CBSE Class 10 Mathematics Areas Related To Circles Worksheet Set 05

Question. If the area of circle is numerically equal to twice its circumference, then the diameter of the circle is
(a) 4 units
(b) 6 units
(c) 8 units
(d) 12 units

Answer: C

Question. The area of a quadrant of a circle whose circumference is 25 cm is
(a) 24 cm²
(b) 28 cm²
(c) 32.5 cm²
(d) 38.5 cm²

Answer: D

Question. If area of quadrant of a circle is 38.5 cm², then its diameter is [Use π = 22/7]
(a) 10 cm
(b) 14 cm
(c) 21 cm
(d) None of the options

Answer: B

Question. The outer diameter and the inner diameter of a circular path are 728m and 700 m respectively. Find the area of the 22 circular path. (Use π = 22/7).   
(a) 45260 m²
(b) 25012 m²
(c) 31416 m²
(d) 19541 m²

Answer: C

Question. If the circumference of a circle is increased by 50%, by what percent will its area be increased?
(a) 75%
(b) 100%
(c) 125%
(d) 150%

Answer: C

Question. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. The length of the arc is
(a) 11 cm
(b) 22 cm
(c) 27 cm
(d) 44 cm

Answer: B

Question. In a circle of diameter 42 cm,if an arc subtends an angle of 60° at the centre where π = 22/ 7 , then what will be the length of arc ?
(a) 11 cm
(b) 20 cm
(c) 22 cm
(d) 28 cm

Answer: C

Question. 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?
(a) 2275
(b) 2650
(c) 3815
(d) 4375

Answer: D

Question. The cost of fencing a circular field at the rate of ₹ 24 per metre is ₹ 5280. The radius of the field is
(a) 15 m
(b) 35 m
(c) 25 m
(d) 30 m

Answer: B

Question. A road which is 7 m wide surrounds a circular park whose circumference is 88 m. The area of the road is
(a) 220 m²
(b) 340 m²
(c) 550 m²
(d) 770 m²

Answer: D

Question. Four equal circles are described about the four corners of a square so that each of them touches two of the others. If each side of the square measures 14 cm, find the area of the remaining portion of the square apart from four circles in cm2 .
(a) 20
(b) 24
(c) 42
(d) 40

Answer: C

Question. The short and long hands of a clock are 4 cm and 6 cm long respectively. The sum of distances travelled by their tips in 2 days is
(a) 1148 cm
(b) 1426.35 cm
(c) 1910.85 cm
(d) None of the options

Answer: C

Question. The area of the circle, the circumference of which is equal to the perimeter of a square of side 11 cm is   
(a) 122 cm²
(b) 144 cm²
(c) 154 cm²
(d) 180 cm²

Answer: C

Question. What is the diameter of a circle whose area is equal to the sum of the areas of the two circles of radii 24 cm and 7 cm?
(a) 20 cm
(b) 30 cm
(c) 50 cm
(d) 80 cm

Answer: C

Question. How many plants will be there in a circular bed whose outer edge measures 30 cm allowing 4 cm2 for each plant?
(a) 18
(b) 750
(d) 24
(d) 120

Answer: A

Question. In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Choose the correct choice as:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.

Question. Assertion (A): If the circumference of a circle is 176 cm, then its radius is 28 cm.
Reason (R): Circumference = 2π × radius.

Answer: A

Question. Assertion (A): If the outer and inner diameter of a circular path is 10 m and 6 m respectively, then area of the path is 16 π m².
Reason (R): If R and r be the radius of outer and inner circular path respectively, then area of circular path = π (R² – r²)

Answer: A

Question. The area of a sector of a circle with radius 6 cm if angle of the sector is 60° is   
(a) 15, 2/ 3 cm²
(b) 16, 1/ 2 cm²
(c) 18, 6/ 7 cm²
(d) 19, 3/8  cm

Answer: C

Question. In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Choose the correct choice as:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.

