
$r_5 = 42 + 10.5 = 52.5$ cm (up to White)

Use $\pi = 22/7$.

Area of Gold (innermost circle) $= \pi r_1^2 = \frac{22}{7} \times 10.5 \times 10.5 = 346.5$ cm$^2$.

Area of Red (annular ring between $r_2$ and $r_1$):

$$\pi(r_2^2 – r_1^2) = \frac{22}{7}(21^2 – 10.5^2) = \frac{22}{7}(441 – 110.25) = \frac{22}{7}\times 330.75 = 1039.5 \text{ cm}^2$$

Area of Blue $= \pi(r_3^2 – r_2^2) = \dfrac{22}{7}(31.5^2 – 21^2) = \dfrac{22}{7}(992.25 – 441) = \dfrac{22}{7}\times 551.25 = 1732.5$ cm$^2$.

Area of Black $= \pi(r_4^2 – r_3^2) = \dfrac{22}{7}(42^2 – 31.5^2) = \dfrac{22}{7}(1764 – 992.25) = \dfrac{22}{7}\times 771.75 = 2425.5$ cm$^2$.

Area of White $= \pi(r_5^2 – r_4^2) = \dfrac{22}{7}(52.5^2 – 42^2) = \dfrac{22}{7}(2756.25 – 1764) = \dfrac{22}{7}\times 992.25 = 3118.5$ cm$^2$.

Q4. The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour?

Answer: Radius $r = 40$ cm. Circumference of one wheel $= 2\pi r = 2\times \dfrac{22}{7}\times 40 = \dfrac{1760}{7}$ cm.

Distance covered in 10 minutes at 66 km/h:

$$\text{Distance} = 66 \times \frac{10}{60} = 11 \text{ km} = 11\times 100000 = 1{,}100{,}000 \text{ cm}$$

Number of revolutions $= \dfrac{\text{Distance}}{\text{Circumference}} = \dfrac{1100000}{1760/7} = \dfrac{1100000\times 7}{1760} = \dfrac{7700000}{1760} = 4375$.

So each wheel makes $\mathbf{4375}$ complete revolutions in 10 minutes.

Q5. Tick the correct answer in the following and justify your choice: If the perimeter and the area of a circle are numerically equal, then the radius of the circle is

(A) 2 units (B) $\pi$ units (C) 4 units (D) 7 units

Answer: Set $2\pi r = \pi r^2 \implies 2 = r$, so $r = 2$ units. Choice (A) is correct.

Exercise 12.2 — Areas of Sector and Segment of a Circle

(Take $\pi = 22/7$ unless stated otherwise.)

Q1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is $60°$.

Answer:

$$A_{\text{sec}} = \frac{\theta}{360°}\pi r^2 = \frac{60}{360}\times \frac{22}{7}\times 6\times 6 = \frac{1}{6}\times \frac{22}{7}\times 36 = \frac{132}{7} \text{ cm}^2$$

So area $= \dfrac{132}{7}$ cm$^2$ $\approx 18.86$ cm$^2$.

Q2. Find the area of a quadrant of a circle whose circumference is 22 cm.

Answer: $2\pi r = 22 \implies r = \dfrac{22\times 7}{2\times 22} = \dfrac{7}{2}$ cm. A quadrant subtends $90°$:

$$A_{\text{quad}} = \frac{1}{4}\pi r^2 = \frac{1}{4}\times \frac{22}{7}\times \left(\frac{7}{2}\right)^2 = \frac{1}{4}\times \frac{22}{7}\times \frac{49}{4} = \frac{77}{8} \text{ cm}^2$$

Area of the quadrant $= \dfrac{77}{8}$ cm$^2 = 9.625$ cm$^2$.

Q3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.

Answer: The minute hand sweeps the full $360°$ in 60 minutes, so in 5 minutes it sweeps $\dfrac{5}{60}\times 360° = 30°$. With $r = 14$ cm:

$$A_{\text{sec}} = \frac{30}{360}\times \frac{22}{7}\times 14\times 14 = \frac{1}{12}\times \frac{22}{7}\times 196 = \frac{22\times 196}{12\times 7} = \frac{4312}{84} = \frac{154}{3} \text{ cm}^2$$

Area swept $= \dfrac{154}{3}$ cm$^2 \approx 51.33$ cm$^2$.

Q4. 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 and (ii) major sector. (Use $\pi = 3.14$.)

Answer: $r = 10$ cm, $\theta = 90°$.

Area of sector $OAB = \dfrac{90}{360}\times 3.14\times 10\times 10 = \dfrac{1}{4}\times 314 = 78.5$ cm$^2$.

Triangle $OAB$ is right-angled at $O$ with legs $10, 10$, so area $= \dfrac{1}{2}\times 10\times 10 = 50$ cm$^2$.

(i) Area of minor segment $= 78.5 – 50 = \mathbf{28.5}$ cm$^2$.

(ii) Area of major sector $= \pi r^2 – $ area of minor sector $= 3.14\times 100 – 78.5 = 314 – 78.5 = \mathbf{235.5}$ cm$^2$.

Q5. 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.

Answer: $r = 21$, $\theta = 60°$, $\pi = 22/7$.

(i) Length of arc:

$$\ell = \frac{60}{360}\times 2\pi r = \frac{1}{6}\times 2\times \frac{22}{7}\times 21 = \frac{1}{6}\times 132 = 22 \text{ cm}$$

(ii) Area of sector:

$$A_{\text{sec}} = \frac{60}{360}\times \pi r^2 = \frac{1}{6}\times \frac{22}{7}\times 441 = \frac{22\times 441}{42} = \frac{9702}{42} = 231 \text{ cm}^2$$

(iii) Triangle $OAB$ has $OA = OB = 21$ cm and $\angle AOB = 60°$, so it is equilateral with side 21. Area $= \dfrac{\sqrt{3}}{4}\times 21^2 = \dfrac{441\sqrt{3}}{4}$ cm$^2$. Therefore

$$A_{\text{seg}} = 231 – \frac{441\sqrt{3}}{4} \text{ cm}^2$$

Q6. 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 $\pi = 3.14$ and $\sqrt{3} = 1.73$.)

Answer: $r = 15$, $\theta = 60°$. Sector area:

$$A_{\text{sec}} = \frac{60}{360}\times 3.14\times 15\times 15 = \frac{1}{6}\times 706.5 = 117.75 \text{ cm}^2$$

Triangle is equilateral with side 15: area $= \dfrac{\sqrt{3}}{4}\times 225 = \dfrac{225\times 1.73}{4} = \dfrac{389.25}{4} = 97.3125$ cm$^2$.

Minor segment $= 117.75 – 97.3125 = \mathbf{20.4375}$ cm$^2 \approx 20.44$ cm$^2$.

Major segment $= \pi r^2 – $ minor segment $= 3.14\times 225 – 20.4375 = 706.5 – 20.4375 = \mathbf{686.0625}$ cm$^2$.

Q7. A chord of a circle of radius 12 cm subtends an angle of $120°$ at the centre. Find the area of the corresponding segment of the circle. (Use $\pi = 3.14$ and $\sqrt{3} = 1.73$.)

Answer: $r = 12$, $\theta = 120°$. Sector area:

$$A_{\text{sec}} = \frac{120}{360}\times 3.14\times 144 = \frac{1}{3}\times 452.16 = 150.72 \text{ cm}^2$$

For triangle $OAB$ with $OA = OB = 12$, $\angle AOB = 120°$: drop perpendicular from $O$ to chord $AB$. Half the chord $= 12\sin 60° = 12\times \dfrac{\sqrt 3}{2} = 6\sqrt 3$, so $AB = 12\sqrt 3$. The perpendicular height $= 12\cos 60° = 6$. Area of triangle $= \dfrac{1}{2}\times 12\sqrt 3 \times 6 = 36\sqrt 3 = 36\times 1.73 = 62.28$ cm$^2$.

Area of segment $= 150.72 – 62.28 = \mathbf{88.44}$ cm$^2$.

Q8. A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope (see figure). Find: (i) the area of that part of the field in which the horse can graze. (ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use $\pi = 3.14$.)

Answer: The horse can graze a quadrant (since the corner of the square gives a $90°$ angle) of radius equal to the rope length.

(i) With $r = 5$, area of quadrant $= \dfrac{1}{4}\pi r^2 = \dfrac{1}{4}\times 3.14\times 25 = \dfrac{78.5}{4} = 19.625$ m$^2$.