Question. Assertion (A): In a circle of radius 6 cm, the angle of a sector is 60°. Then the area of the sector is 18, 6/ 7 cm².
Reason (R): Area of the circle with radius r is πr².

Answer: B

Question. Assertion (A): The length of the minute hand of a clock is 7 cm. Then the area swept by the minute hand in 5 minute is 12, 5/ 6 cm².
Reason (R): The length of an arc of a sector of angle q and radius r is given by l = θ / 360 ° × 2πr 

Answer: B

Question. The cost of fencing a circular field at the rate of ₹ 24 per metre is ₹ 5280. The field is to be ploughed at the rate of ₹ 0.50 per m². The cost of ploughing the field is [Take π = 22/ 7]
(a) ₹ 1925
(b) ₹ 1650
(c) ₹ 2010
(d) ₹ 2525

Answer: A

Question. A piece of wire 22 cm long is bent into the form of an arc of a circle subtending an angle of 60° at its centre. The radius of the circle is [Take π = 7/22 ]
(a) 7 cm
(b) 14 cm
(c) 21 cm
(d) 28 cm

Answer: C

Question. The area of a ring shaped region enclosed between two concentric circles of radii20 cm and 15 cm is   
(a) 330 cm²
(b) 415 cm²
(c) 520 cm²
(d) 550 cm²

Answer: D

Question. Each wheel of a car makes 5 revolutions per second. If the diameter of a wheel is 84 cm, find the speed of the car in km/h.
(a) 48 km / h
(b) 32 km / h
(c) 41 km / h
(d) 25 km / h

Answer: A

Question. All the vertices of a rhombus lie on a circle. The area of the rhombus, if the area of the circle is 1256 cm2 is [Use π = 3.14]
(a) 300 cm²
(b) 600 cm²
(c) 800 cm²
(d) 900 cm²

Answer: C

Question. If the perimeter of a circle is equal to that of a square, then the ratio of their areas is
(a) 22 : 7
(b) 14 : 11
(c) 7 : 22
(d) 11 : 14

Answer: B

Question. The angle described by a minute hand in 5 minutes is
(a) 30°
(b) 60°
(c) 90°
(d) None of the options

Answer: A

Question. The area of circle whose circumference is 22 cm is
(a) 32/ 2 cm²
(b) 45/ 2 cm²
(c) 55/ 2 cm²
(d) 77/2 cm²

Answer: D

Question. In the given figure, PQ = 24 cm, PR = 7 cm and O is the centre of the circle. The area of the shaded portion is   
(a) 132.58 cm²
(b) 148.20 cm²
(c) 154.36 cm²
(d) 161.54 cm²

Answer: D

Question 1. If the diameter of circle be d, then area of circle is
(A) \( \pi d^2 \)
(B) \( \pi d/2^2 \)
(C) \( \frac{\pi d^2}{4} \)
(D) None of these
Answer: (C) \( \frac{\pi d^2}{4} \)
In simple words: The area of a circle is normally \( \pi r^2 \). Since the radius is half of the diameter \( \left(r = \frac{d}{2}\right) \), substituting this gives the area as \( \frac{\pi d^2}{4} \).

Exam Tip: Remember to square both the numerator and the denominator when substituting \( \frac{d}{2} \) for \( r \).

Question 2. Radius of a circle is 42 cm. An arc subtends an angle of 60°. What is the length of arc ?
(A) 22 cm
(B) 44 cm
(C) 66 cm
(D) 88 cm
Answer: (B) 44 cm
In simple words: The length of the arc is \( \frac{60}{360} \) of the circle's full perimeter, which works out to exactly 44 cm.

Exam Tip: Use the standard formula \( \text{Arc Length} = \frac{\theta}{360^\circ} \times 2\pi r \) and simplify the fraction first before multiplying the remaining terms.

Question 3. Radii of two concentric circle is 7 cm and 14 cm. The shaded area between them is equal to
(A) \( 42\pi \)
(B) \( 84\pi \)
(C) \( 21\pi \)
(D) None of these
Answer: (D) None of these

In simple words: The area of the outer larger circle is \( 196\pi \) and the inner circle is \( 49\pi \). Subtracting the smaller circle's area leaves \( 147\pi \), which is not listed in the choices.

Exam Tip: The area of a ring between concentric circles is calculated using the formula \( \text{Area} = \pi(R^2 - r^2) \).

Question 4. The area of a sector of a circle is \( \frac{1}{6} \) th of area of circle what is angle made by sector?
(A) 80°
(B) 70°
(C) 60°
(D) None of these
Answer: (C) 60°
In simple words: Since a complete circle is 360 degrees, a sector that is \( \frac{1}{6} \) of the total area must have a central angle of \( 60^\circ \).

Exam Tip: Equate \( \frac{\theta}{360^\circ} = \frac{1}{6} \) and solve directly to get the angle of the sector.

Question 5. The circumference of a circle exceedsits diameter by 16.8 cm. Find the radius of circle.
Answer: Let \( r \) be the radius of the circle.
The circumference of the circle is \( 2\pi r \) and its diameter is \( 2r \).
According to the given condition:
\( 2\pi r - 2r = 16.8 \)
\( 2r(\pi - 1) = 16.8 \)
\( 2r\left(\frac{22}{7} - 1\right) = 16.8 \)
\( 2r\left(\frac{15}{7}\right) = 16.8 \)
\( \frac{30r}{7} = 16.8 \)
\( r = \frac{16.8 \times 7}{30} = 3.92\text{ cm} \)
Therefore, the radius of the circle is 3.92 cm.
In simple words: We set up the relation between the perimeter and diameter to solve for the radius, yielding 3.92 cm.

Exam Tip: Keep \( \pi \) as \( \frac{22}{7} \) to simplify fractional terms when solving linear equations for the radius.

Question 6. A sector is cut off from a circle of radius 28 cm. The angle of the sector is 120 degrees.Find the length of arc and area.
Answer: Let the radius of the circle be \( r = 28\text{ cm} \) and the sector angle be \( \theta = 120^\circ \).
The length of the arc is:
\( \text{Length of arc} = \frac{\theta}{360^\circ} \times 2\pi r \)
\( \text{Length of arc} = \frac{120^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 28 \)
\( \text{Length of arc} = \frac{1}{3} \times 176 = \frac{176}{3} \approx 58.67\text{ cm} \)
The area of the sector is:
\( \text{Area} = \frac{\theta}{360^\circ} \times \pi r^2 \)
\( \text{Area} = \frac{120^\circ}{360^\circ} \times \frac{22}{7} \times 28 \times 28 \)
\( \text{Area} = \frac{1}{3} \times 22 \times 4 \times 28 = \frac{2464}{3} \approx 821.33\text{ cm}^2 \)
In simple words: The arc length is approximately 58.67 cm, and the sector's total area is around 821.33 sq. cm.

Exam Tip: Always state your final answers with both fractional values and their decimal approximations to secure complete marks.

Question 7. The length of minute hand of a clock is 21 cm. Find the area swept by the clock in 2 minutes.
Answer: The length of the minute hand is the radius \( r = 21\text{ cm} \).
In 60 minutes, the minute hand completes \( 360^\circ \).
In 1 minute, the angle swept is \( \frac{360^\circ}{60} = 6^\circ \).
In 2 minutes, the angle swept is \( \theta = 2 \times 6^\circ = 12^\circ \).
The area swept is:
\( \text{Area} = \frac{\theta}{360^\circ} \times \pi r^2 \)
\( \text{Area} = \frac{12^\circ}{360^\circ} \times \frac{22}{7} \times 21 \times 21 \)
\( \text{Area} = \frac{1}{30} \times 22 \times 3 \times 21 = \frac{1386}{30} = 46.2\text{ cm}^2 \)
Therefore, the area swept is 46.2 sq. cm.
In simple words: The minute hand turns by 12 degrees in 2 minutes, covering a surface area of 46.2 sq. cm.