(ii) With $r = 10$, area of quadrant $= \dfrac{1}{4}\times 3.14\times 100 = 78.5$ m$^2$. Increase $= 78.5 – 19.625 = \mathbf{58.875}$ m$^2$.

Q9. A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in figure. Find: (i) the total length of the silver wire required; (ii) the area of each sector of the brooch.

Answer: Diameter $= 35$ mm, so radius $r = \dfrac{35}{2}$ mm. Circumference $= 2\pi r = 2\times \dfrac{22}{7}\times \dfrac{35}{2} = 110$ mm. Five diameters add $5\times 35 = 175$ mm.

(i) Total wire $= 110 + 175 = \mathbf{285}$ mm.

(ii) Each of the 10 sectors has angle $36°$. Area of one sector:

$$A_{\text{sec}} = \frac{36}{360}\times \pi r^2 = \frac{1}{10}\times \frac{22}{7}\times \frac{35}{2}\times \frac{35}{2} = \frac{22\times 1225}{10\times 7\times 4} = \frac{26950}{280} = \frac{385}{4} \text{ mm}^2$$

Area of each sector $= \dfrac{385}{4}$ mm$^2 = 96.25$ mm$^2$.

Q10. An umbrella has 8 ribs which are equally spaced (see figure). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

Answer: Each of the 8 sectors has central angle $\dfrac{360°}{8} = 45°$, $r = 45$ cm.

$$A = \frac{45}{360}\times \pi r^2 = \frac{1}{8}\times \frac{22}{7}\times 2025 = \frac{22\times 2025}{56} = \frac{44550}{56} = \frac{22275}{28} \text{ cm}^2$$

Area between two consecutive ribs $= \dfrac{22275}{28}$ cm$^2 \approx 795.54$ cm$^2$.

Q11. 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.

Answer: Each wiper sweeps a sector with $r = 25$, $\theta = 115°$:

$$A_{\text{sec}} = \frac{115}{360}\times \frac{22}{7}\times 625 = \frac{115\times 22\times 625}{360\times 7} = \frac{1581250}{2520} = \frac{158125}{252} \text{ cm}^2$$

Two wipers (no overlap): total $= 2\times \dfrac{158125}{252} = \dfrac{158125}{126}$ cm$^2 \approx 1254.96$ cm$^2$.

Q12. To warn ships for 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 $\pi = 3.14$.)

Answer: $r = 16.5$ km, $\theta = 80°$:

$$A_{\text{sec}} = \frac{80}{360}\times 3.14\times 16.5\times 16.5 = \frac{2}{9}\times 3.14\times 272.25 = \frac{2\times 854.865}{9} = \frac{1709.73}{9}$$

$= 189.97$ km$^2$ (approx). Area warned $\approx \mathbf{189.97}$ km$^2$.

Q13. A round table cover has six equal designs as shown in figure. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of Rs 0.35 per cm$^2$. (Use $\sqrt{3} = 1.7$.)

Answer: Each design is a circular segment with central angle $60°$ (six equal designs round a circle). $r = 28$ cm.

Sector area $= \dfrac{60}{360}\times \dfrac{22}{7}\times 784 = \dfrac{1}{6}\times \dfrac{22\times 784}{7} = \dfrac{22\times 784}{42} = \dfrac{17248}{42} = \dfrac{1232}{3}$ cm$^2$.

Triangle: equilateral, side 28; area $= \dfrac{\sqrt{3}}{4}\times 784 = 196\sqrt{3} = 196\times 1.7 = 333.2$ cm$^2$.

Area of one segment $= \dfrac{1232}{3} – 333.2 = 410.667 – 333.2 = 77.467$ cm$^2$.

Total area of 6 designs $= 6\times 77.467 = 464.8$ cm$^2$. Cost $= 464.8\times 0.35 = \mathbf{Rs\;162.68}$.

Q14. Tick the correct answer in the following: Area of a sector of angle $p$ (in degrees) of a circle with radius $R$ is

(A) $\dfrac{p}{180}\times 2\pi R$ (B) $\dfrac{p}{180}\times \pi R^2$ (C) $\dfrac{p}{360}\times 2\pi R$ (D) $\dfrac{p}{720}\times 2\pi R^2$

Answer: Standard formula is $\dfrac{p}{360}\pi R^2$. Multiply numerator and denominator by 2 to get $\dfrac{p}{720}\times 2\pi R^2$. Correct option (D).

Exercise 12.3 — Areas of Combinations of Plane Figures

(Take $\pi = 22/7$ unless stated otherwise.)

Q1. Find the area of the shaded region in the figure, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle. (PQR is inscribed with $\angle P = 90°$ between PR and PQ.)

Answer: Since $\angle RPQ = 90°$, $RQ$ is a diameter (angle in a semicircle). $RQ = \sqrt{PR^2 + PQ^2} = \sqrt{49+576} = \sqrt{625} = 25$ cm. Radius $r = 12.5$ cm.

Area of semicircle $= \dfrac{1}{2}\pi r^2 = \dfrac{1}{2}\times \dfrac{22}{7}\times 156.25 = \dfrac{3437.5}{14} = \dfrac{1718.75}{7}$ cm$^2 = 245.535\ldots$ cm$^2$.

Area of triangle $PQR = \dfrac{1}{2}\times 7\times 24 = 84$ cm$^2$.

Shaded area $= 245.54 – 84 = \mathbf{161.54}$ cm$^2$ (approximately).

Q2. Find the area of the shaded region in the figure, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and $\angle AOC = 40°$.

Answer: Shaded region $= $ (sector AOC of larger circle) $-$ (sector BOD of smaller circle), with same angle $40°$.

$$A = \frac{40}{360}\pi(R^2 – r^2) = \frac{1}{9}\times \frac{22}{7}\times (196 – 49) = \frac{1}{9}\times \frac{22}{7}\times 147 = \frac{22\times 147}{63} = \frac{3234}{63} = \frac{154}{3}$$

Shaded area $= \dfrac{154}{3}$ cm$^2 \approx 51.33$ cm$^2$.

Q3. Find the area of the shaded region in the figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles.

Answer: Two semicircles (each with diameter equal to the side of the square, 14 cm) lie inside the square along sides AD and BC.

Diameter $= 14 \implies r = 7$ cm. Area of one semicircle $= \dfrac{1}{2}\pi r^2 = \dfrac{1}{2}\times \dfrac{22}{7}\times 49 = 77$ cm$^2$. Two semicircles $= 154$ cm$^2$.

Area of square $= 14\times 14 = 196$ cm$^2$. Shaded area (the part of the square outside both semicircles) $= 196 – 154 = \mathbf{42}$ cm$^2$.

Q4. Find the area of the shaded region in the figure, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre.

Answer: Shaded $= $ (area of triangle $OAB$) $+$ (area of major sector with centre $O$, angle $360°-60°=300°$, radius 6) $-$ (the part of the triangle inside the small circle, which equals the $60°$ sector of radius 6 since the angle at $O$ is $60°$ in the triangle).

Or, using the simple form: Shaded $= $ Area of triangle $+$ Area of circle $-$ 2 $\times$ ($60°$ sector at $O$).

Triangle of side 12: $\dfrac{\sqrt{3}}{4}\times 144 = 36\sqrt{3}$ cm$^2$.

Full circle of radius 6: $\pi\times 36 = \dfrac{22}{7}\times 36 = \dfrac{792}{7}$ cm$^2$.

$60°$ sector at $O$ of radius 6: $\dfrac{1}{6}\times \dfrac{792}{7} = \dfrac{132}{7}$ cm$^2$.

Shaded area $= 36\sqrt{3} + \dfrac{792}{7} – \dfrac{132}{7} = 36\sqrt{3} + \dfrac{660}{7}$ cm$^2$.

Q5. From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut. Also, a circle of diameter 2 cm is cut as shown in figure. Find the area of the remaining (shaded) portion of the square.

Answer: Area of square $= 16$ cm$^2$. Four quadrants of radius 1 form one full circle of radius 1: area $= \pi(1)^2 = \dfrac{22}{7}$ cm$^2$. Central circle of diameter 2 has radius 1, area $= \pi(1)^2 = \dfrac{22}{7}$ cm$^2$.

$$\text{Shaded} = 16 – \frac{22}{7} – \frac{22}{7} = 16 – \frac{44}{7} = \frac{112 – 44}{7} = \frac{68}{7} \text{ cm}^2$$

Shaded area $= \dfrac{68}{7}$ cm$^2 \approx 9.71$ cm$^2$.

Q6. In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown. Find the area of the design (shaded region).

Answer: Side of equilateral triangle inscribed in a circle of radius $r$ is $a = r\sqrt 3 = 32\sqrt 3$ cm.

Area of triangle $= \dfrac{\sqrt 3}{4}(32\sqrt 3)^2 = \dfrac{\sqrt 3}{4}\times 32^2 \times 3 = \dfrac{3\sqrt 3}{4}\times 1024 = 768\sqrt 3$ cm$^2$.

Area of circle $= \dfrac{22}{7}\times 32\times 32 = \dfrac{22528}{7}$ cm$^2$.

Shaded area $= \dfrac{22528}{7} – 768\sqrt 3$ cm$^2$.

Q7. In the figure, ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touches externally two of the remaining three circles. Find the area of the shaded region.

Answer: Each circle has radius equal to half the side, $r = 7$ cm. Inside the square, each corner contains a $90°$ sector of radius 7. Four such sectors put together form one full circle of radius 7 (since $4\times 90° = 360°$).

Area covered by sectors $= \pi r^2 = \dfrac{22}{7}\times 49 = 154$ cm$^2$.

Area of square $= 14\times 14 = 196$ cm$^2$. Shaded area $= 196 – 154 = \mathbf{42}$ cm$^2$.

Q8. The figure depicts a racing track whose left and right ends are semicircular. The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find: (i) the distance around the track along its inner edge; (ii) the area of the track.

Answer: Inner straight length $= 106$ m each (two segments). Inner radius of each semicircle $= \dfrac{60}{2} = 30$ m.

(i) Inner perimeter $= 2\times 106 + 2\times (\pi r) = 212 + 2\times \dfrac{22}{7}\times 30 = 212 + \dfrac{1320}{7} = \dfrac{1484 + 1320}{7} = \dfrac{2804}{7}$ m $= 400.57$ m.

(ii) Area of track $= $ area of two rectangles $+$ area of annular ring (outer circle of radius $40$ minus inner circle of radius $30$). Rectangles: $2\times (106\times 10) = 2120$ m$^2$. Annular ring: $\pi(40^2 – 30^2) = \dfrac{22}{7}\times (1600 – 900) = \dfrac{22}{7}\times 700 = 2200$ m$^2$.

Total area $= 2120 + 2200 = \mathbf{4320}$ m$^2$.

Q9. In the figure, AB and CD are two diameters of a circle (perpendicular to each other) and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.

Answer: Larger circle: radius $7$, area $= \dfrac{22}{7}\times 49 = 154$ cm$^2$.

Smaller circle: diameter $OD = 7$, radius $3.5$, area $= \dfrac{22}{7}\times 12.25 = 38.5$ cm$^2$.

Triangle $ABC$ is right-angled at $C$ (with base $AB = 14$ and height $OC = 7$): area $= \dfrac{1}{2}\times 14\times 7 = 49$ cm$^2$.

Shaded $= $ Half of larger circle $+$ smaller circle $-$ triangle $ABC$ $= 77 + 38.5 – 49 = \mathbf{66.5}$ cm$^2$.

Q10. The area of an equilateral triangle ABC is 17320.5 cm$^2$. With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle. Find the area of the shaded region. (Use $\pi = 3.14$ and $\sqrt{3} = 1.73205$.)

Answer: $\dfrac{\sqrt 3}{4}a^2 = 17320.5 \implies a^2 = \dfrac{17320.5\times 4}{1.73205} = \dfrac{69282}{1.73205} = 40000$, so $a = 200$ cm. Radius of each circle $= 100$ cm.

Each angle of equilateral triangle is $60°$. Three sectors at the three vertices together have angle $3\times 60° = 180°$ — i.e. half a full circle of radius 100.

Sectors area $= \dfrac{1}{2}\pi r^2 = \dfrac{1}{2}\times 3.14\times 10000 = 15700$ cm$^2$.

Shaded area (inside triangle, outside the sectors) $= 17320.5 – 15700 = \mathbf{1620.5}$ cm$^2$.

Q11. On a square handkerchief, nine circular designs each of radius 7 cm are made. Find the area of the remaining portion of the handkerchief.

Answer: Nine circles in a $3\times 3$ grid. Each circle has diameter 14, so the side of the square is $3\times 14 = 42$ cm.

Square area $= 42^2 = 1764$ cm$^2$. Total area of 9 circles $= 9\pi r^2 = 9\times \dfrac{22}{7}\times 49 = 9\times 154 = 1386$ cm$^2$.

Remaining area $= 1764 – 1386 = \mathbf{378}$ cm$^2$.

Q12. In the figure, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the (i) quadrant OACB, (ii) shaded region.

Answer: (i) Quadrant OACB: $r = 3.5$ cm, $\theta = 90°$:

$$A_{\text{quad}} = \frac{1}{4}\pi r^2 = \frac{1}{4}\times \frac{22}{7}\times 12.25 = \frac{269.5}{28} = \frac{77}{8} \text{ cm}^2$$

(ii) Triangle $OBD$: right-angled at $O$ with legs $OB = 3.5$ and $OD = 2$. Area $= \dfrac{1}{2}\times 3.5\times 2 = 3.5$ cm$^2$. Shaded $= \dfrac{77}{8} – 3.5 = 9.625 – 3.5 = \mathbf{6.125}$ cm$^2$.

Q13. In the figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use $\pi = 3.14$.)

Answer: Diagonal $OB$ of the square is the radius of the quadrant. $OB = 20\sqrt 2$ cm, so radius $r = 20\sqrt 2$.

Area of quadrant $= \dfrac{1}{4}\pi r^2 = \dfrac{1}{4}\times 3.14\times 800 = 628$ cm$^2$.

Area of square $= 20\times 20 = 400$ cm$^2$. Shaded area $= 628 – 400 = \mathbf{228}$ cm$^2$.

Q14. AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see figure). If $\angle AOB = 30°$, find the area of the shaded region.

Answer: Shaded region $= $ sector of larger circle $-$ sector of smaller circle, both with $\theta = 30°$:

$$A = \frac{30}{360}\pi(R^2 – r^2) = \frac{1}{12}\times \frac{22}{7}\times (441 – 49) = \frac{1}{12}\times \frac{22}{7}\times 392 = \frac{22\times 392}{84} = \frac{8624}{84} = \frac{308}{3}$$

Shaded area $= \dfrac{308}{3}$ cm$^2 \approx 102.67$ cm$^2$.

Q15. In the figure, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.

Answer: Quadrant $ABC$ has radius 14 with right angle at $A$. So $AB = AC = 14$, and $BC = 14\sqrt 2$. Radius of semicircle on $BC$ = $7\sqrt 2$.

Area of quadrant $= \dfrac{1}{4}\pi\times 14^2 = \dfrac{1}{4}\times \dfrac{22}{7}\times 196 = 154$ cm$^2$.

Area of right triangle $ABC$ $= \dfrac{1}{2}\times 14\times 14 = 98$ cm$^2$.

Area of semicircle on $BC$: $\dfrac{1}{2}\pi(7\sqrt 2)^2 = \dfrac{1}{2}\times \dfrac{22}{7}\times 98 = 154$ cm$^2$.

Now the shaded region = (semicircle on $BC$) $-$ (quadrant $-$ triangle) $= 154 – (154 – 98) = 154 – 56 = \mathbf{98}$ cm$^2$.

Q16. Calculate the area of the designed region in the figure common between the two quadrants of circles of radius 8 cm each.

Answer: Two quadrants of radius 8 are drawn at opposite corners of a square of side 8. Each quadrant has area $\dfrac{1}{4}\pi r^2 = \dfrac{1}{4}\times \dfrac{22}{7}\times 64 = \dfrac{352}{7}$ cm$^2$.

Sum of the two quadrants $= 2\times \dfrac{352}{7} = \dfrac{704}{7}$ cm$^2$.

Designed region $= $ sum of two quadrants $-$ area of square (the overlapping region equals the square once):

$$\text{Designed area} = \frac{704}{7} – 64 = \frac{704 – 448}{7} = \frac{256}{7} \text{ cm}^2 \approx 36.57 \text{ cm}^2$$

Worked Example: A Race-Track Combination

Race-tracks are an ASSEB favourite. The figure below shows a typical race-track: a rectangle of length $L$ and width $W$ (the inner straight portion is the rectangle), with a semicircle of diameter $W$ glued to each shorter side. Then the entire figure is bordered by a uniform width $w$ to form the outer boundary.

Total perimeter (outer boundary) $= 2L + 2\pi(W/2 + w) = 2L + \pi(W + 2w)$.