Exam Tip: Note that every minute on a clock dial corresponds to a central angle sweep of exactly \( 6^\circ \).

Question 8. A sheet of paper is in the form of rectangle ABCD in which AB =40 cm and AD = 28 cm . A semi circle portion with BC as diameter is cut off. Find the area of the remaining part of the rectangle.
Answer: The rectangle has dimensions \( AB = 40\text{ cm} \) and \( AD = 28\text{ cm} \).
Area of the rectangle \( ABCD = AB \times AD = 40 \times 28 = 1120\text{ cm}^2 \).
Since \( BC = AD = 28\text{ cm} \), the semicircle has a diameter of 28 cm, making its radius \( r = 14\text{ cm} \).
Area of the semicircle to be cut off is:
\( \text{Area of semicircle} = \frac{1}{2} \pi r^2 = \frac{1}{2} \times \frac{22}{7} \times 14 \times 14 = 11 \times 28 = 308\text{ cm}^2 \).
The area of the remaining part is:
\( \text{Remaining Area} = \text{Area of rectangle} - \text{Area of semicircle} = 1120 - 308 = 812\text{ cm}^2 \).
Therefore, the remaining area is 812 sq. cm.
In simple words: Subtracting the semicircle area (308 sq. cm) from the rectangle's total area (1120 sq. cm) leaves a remaining area of 812 sq. cm.

Exam Tip: Correctly identify which side of the rectangle serves as the diameter of the semicircle to establish the correct radius.

Question 9. A play ground has the shape of a rectangle, with two semi-circles on its smaller sides as diameter, added to its outside. If the sides of the rectangle are 36m and 24m, find the area of the playground.
Answer: The rectangle has length \( 36\text{ m} \) and width \( 24\text{ m} \).
The smaller sides of the rectangle are \( 24\text{ m} \) in length, which serve as the diameters of the two semicircles.
Radius of each semicircle is \( r = \frac{24}{2} = 12\text{ m} \).
The two semicircles combined form one complete circle of radius \( 12\text{ m} \).
Area of the rectangle \( = 36 \times 24 = 864\text{ m}^2 \).
Area of the two semicircles \( = \pi r^2 = 3.14 \times 12^2 = 3.14 \times 144 = 452.16\text{ m}^2 \).
Total area of the playground is:
\( \text{Total Area} = \text{Area of rectangle} + \text{Area of semicircles} = 864 + 452.16 = 1316.16\text{ m}^2 \).
(Or, using \( \pi = \frac{22}{7} \), the area is \( 864 + 452.57 = 1316.57\text{ m}^2 \)).

In simple words: The playground is made up of a central rectangle plus a full circle split on both ends, which gives a total combined area of 1316.16 sq. m.

Exam Tip: Combining two identical semicircles into one full circle simplifies the calculation of their total area.

Question 10. A chord of a circle of radius 6 cm, subtends an angle 60° with the center A then the area of corresponding segment of the circle is
(A) \( 3.27\text{ cm}^2 \)
(B) \( 6\text{ cm}^2 \)
(C) \( 6.8\text{ cm}^2 \)
(D) None of these
Answer: (A) \( 3.27\text{ cm}^2 \)
In simple words: Subtracting the area of the equilateral triangle formed at the center (\( 15.59\text{ cm}^2 \)) from the sector's total area (\( 18.85\text{ cm}^2 \)) leaves the segment area as \( 3.27\text{ cm}^2 \).

Exam Tip: Remember that when the central angle is \( 60^\circ \), the triangle formed by the radii and the chord is always an equilateral triangle with area \( \frac{\sqrt{3}}{4}r^2 \).

Question 11. A car has two wipers which do not overlap. Each wiper has a blade of length, 25 cm sweeping through an angle of 115º. Find the total area cleaned at each sweep of the blades
(A) \( 478.29\text{ cm}^2 \)
(B) \( 627.48\text{ cm}^2 \)
(C) \( 444\text{ cm}^2 \)
(D) \( 528\text{ cm}^2 \)
Answer: (D) None of these (or B for a single wiper)
In simple words: The area swept by one blade is \( 627.48\text{ cm}^2 \). For both non-overlapping blades, the total cleaned area is \( 2 \times 627.48 = 1254.96\text{ cm}^2 \). Since this is not in the options, the correct choice is None of these.