Area of the track itself $= 2(L\cdot w) + \pi\left[(W/2 + w)^2 – (W/2)^2\right]$.

Substituting the standard textbook values $L = 106$ m, $W = 60$ m, $w = 10$ m gives outer perimeter $= 212 + \pi(60 + 20) = 212 + \dfrac{22}{7}\times 80 = 212 + \dfrac{1760}{7} = \dfrac{1484+1760}{7} = \dfrac{3244}{7} \approx 463.4$ m, and track area $= 2\times 1060 + \dfrac{22}{7}(40^2 – 30^2) = 2120 + 2200 = 4320$ m$^2$.

Worked Example: Annular Ring of a Path

Many gardens or sports fields have a circular flowerbed surrounded by a path. If the flowerbed has radius $r$ and the path width is $w$, the outer radius is $R = r + w$ and the area of the path is the annular ring $\pi(R^2 – r^2)$.

Example: $r = 10$ m, $w = 2.1$ m gives $R = 12.1$ m and path area $= \dfrac{22}{7}(146.41 – 100) = \dfrac{22}{7}\times 46.41 = \dfrac{1021.02}{7} = 145.86$ m$^2$.

Worked Example: Square Inscribed Inside / Outside a Circle

This combination appears in dozens of textbook problems. Memorise these two relationships:

Square inscribed in a circle of radius $r$. Diagonal of square $= 2r$, so side $= r\sqrt 2$ and area of square $= 2r^2$. Area of the four “corner segments” (circle $-$ square) $= \pi r^2 – 2r^2 = r^2(\pi – 2)$.
Circle inscribed in a square of side $a$. Diameter of circle $= a$, so radius $= a/2$ and area of circle $= \pi a^2/4$. Area outside the circle but inside the square (four “L-shaped” pieces, one in each corner) $= a^2 – \pi a^2/4 = a^2(1 – \pi/4)$.

Additional Questions and Answers

A. Multiple-Choice Questions (1 mark)

1. The circumference of a circle of radius 5 cm is

(A) 25π cm (B) 10π cm (C) 5π cm (D) $\pi/5$ cm

Answer: $C = 2\pi r = 2\pi\times 5 = 10\pi$ cm. (B).

2. The area of a circle whose circumference is 22 cm is

(A) 38.5 cm$^2$ (B) 77 cm$^2$ (C) 49 cm$^2$ (D) 154 cm$^2$

Answer: $2\pi r = 22 \implies r = 7/2$. Area $= \dfrac{22}{7}\times \dfrac{49}{4} = \dfrac{77}{2} = 38.5$ cm$^2$. (A).

3. The area of the sector of angle $\theta$ of a circle with radius $r$ is

(A) $\dfrac{\theta}{180}\pi r^2$ (B) $\dfrac{\theta}{360}\pi r^2$ (C) $\dfrac{\theta}{180}2\pi r$ (D) $\dfrac{\theta}{360}2\pi r$

Answer: (B).

4. The length of an arc that subtends an angle of $90°$ at the centre of a circle of radius 14 cm is

(A) 11 cm (B) 22 cm (C) 14 cm (D) 7 cm

Answer: $\ell = \dfrac{90}{360}\times 2\pi r = \dfrac{1}{4}\times 2\times \dfrac{22}{7}\times 14 = 22$ cm. (B).

5. If the radius of a circle is doubled, its area becomes

(A) 2 times (B) 3 times (C) 4 times (D) 8 times

Answer: $\pi(2r)^2 = 4\pi r^2$. (C).

6. The perimeter of a quadrant of a circle of radius $r$ is

(A) $\dfrac{\pi r}{2}$ (B) $2\pi r$ (C) $\dfrac{\pi r}{2} + 2r$ (D) $\pi r + 2r$

Answer: Perimeter $= $ arc $+$ two radii $= \dfrac{1}{4}\times 2\pi r + 2r = \dfrac{\pi r}{2} + 2r$. (C).

7. The area of the circle that can be inscribed in a square of side 6 cm is

(A) $36\pi$ cm$^2$ (B) $18\pi$ cm$^2$ (C) $12\pi$ cm$^2$ (D) $9\pi$ cm$^2$

Answer: The inscribed circle has diameter equal to the side. Radius $= 3$. Area $= 9\pi$. (D).

8. If the perimeter and the area of a circle are numerically equal, then the radius of the circle is

(A) $\pi$ units (B) 2 units (C) 4 units (D) 7 units

Answer: $2\pi r = \pi r^2 \implies r = 2$. (B).

9. The number of revolutions made by a circular wheel of radius 0.7 m in rolling a distance of 176 m is

(A) 22 (B) 24 (C) 75 (D) 40

Answer: Circumference $= 2\pi r = 2\times \dfrac{22}{7}\times 0.7 = 4.4$ m. Revolutions $= 176/4.4 = 40$. (D).

10. If the area of a circle is 154 cm$^2$, its radius is

(A) 14 cm (B) 7 cm (C) 11 cm (D) 22 cm

Answer: $\pi r^2 = 154 \implies r^2 = 49 \implies r = 7$ cm. (B).

11. The area swept by the minute hand of a clock of length 14 cm in 30 minutes is

(A) 308 cm$^2$ (B) 154 cm$^2$ (C) 77 cm$^2$ (D) 616 cm$^2$

Answer: 30 min = $180°$. $\dfrac{180}{360}\times \dfrac{22}{7}\times 196 = \dfrac{1}{2}\times 616 = 308$ cm$^2$. (A).

12. The circumference of a circle exceeds its diameter by 16.8 cm. The diameter is

(A) 7.84 cm (B) 15.4 cm (C) 7.7 cm (D) 9.6 cm

Answer: $\pi d – d = 16.8 \implies d(\pi – 1) = 16.8 \implies d = \dfrac{16.8}{22/7 – 1} = \dfrac{16.8}{15/7} = \dfrac{16.8\times 7}{15} = \dfrac{117.6}{15} = 7.84$ cm. (A).

B. Fill in the Blanks

Area of a sector of central angle $\theta$ is $\dfrac{\theta}{360}\times \underline{\pi r^2}$.
Length of an arc subtending angle $\theta$ at the centre of a circle of radius $r$ is $\underline{\dfrac{\theta}{360}\times 2\pi r}$.
The area of a circle inscribed in an equilateral triangle of side $a$ is $\underline{\pi a^2/12}$.
If $\theta = 360°$, the area of the sector equals the $\underline{\text{area of the circle}}$.
If $\theta = 180°$, the sector is called a $\underline{\text{semicircle}}$.
The area of a quadrant of a circle of radius 7 cm is $\underline{38.5}$ cm$^2$.
If the diameter of a wheel is 28 cm, its circumference is $\underline{88}$ cm.
The area between two concentric circles of radii $R$ and $r$ ($R>r$) is $\underline{\pi(R^2 – r^2)}$.

C. True or False

The area of a sector is always less than the area of the corresponding circle. True (unless $\theta = 360°$, in which case they are equal).
The chord of a segment is the boundary of the segment. False — the boundary of the segment also includes the arc.
The angle described by a minute hand in one hour is $360°$. True.
If two circles have equal areas, their circumferences are equal. True.
An arc of length equal to the radius subtends a central angle of $1$ radian. True.

D. Short-Answer Questions

1. Find the area of the largest circle that can be drawn inside a rectangle of length 14 cm and breadth 7 cm.

Answer: The largest such circle has diameter equal to the smaller side $= 7$ cm, so $r = 3.5$ cm. Area $= \dfrac{22}{7}\times 12.25 = 38.5$ cm$^2$.

2. Find the area of a sector whose arc length is 11 cm and radius is 7 cm.

Answer: Use $A_{\text{sec}} = \dfrac{1}{2}r\ell = \dfrac{1}{2}\times 7\times 11 = 38.5$ cm$^2$.

3. The radii of two circles are in the ratio $3:4$. Find the ratio of their areas.

Answer: $A_1 : A_2 = \pi r_1^2 : \pi r_2^2 = 9 : 16$.

4. A piece of wire of length 22 cm is bent first into a circle and then into a square. Find the difference between the two enclosed areas.

Answer: Circle: $2\pi r = 22 \implies r = 7/2$, area $= 38.5$ cm$^2$. Square: side $= 22/4 = 5.5$ cm, area $= 30.25$ cm$^2$. Difference $= 38.5 – 30.25 = 8.25$ cm$^2$.

5. A pendulum swings through an angle of $30°$ and describes an arc of length 8.8 cm. Find the length of the pendulum.

Answer: $\ell = \dfrac{\theta}{360}\times 2\pi r \implies 8.8 = \dfrac{30}{360}\times 2\times \dfrac{22}{7}\times r = \dfrac{1}{12}\times \dfrac{44}{7}r = \dfrac{44 r}{84} = \dfrac{11 r}{21}$. So $r = \dfrac{8.8\times 21}{11} = \dfrac{184.8}{11} = 16.8$ cm.

6. A circular flower-bed of radius 14 m is surrounded by a path 1.5 m wide. Find the area of the path.

Answer: Outer radius $R = 15.5$ m, inner $r = 14$ m. Area of path $= \pi(R^2 – r^2) = \dfrac{22}{7}(240.25 – 196) = \dfrac{22}{7}\times 44.25 = \dfrac{973.5}{7} = 139.07$ m$^2$ (approx).

7. The wheel of a motor cycle has a diameter of 70 cm. How many revolutions per minute must the wheel make to keep a speed of 66 km/h?

Answer: Circumference $= \pi d = \dfrac{22}{7}\times 70 = 220$ cm $= 2.2$ m. Speed $= 66$ km/h $= 1100$ m/min. Revolutions per minute $= 1100/2.2 = 500$.

8. A chord of a circle of radius 14 cm subtends a right angle at the centre. Find the area of the minor segment. (Use $\pi = 22/7$.)

Answer: Sector area $= \dfrac{90}{360}\times \dfrac{22}{7}\times 196 = \dfrac{1}{4}\times 616 = 154$ cm$^2$. Triangle (right-isosceles legs 14): area $= \dfrac{1}{2}\times 14\times 14 = 98$ cm$^2$. Minor segment $= 154 – 98 = 56$ cm$^2$.

9. Find the area of the shaded region: a semicircle of diameter 14 cm with two equal smaller semicircles of diameter 7 cm inscribed inside on the diameter.

Answer: Big semicircle radius 7: area $= \dfrac{1}{2}\pi(49) = 77$ cm$^2$. Each small semicircle radius 3.5: area $= \dfrac{1}{2}\pi(12.25) = 19.25$ cm$^2$. Two small semicircles: $38.5$ cm$^2$. Shaded $= 77 – 38.5 = 38.5$ cm$^2$.

10. The minute hand of a clock is 12 cm long. Find the area swept by it in 35 minutes. (Use $\pi = 22/7$.)

Answer: 35 min corresponds to $\dfrac{35}{60}\times 360° = 210°$. Area $= \dfrac{210}{360}\times \dfrac{22}{7}\times 144 = \dfrac{7}{12}\times \dfrac{22\times 144}{7} = \dfrac{22\times 144}{12} = 22\times 12 = 264$ cm$^2$.

E. Long-Answer Questions

1. The inner circumference of a circular track is 264 m and width of the track is 7 m. Find the cost of putting up a fence along the outer boundary of the track at the rate of Rs 50 per metre.

Answer: Inner radius: $2\pi r = 264 \implies r = \dfrac{264\times 7}{2\times 22} = 42$ m. Outer radius $R = 42 + 7 = 49$ m. Outer circumference $= 2\pi R = 2\times \dfrac{22}{7}\times 49 = 308$ m. Cost $= 308\times 50 = $ Rs 15,400.

2. A round park of diameter 280 m has a 7 m wide path running around it. Find the cost of paving this path at Rs 30 per square metre.

Answer: Inner radius $r = 140$ m, outer radius $R = 147$ m. Area of path $= \pi(R^2 – r^2) = \dfrac{22}{7}(147^2 – 140^2) = \dfrac{22}{7}\times (21609 – 19600) = \dfrac{22}{7}\times 2009 = \dfrac{44198}{7} = 6314$ m$^2$. Cost $= 6314\times 30 = $ Rs 1,89,420.

3. A square ABCD is inscribed in a circle of radius 10 cm. Find (i) the area of the circle (ii) the area of the square (iii) the area of the four shaded segments outside the square but inside the circle.

Answer: (i) Circle: $\pi r^2 = 3.14\times 100 = 314$ cm$^2$.

(ii) Diagonal of square $= $ diameter $= 20$ cm; side $= 20/\sqrt 2 = 10\sqrt 2$. Area of square $= (10\sqrt 2)^2 = 200$ cm$^2$.

(iii) Four segments together $= $ circle $-$ square $= 314 – 200 = 114$ cm$^2$.

4. A grass plot is in the form of a circle of radius 14 m. A horse is tethered to a peg at the centre with a 7 m rope. (i) What area can the horse graze? (ii) What is the area of the part of the plot the horse cannot graze?

Answer: (i) Inner circle of radius 7: area $= \dfrac{22}{7}\times 49 = 154$ m$^2$. (ii) Plot area $= \dfrac{22}{7}\times 196 = 616$ m$^2$. Ungrazed area $= 616 – 154 = 462$ m$^2$.

5. The wheel of a railway engine has a radius of 0.5 m. How many revolutions does it make to cover a distance of 9.9 km? (Use $\pi = 22/7$.)

Answer: Circumference $= 2\pi r = 2\times \dfrac{22}{7}\times 0.5 = \dfrac{22}{7}$ m. Distance $= 9900$ m. Revolutions $= \dfrac{9900\times 7}{22} = \dfrac{69300}{22} = 3150$.

6. A square OABC of side 7 cm is inscribed in a quadrant of a circle of centre O. Find the area outside the square but inside the quadrant. (Use $\pi = 22/7$.)

Answer: Diagonal $OB = 7\sqrt 2$ cm = radius of quadrant. Quadrant area $= \dfrac{1}{4}\pi r^2 = \dfrac{1}{4}\times \dfrac{22}{7}\times 98 = 77$ cm$^2$. Square area $= 49$ cm$^2$. Required area $= 77 – 49 = 28$ cm$^2$.

7. Three circles each of radius 3.5 cm are drawn so that each touches the other two. Find the area enclosed between them.

Answer: Centres form an equilateral triangle of side $2r = 7$ cm. Triangle area $= \dfrac{\sqrt 3}{4}\times 49 = \dfrac{49\sqrt 3}{4}$. Sectors at each vertex have angle $60°$; total sectors area $= 3\times \dfrac{60}{360}\pi r^2 = \dfrac{1}{2}\pi(12.25) = \dfrac{1}{2}\times \dfrac{22}{7}\times 12.25 = 19.25$ cm$^2$. Enclosed (curved-triangle) area $= \dfrac{49\sqrt 3}{4} – 19.25 \approx 21.22 – 19.25 = 1.97$ cm$^2$.

8. A piece of wire 20 cm long is bent into the form of an arc of a circle subtending an angle of $60°$ at its centre. Find the radius of the circle.

Answer: $\ell = \dfrac{60}{360}\times 2\pi r = \dfrac{\pi r}{3}$. So $\dfrac{\pi r}{3} = 20 \implies r = \dfrac{60}{\pi} = \dfrac{60\times 7}{22} = \dfrac{420}{22} = \dfrac{210}{11}$ cm $\approx 19.09$ cm.

9. A pulley wheel of radius 14 cm has a belt wrapped around it making half a turn. Find the length of the belt in contact with the pulley.

Answer: Half a turn corresponds to $180°$. $\ell = \dfrac{180}{360}\times 2\pi r = \pi r = \dfrac{22}{7}\times 14 = 44$ cm.

10. The diameter of the front and rear wheels of a tractor are 80 cm and 200 cm respectively. Find the number of revolutions that a rear wheel makes to cover the distance which the front wheel covers in 1400 revolutions.

Answer: Front wheel circumference $= \pi\times 80 = 80\pi$ cm. Distance covered $= 1400\times 80\pi = 112000\pi$ cm. Rear wheel circumference $= 200\pi$ cm. Revolutions $= \dfrac{112000\pi}{200\pi} = 560$.

11. A path 5 m wide runs around the outside of a rectangular field 65 m long and 25 m wide. The corners of the path have been replaced by quadrants of circles of radius 5 m. Find the total area of the path.

Answer: Outer rectangle (without rounded corners) is $75\times 35 = 2625$ m$^2$, inner rectangle $65\times 25 = 1625$ m$^2$. The straight parts of the path have area $2625 – 1625 = 1000$ m$^2$ if corners were square. Replacing the four square corners (each $5\times 5 = 25$ m$^2$) with four quadrants together forming a full circle of radius 5 (area $\pi\times 25 = \dfrac{22\times 25}{7} = \dfrac{550}{7} \approx 78.57$ m$^2$) reduces the path area by $4\times 25 – \dfrac{550}{7} = 100 – 78.57 = 21.43$ m$^2$. So path area $\approx 1000 – 21.43 = 978.57$ m$^2$.