Exam Tip: Read carefully whether the question asks for the area of a single wiper or the total area of both wipers combined.

Question 12. The wheel of car is of radius 40 cm each. How many complete revolution does each wheel makes in 20 sec. If the speed of car is 22m/s ?
(A) 500
(B) 150
(C) 175
(D) None of these
Answer: (C) 175
In simple words: The car covers 44,000 cm in 20 seconds. Dividing this by the wheel's circumference shows it takes exactly 175 full turns.

Exam Tip: Ensure all variables are converted to a single unit system (such as meters to centimeters) before performing division.

Question 13. The length of the minute hand of a clock is 28 cm, and then what is the area swept by the minute hand in 15 minutes?
(A) \( 308\text{ cm}^2 \)
(B) \( 784\text{ cm}^2 \)
(C) \( 896\text{ cm}^2 \)
(D) \( 616\text{ cm}^2 \)
Answer: (D) \( 616\text{ cm}^2 \)
In simple words: A quarter-hour sweep is a perfect 90-degree quadrant. Calculating its area with a 28 cm radius gives exactly 616 sq. cm.

Exam Tip: A 15-minute interval always corresponds to exactly one quadrant (\( \frac{1}{4} \)) of a complete circular clock face.

Question 14. The area of sector OAPB is
(A) \( \frac{\pi r^{2\theta}}{360^\circ} \)
(B) \( \frac{2\pi r^{2\theta}}{180^\circ} \)
(C) \( \frac{\pi r^2 \theta}{360^\circ} \)
(D) None of these
Answer: (C) \( \frac{\pi r^2 \theta}{360^\circ} \)

In simple words: The standard formula for the area of a sector with central angle \( \theta \) is \( \frac{\theta}{360^\circ} \pi r^2 \), which is given by Option (C).

Exam Tip: Be careful not to confuse the formulas for sector area and arc length during exams.

Question 15. The length of arc APB is
(A) \( \frac{2\pi r \theta}{360^\circ} \)
(B) \( \frac{\pi r^2 \theta}{360^\circ} \)
(C) \( \frac{2\pi r^2 \theta}{360^\circ} \)
(D) None of these
Answer: (A) \( \frac{2\pi r \theta}{360^\circ} \)
In simple words: The perimeter portion is a fraction \( \frac{\theta}{360^\circ} \) of the total circumference \( 2\pi r \), which is represented by Option (A).

Exam Tip: Arc length is a linear dimension (with \( r \)), whereas area is a squared dimension (with \( r^2 \)).

Question 16. The cost of fencing a circular field at the rate of Rs 12 meter is Rs 2640 find the area of circle
(A) \( 3300\text{ m}^2 \)
(B) \( 3850\text{ m}^2 \)
(C) \( 4400\text{ m}^2 \)
(D) \( 5500\text{ m}^2 \)
Answer: (B) \( 3850\text{ m}^2 \)
In simple words: Dividing the total cost by the rate gives a perimeter of 220 m. This corresponds to a 35 m radius, yielding a total circle area of 3850 sq. m.

Exam Tip: Always work out the perimeter from the total fencing cost first to find the radius of the circular boundary.

Question 17. If the radius of the given figure is 4 cm and angle is 60° the area of segment OAPB is
(A) \( \frac{2}{5}\pi \)
(B) \( \frac{8}{3}\pi \)
(C) \( \frac{4}{3}\pi \)
(D) \( 4\pi \)
Answer: (B) \( \frac{8}{3}\pi \)
In simple words: The area of the sector is \( \frac{60}{360} \times \pi \times 4^2 = \frac{8}{3}\pi \), which is given by Option (B). Note that "segment" in the question text is a typo for "sector".