12. Two circles touch externally. The sum of their areas is $130\pi$ sq cm and the distance between their centres is 14 cm. Find the radii of the circles.

Answer: Let radii be $r_1, r_2$. Then $r_1 + r_2 = 14$ (touching externally) and $\pi(r_1^2 + r_2^2) = 130\pi$ so $r_1^2 + r_2^2 = 130$. From $(r_1+r_2)^2 = r_1^2 + r_2^2 + 2r_1 r_2$: $196 = 130 + 2r_1 r_2 \implies r_1 r_2 = 33$. So $r_1, r_2$ are roots of $x^2 – 14x + 33 = 0 \implies x = \dfrac{14\pm\sqrt{196 – 132}}{2} = \dfrac{14\pm 8}{2} = 11$ or $3$. So radii are 11 cm and 3 cm.

13. The boundary of the shaded region in the figure consists of three semicircular arcs, the smallest two being equal. If the diameter of the largest is 14 cm, find the perimeter and area of the shaded region.

Answer: Largest semicircle diameter 14 (radius 7); the two smaller equal semicircles fit on its diameter, each of diameter 7 (radius 3.5).

Perimeter $= \pi(7) + 2\times \pi(3.5) = 7\pi + 7\pi = 14\pi = 14\times \dfrac{22}{7} = 44$ cm.

Area $= \dfrac{1}{2}\pi(7)^2 – 2\times \dfrac{1}{2}\pi(3.5)^2 = \dfrac{49\pi}{2} – \dfrac{12.25\pi}{1} = \dfrac{49\pi – 24.5\pi}{2} = \dfrac{24.5\pi}{2}$. With $\pi = 22/7$: $= \dfrac{24.5\times 22}{14} = \dfrac{539}{14} = 38.5$ cm$^2$.

14. The minute hand of a clock is 10 cm long. Find the area swept by the minute hand between 9:00 a.m. and 9:35 a.m. (Use $\pi = 3.14$.)

Answer: 35 minutes corresponds to angle $\dfrac{35}{60}\times 360° = 210°$. Area $= \dfrac{210}{360}\times 3.14\times 100 = \dfrac{7}{12}\times 314 = \dfrac{2198}{12} = 183.17$ cm$^2$ approx.

15. Three semicircles each of diameter 3 cm, a circle of diameter 4.5 cm and a semicircle of radius 4.5 cm are drawn in a given figure. Find the area of the shaded region.

Answer: Big semicircle radius 4.5: area $= \dfrac{1}{2}\pi(4.5)^2 = \dfrac{1}{2}\times 3.14\times 20.25 = 31.7925$ cm$^2$. Three small semicircles (radius 1.5 each): total area $= 3\times \dfrac{1}{2}\pi(1.5)^2 = \dfrac{3}{2}\times 3.14\times 2.25 = 10.6$ cm$^2$ approx. Circle of diameter 4.5 (radius 2.25): area $= 3.14\times 5.0625 = 15.896$ cm$^2$ approx.

Shaded $\approx 31.79 + 10.6 – 15.9 \approx 26.49$ cm$^2$.

Common Mistakes Students Make

Confusing radius and diameter. If a question gives the diameter, halve it before plugging into $\pi r^2$. Many students compute $\pi d^2$ by mistake — that is FOUR times the actual area.
Forgetting the angle fraction. Sector area is $(\theta/360°) \pi r^2$, NOT $\pi r^2$. Always check the central angle.
Mixing up sector and segment. A sector is the pie-slice (two radii + arc). A segment is the chord-cap (chord + arc). They are different regions; the segment is found by subtracting a triangle from the sector.
Wrong choice of $\pi$. Use $22/7$ when radii are multiples of 7 (cleaner arithmetic). Use $3.14$ when the question explicitly states it. Never mix.
Inconsistent units. Speed in km/h must be converted to cm/min (or m/min) before computing wheel revolutions. Always check that lengths and circumferences use the same unit.
Not closing the perimeter of a sector. The perimeter of a sector is arc length PLUS the two radii — not just the arc.
Inscribed-square trap. The diagonal of an inscribed square equals the diameter (not the side). The side is therefore $r\sqrt 2$, not $2r$.
Wrong triangle in segment. When computing area of a segment with central angle $\theta$, the triangle has TWO equal sides of length $r$ (the radii), not one side equal to the chord. Use $\dfrac{1}{2}r^2\sin\theta$.

Quick-Recall Tables of Standard Values

Common circumferences and areas (with $\pi = 22/7$):

Radius	Diameter	Circumference	Area
3.5 cm	7 cm	22 cm	38.5 cm$^2$
7 cm	14 cm	44 cm	154 cm$^2$
10.5 cm	21 cm	66 cm	346.5 cm$^2$
14 cm	28 cm	88 cm	616 cm$^2$
17.5 cm	35 cm	110 cm	962.5 cm$^2$
21 cm	42 cm	132 cm	1386 cm$^2$
28 cm	56 cm	176 cm	2464 cm$^2$
35 cm	70 cm	220 cm	3850 cm$^2$

Sector areas (radius $r$, angle $\theta$) — quick fractions:

Angle $\theta$	Fraction of circle	Sector area formula	Common name
30°	1/12	$\dfrac{\pi r^2}{12}$	—
45°	1/8	$\dfrac{\pi r^2}{8}$	—
60°	1/6	$\dfrac{\pi r^2}{6}$	—
90°	1/4	$\dfrac{\pi r^2}{4}$	quadrant
120°	1/3	$\dfrac{\pi r^2}{3}$	—
180°	1/2	$\dfrac{\pi r^2}{2}$	semicircle
240°	2/3	$\dfrac{2\pi r^2}{3}$	—
270°	3/4	$\dfrac{3\pi r^2}{4}$	three-quarter
360°	1	$\pi r^2$	full circle

Worked Example: Combination of Square and Four Quadrants

Consider a square of side $a$. From each corner of the square draw a quadrant of radius $a/2$, so that the four quadrants meet at the centre of each side. The four quadrants together cover one full circle of radius $a/2$.

Area of square $= a^2$. Combined area of the four quadrants $= 4\times \dfrac{1}{4}\pi(a/2)^2 = \pi a^2/4$.

Area outside all four quadrants but inside the square (the curved central region) $= a^2 – \pi a^2/4 = a^2\left(1 – \pi/4\right)$.

For $a = 14$: square area $= 196$, four quadrants $= \dfrac{22}{7}\times 49 = 154$, central curved region $= 196 – 154 = 42$ cm$^2$.

Worked Example: Annular Ring with Two Concentric Circles

An annular ring is the region bounded by two concentric circles of radii $R$ (outer) and $r$ (inner). Three quantities of interest:

Area of ring: $\pi(R^2 – r^2)$.
Width of ring: $R – r$.
Mean (average) circumference: $\pi(R + r)$. (Useful in finding length of path along the centreline.)

If the ring corresponds to a sector of angle $\theta$ rather than a full circle (e.g. the area between two concentric arcs), use:

$$A = \frac{\theta}{360°}\pi(R^2 – r^2)$$

This appears in textbook Q14 of Exercise 12.3 with $\theta = 30°$, $R = 21$, $r = 7$.

Worked Example: Circle Inscribed in an Equilateral Triangle

If an equilateral triangle has side $a$, the radius of its inscribed circle (incircle) is $r = \dfrac{a}{2\sqrt 3}$ and the radius of its circumscribed circle is $R = \dfrac{a}{\sqrt 3}$. Note $R = 2r$.

Verification with $a = 12$ cm: incircle radius $r = \dfrac{12}{2\sqrt 3} = \dfrac{6}{\sqrt 3} = 2\sqrt 3 \approx 3.46$ cm. Circumcircle radius $R = \dfrac{12}{\sqrt 3} = 4\sqrt 3 \approx 6.93$ cm.

Area of equilateral triangle: $\dfrac{\sqrt 3}{4}a^2$. Area of incircle: $\pi r^2 = \pi\dfrac{a^2}{12}$. So fraction of triangle covered by incircle $= \dfrac{\pi a^2/12}{\sqrt 3 a^2/4} = \dfrac{\pi}{3\sqrt 3} \approx 0.6046$ — roughly 60% of the triangle.

Why the Formula $A_{\text{sec}} = \frac{1}{2}r\ell$ Works

Take a sector with radius $r$ and arc length $\ell$. As the central angle $\theta$ varies, both the area $A_{\text{sec}}$ and the arc length $\ell$ scale linearly with $\theta$:

$$\ell = \frac{\theta}{360°}\cdot 2\pi r, \qquad A_{\text{sec}} = \frac{\theta}{360°}\cdot \pi r^2$$

Dividing the second equation by the first:

$$\frac{A_{\text{sec}}}{\ell} = \frac{\pi r^2}{2\pi r} = \frac{r}{2} \quad\Longrightarrow\quad A_{\text{sec}} = \frac{1}{2}\, r\, \ell$$

This formula is helpful when the arc length is given directly, instead of the angle.

HSLC Examination Tips

Read carefully. “Find the perimeter of the sector” is different from “find the length of the arc.” Perimeter includes the two radii.
Always draw the figure. Even if the question gives a figure, redraw a rough copy in your answer sheet and label the radii, angles and chord lengths.
Express the answer in the units the question asks for. If the radii are in cm, the area is in cm$^2$; if in m, area in m$^2$. For “cost per cm$^2$,” compute area in cm$^2$, then multiply by the rate.
Decompose, don’t memorise. For shaded regions in combinations, write the area as a sum / difference of basic shapes. Then plug in the formulas.
Check by approximate $\pi$. If your answer comes out odd, redo with $\pi \approx 3.14$ to verify the order of magnitude.
Mark allocation. ASSEB Class 10 Mathematics typically gives 3-4 marks for sector / segment problems and 4-6 marks for combination problems. Show every step.
Time management. Spend at most 5-7 minutes per long question. If the calculation gets messy, reread the problem — you may have misread the angle.

Mixed Conceptual Questions

C1. Why is the area of a sector smaller than the area of the corresponding circle, except when $\theta = 360°$?

Answer: Because the sector covers only a fraction $\theta/360°$ of the full circle. As $\theta \to 360°$, the sector becomes the full circle. For $\theta < 360°$ the fraction is strictly less than 1, so the area is less.

C2. If two sectors of two different circles have the same arc length but different radii, what can you say about their areas?

Answer: Using $A_{\text{sec}} = \dfrac{1}{2}r\ell$, when $\ell$ is fixed, the area is proportional to $r$. So the sector with the larger radius has the larger area.

C3. The minute hand and hour hand of a clock are 14 cm and 6 cm long. In one full hour the minute hand sweeps how much more area than the hour hand?

Answer: Minute hand in 1 hour sweeps a full circle: area $= \pi(14)^2 = \dfrac{22}{7}\times 196 = 616$ cm$^2$. Hour hand in 1 hour sweeps $30°$: area $= \dfrac{30}{360}\pi(6)^2 = \dfrac{1}{12}\times \dfrac{22}{7}\times 36 = \dfrac{66}{7}$ cm$^2 \approx 9.43$ cm$^2$. Difference $= 616 – 9.43 = 606.57$ cm$^2$ approximately.

C4. Two circles of radii $r_1$ and $r_2$ have circumferences in the ratio $3:5$. What is the ratio of their areas?

Answer: Circumferences in ratio $3:5$ means radii in ratio $3:5$. Areas are in ratio $r_1^2 : r_2^2 = 9 : 25$.

C5. A wire bent into a circle of radius $r$ is straightened and bent into a square. Compare the enclosed areas.

Answer: Wire length $= 2\pi r$. Square side $= 2\pi r/4 = \pi r/2$. Square area $= \pi^2 r^2/4 \approx 2.467 r^2$. Circle area $= \pi r^2 \approx 3.142 r^2$. Hence circle encloses more area than the square (a general fact: among shapes of equal perimeter, the circle has the largest area — the isoperimetric inequality).

C6. State whether the following statement is true: “If a square is inscribed in a circle, the area of the square equals half the area of the circle.”

Answer: Square inscribed in circle of radius $r$ has diagonal $2r$ so side $r\sqrt 2$ and area $2r^2$. Half the circle area $= \pi r^2/2 \approx 1.571 r^2$. So the square area ($2r^2$) is bigger than half of the circle area. Statement: FALSE.

C7. The radius of a circle is increased by 10%. By what percentage does its area increase?

Answer: New radius $= 1.1 r$. New area $= \pi(1.1r)^2 = 1.21\pi r^2$. Percentage increase $= 21\%$.

C8. The circumference of a circle exceeds its diameter by $30$ cm. Find the diameter.

Answer: $\pi d – d = 30 \implies d(\pi – 1) = 30 \implies d = \dfrac{30}{22/7 – 1} = \dfrac{30\times 7}{15} = 14$ cm.

Practice Problem Set (with Solutions)

Below are 20 additional problems modelled on past ASSEB / HSLC papers. Try each one yourself before checking the solution.

P1. Find the area of a sector of a circle whose radius is 14 cm and the corresponding arc length is 22 cm.

Answer: $A_{\text{sec}} = \dfrac{1}{2}r\ell = \dfrac{1}{2}\times 14\times 22 = 154$ cm$^2$.

P2. The perimeter of a semicircular protractor is 36 cm. Find its diameter.

Answer: Perimeter $= \pi r + 2r = r(\pi + 2) = 36$. So $r = \dfrac{36}{22/7 + 2} = \dfrac{36}{36/7} = 7$ cm. Diameter = 14 cm.

P3. Two circles touch internally. The sum of their areas is $116\pi$ cm$^2$ and the distance between their centres is 6 cm. Find the radii of the two circles.

Answer: Internal touch: $r_1 – r_2 = 6$ (with $r_1 > r_2$). And $r_1^2 + r_2^2 = 116$. Substituting $r_1 = r_2 + 6$: $(r_2+6)^2 + r_2^2 = 116 \implies 2r_2^2 + 12r_2 + 36 = 116 \implies r_2^2 + 6r_2 – 40 = 0$. So $r_2 = \dfrac{-6 + \sqrt{36+160}}{2} = \dfrac{-6+14}{2} = 4$. Hence $r_1 = 10$, $r_2 = 4$ cm.

P4. The wheel of a bicycle has a diameter of 42 cm. How much distance will it cover in 200 revolutions?

Answer: Circumference $= \pi d = \dfrac{22}{7}\times 42 = 132$ cm. Distance $= 200\times 132 = 26400$ cm $= 264$ m.

P5. Find the area of the segment AYB shown in the figure, if radius of the circle is 21 cm and $\angle AOB = 120°$. (Use $\pi = 22/7$ and $\sqrt 3 = 1.73$.)

Answer: Sector area $= \dfrac{120}{360}\times \dfrac{22}{7}\times 441 = \dfrac{1}{3}\times \dfrac{9702}{7} = \dfrac{3234}{7} = 462$ cm$^2$. Triangle: $OA = OB = 21$, $\angle AOB = 120°$. Drop perpendicular from $O$: half-chord = $21\sin 60°= 21\times \dfrac{\sqrt 3}{2}$, so $AB = 21\sqrt 3$; height = $21\cos 60° = 10.5$. Area = $\dfrac{1}{2}\times 21\sqrt 3\times 10.5 = \dfrac{21\times 10.5\sqrt 3}{2} = \dfrac{220.5\times 1.73}{2} = \dfrac{381.465}{2} = 190.7$ cm$^2$. Segment $= 462 – 190.7 = 271.3$ cm$^2$.

P6. A copper wire when bent in the form of a square encloses an area of 484 cm$^2$. If the same wire is bent in the form of a circle, find the area enclosed by it. (Use $\pi = 22/7$.)

Answer: Side of square $= \sqrt{484} = 22$ cm; perimeter $= 88$ cm. Circle of circumference 88: $r = \dfrac{88\times 7}{2\times 22} = 14$ cm. Area $= \dfrac{22}{7}\times 196 = 616$ cm$^2$.

P7. Find the area of the largest triangle that can be inscribed in a semicircle of radius $r$.

Answer: Largest triangle has its base as the diameter of the semicircle ($= 2r$) and its apex on the semicircle. Maximum height equals $r$. Area $= \dfrac{1}{2}\times 2r\times r = r^2$.

P8. Find the area of a circle whose circumference is 44 m.

Answer: $2\pi r = 44 \implies r = 7$ m. Area $= \dfrac{22}{7}\times 49 = 154$ m$^2$.

P9. The diameters of the inner and outer circles of a circular ring are 14 cm and 21 cm. Find the area of the ring.

Answer: $r = 7$, $R = 10.5$. Area $= \pi(R^2 – r^2) = \dfrac{22}{7}(110.25 – 49) = \dfrac{22}{7}\times 61.25 = \dfrac{1347.5}{7} = 192.5$ cm$^2$.

P10. The hour hand of a clock is 6 cm long. What is the area swept by it in 5 hours?

Answer: Hour hand sweeps $360°$ in 12 hours, so in 5 hours it sweeps $\dfrac{5}{12}\times 360° = 150°$. Area $= \dfrac{150}{360}\times \dfrac{22}{7}\times 36 = \dfrac{5}{12}\times \dfrac{792}{7} = \dfrac{3960}{84} = \dfrac{330}{7}$ cm$^2 \approx 47.14$ cm$^2$.

P11. The area of a sector is one-sixth of the area of the circle. Find the angle of the sector.

Answer: $\dfrac{\theta}{360°} = \dfrac{1}{6} \implies \theta = 60°$.

P12. A pendulum 21 cm long swings $30°$ on either side of its mean position. Find the total length of the path described by its tip.

Answer: Total swing $= 60°$. Arc length $= \dfrac{60}{360}\times 2\pi r = \dfrac{1}{6}\times 2\times \dfrac{22}{7}\times 21 = \dfrac{1}{6}\times 132 = 22$ cm.

P13. The area enclosed between two concentric circles is $770$ cm$^2$. If the radius of the outer circle is 21 cm, find the radius of the inner circle.

Answer: $\pi(R^2 – r^2) = 770 \implies R^2 – r^2 = \dfrac{770\times 7}{22} = 245$. With $R = 21$: $441 – r^2 = 245 \implies r^2 = 196 \implies r = 14$ cm.

P14. The radius of a circular wheel is 1.75 m. How many revolutions will it make to cover 11 km?

Answer: Circumference $= 2\pi r = 2\times \dfrac{22}{7}\times 1.75 = 11$ m. Revolutions $= \dfrac{11000}{11} = 1000$.

P15. Four equal circles each of radius 7 cm touch each other (externally). Find the area between them (the curved-square region in the middle).

Answer: Centres form a square of side 14 (each side = sum of two radii). Area of square $= 196$ cm$^2$. Inside the square at each corner is a $90°$ sector of radius 7; together they form one full circle of radius 7. Area of full circle $= \dfrac{22}{7}\times 49 = 154$ cm$^2$. Curved-square area $= 196 – 154 = 42$ cm$^2$.

P16. A wire is bent into the shape of an equilateral triangle of side 6.6 cm. The same wire is rebent into a circular form. Find the area of the circle (in terms of $\pi$).

Answer: Length of wire $= 3\times 6.6 = 19.8$ cm. Circumference $= 19.8 \implies r = \dfrac{19.8}{2\pi} = \dfrac{9.9}{\pi}$. Area $= \pi r^2 = \pi\times \dfrac{98.01}{\pi^2} = \dfrac{98.01}{\pi}$ cm$^2$. Numerically $= \dfrac{98.01\times 7}{22} = \dfrac{686.07}{22} \approx 31.19$ cm$^2$.

P17. The wheels of a car make 1000 revolutions in covering 88 km. Find the radius of the wheels.

Answer: Distance per revolution $= \dfrac{88\times 1000}{1000} = 88$ m. So circumference $= 88$ m. $2\pi r = 88 \implies r = 14$ m.

P18. The radius of a circular garden is 56 m. Find the cost of fencing it at the rate of Rs 5 per metre.

Answer: Circumference $= 2\pi r = 2\times \dfrac{22}{7}\times 56 = 352$ m. Cost $= 352\times 5 = $ Rs 1760.

P19. A round metal sheet of diameter 28 cm is cut into 16 equal sectors. Find the area of each sector.

Answer: Total area $= \dfrac{22}{7}\times 14^2 = 616$ cm$^2$. Each sector area $= \dfrac{616}{16} = 38.5$ cm$^2$.

P20. The minute hand of a clock is 14 cm long. Through what total distance does its tip move from 6:00 a.m. to 6:30 a.m.?

Answer: 30 min $= 180°$. Arc $= \dfrac{180}{360}\times 2\pi r = \pi r = \dfrac{22}{7}\times 14 = 44$ cm.

Glossary / Key Terms

Term	Definition
Circle	The set of points in a plane at a fixed distance (radius) from a fixed point (centre).
Radius ($r$)	The distance from the centre of the circle to any point on the circle.
Diameter ($d$)	A chord passing through the centre; equal to $2r$.
Chord	A line segment with both endpoints on the circle.
Arc	A connected portion of the circle.
Sector	The region bounded by two radii and the arc joining their endpoints.
Segment	The region bounded by a chord and an arc.
Minor / Major	The smaller / larger of the two regions formed by an arc or a chord.
Semicircle	Half of a circle (sector or segment with $\theta = 180°$).
Quadrant	One quarter of a circle ($\theta = 90°$).
Annular ring	The region between two concentric circles.
Concentric circles	Circles having the same centre but different radii.
Circumference	The total length around the circle, $C = 2\pi r$.
Pi ($\pi$)	The ratio circumference / diameter; $\pi \approx 22/7 \approx 3.14159$.
Length of arc	$\ell = (\theta/360°)\cdot 2\pi r$.
Area of sector	$A = (\theta/360°)\cdot \pi r^2 = \tfrac{1}{2}r\ell$.
Combination of figures	A region built by adding or subtracting basic shapes (circle, square, triangle, etc.).
Race-track	A combination of a rectangle and two semicircles attached to opposite shorter sides.
Inscribed circle	A circle drawn inside a polygon, touching all the sides.
Circumscribed circle	A circle drawn outside a polygon, passing through all the vertices.

Quick Reference Card

If you know	You can find	Formula
Radius $r$	Diameter $d$	$d = 2r$
Diameter $d$	Radius $r$	$r = d/2$
Radius $r$	Circumference $C$	$C = 2\pi r$
Diameter $d$	Circumference $C$	$C = \pi d$
Circumference $C$	Radius $r$	$r = C/(2\pi)$
Radius $r$	Area $A$	$A = \pi r^2$
Area $A$	Radius $r$	$r = \sqrt{A/\pi}$
$r$ and $\theta$	Arc length $\ell$	$\ell = (\theta/360°)\,2\pi r$
$r$ and $\theta$	Sector area	$A_{\text{sec}} = (\theta/360°)\,\pi r^2$
$r$ and $\ell$	Sector area	$A_{\text{sec}} = \tfrac{1}{2}r\ell$
Sector and triangle	Segment area	$A_{\text{seg}} = A_{\text{sec}} – A_\triangle$
$R$ and $r$	Annular ring area	$\pi(R^2 – r^2)$
$r$ (incircle)	Side of equilateral $\triangle$	$a = 2\sqrt 3 \, r$
$R$ (circumcircle)	Side of equilateral $\triangle$	$a = R\sqrt 3$
Side $a$ of square	Diagonal	$a\sqrt 2$
Diagonal $D$ of square	Side	$a = D/\sqrt 2$

Bridging to Chapter 13

The two formulas $C = 2\pi r$ and $A = \pi r^2$ that you have practised throughout this chapter reappear in nearly every problem of Chapter 13 — Surface Areas and Volumes:

Cylinder: Curved surface = $2\pi r h$ (circumference times height); Total surface = $2\pi r(h + r)$; Volume = $\pi r^2 h$ (area of base times height).
Cone: Curved surface = $\pi r \ell$ (where $\ell$ is the slant height — same letter as arc length, but a different quantity); Total surface = $\pi r(\ell + r)$; Volume = $\dfrac{1}{3}\pi r^2 h$.
Sphere: Surface area = $4\pi r^2$ (exactly four times the area of a “great circle”); Volume = $\dfrac{4}{3}\pi r^3$.
Hemisphere: Curved surface = $2\pi r^2$; Total surface = $3\pi r^2$; Volume = $\dfrac{2}{3}\pi r^3$.

So the time you invest in mastering Chapter 12 pays back doubly when you reach Chapter 13. Make sure you can recall every formula in this Quick Reference card without looking.

End of Chapter 12 — Areas Related to Circles. For Chapter 13 (Surface Areas and Volumes) the same circle formulas $C = 2\pi r$, $A = \pi r^2$ reappear inside cylinders, cones and spheres, so this chapter is a direct stepping-stone. Practice every solved example, verify the units (cm$^2$ vs m$^2$ vs km$^2$), and remember to use $\pi = 22/7$ unless the question specifies $\pi = 3.14$. Wishing you the very best for your HSLC examination from all of us at HSLC Guru.