Exam Tip: If the question asks for the sector's total area, calculate it directly; the minor segment itself would be \( \frac{8}{3}\pi - 4\sqrt{3} \).

Question 18. The radii of two circle are 3 cm and 4cm. Then the radius of the circle having area equal to the sum of the areas of two circles is equal to
(A) 5 cm
(B) 7cm
(C) 8 cm
(D) none of these
Answer: (A) 5 cm
In simple words: According to Pythagoras-like relation for circle areas, \( R^2 = 3^2 + 4^2 = 25 \), so the final radius is 5 cm.

Exam Tip: When the sum of areas of two circles is equal to a third circle's area, the radii satisfy the relation \( R^2 = r_1^2 + r_2^2 \).

Question 19. The ratio of the areas of the incircle and circumcircle of a square is
(A) 1: \( \sqrt{2} \)
(B) 1: \( \sqrt{3} \)
(C) 1:9
(D) 1:2
Answer: (D) 1:2
In simple words: The ratio of the square of the inradius to the square of the circumradius is exactly \( \frac{1}{2} \), which gives an area ratio of 1:2.

Exam Tip: The diagonal of any square is \( \sqrt{2} \) times its side length. This factor of \( \sqrt{2} \) yields an exact doubling of the area of the outer circle compared to the inner one.

Question 20. A horse is tied to a peg at center of a field by 10m long rope, then what is the area of grass grazed by the horse ?
(A) \( 50 \pi\text{ m}^2 \)
(B) \( 75 \pi\text{ m}^2 \)
(C) \( 100 \pi\text{ m}^2 \)
(D) \( 150 \pi\text{ m}^2 \)
Answer: (C) \( 100 \pi\text{ m}^2 \)
In simple words: The horse can graze over a full circle of radius 10 m, covering a total surface area of \( 100\pi \) sq. m.

Exam Tip: Since the peg is situated exactly at the center of the field, the grazing path forms a complete circle rather than a quadrant sector.

Question 21. Find the area of shaded part of given circle if one side of given square be 10 cm. [take \( \pi \) = 3.14]
(A) 57 cm²
(B) 114cm²
(C) 171 cm²
(D) 228 cm²
Answer: (A) 57 cm²

In simple words: The diagonal of the 10 cm square is the diameter of the circle, making the circle area 157 sq. cm. Subtracting the square's area (100 sq. cm) leaves 57 sq. cm.

Exam Tip: The diagonal of an inscribed square of side \( a \) is always equal to the diameter \( 2r \) of the surrounding circle.

Question 22. In the given figure, if the side of square ABCD is 2 cm. Then what is the area of shaded region ?
(A) 2 - \( \pi \)
(B) 7 - \( \pi \)
(C) 4 - \( \pi \)
(D) 8 - \( \pi \)
Answer: (C) 4 - \( \pi \)

In simple words: The area of the 2 cm square is 4 sq. cm. Subtracting the area of the inscribed circle of radius 1 cm (\( \pi \) sq. cm) leaves exactly \( 4 - \pi \) sq. cm.

Exam Tip: The diameter of a circle inscribed inside a square is equal to the side length of the square.

Question 23. An umbrella has 8 ribs which are equally spaced. Assuming umbrella to be a flat circle of radius 42 cm. The area between the two consecutive ribs of the umbrella is equal to
(A) 660 cm²
(B) 693 cm²
(C) 770 cm²
(D) 774 cm²
Answer: (B) 693 cm²
In simple words: The total area of the flat circle is 5544 sq. cm. Dividing this into 8 equal sectors yields exactly 693 sq. cm for each section.

Exam Tip: Since there are 8 ribs, the flat circle is divided into 8 equal sectors. Solve by dividing the total area \( \pi r^2 \) by 8.

Question 24. What is the area between two concentric circles of radii 21 cm and 7cm?
(A) 1156 cm²
(B) 1232 cm²
(C) 1330 cm²
(D) 1460 cm²
Answer: (B) 1232 cm²
In simple words: The area of the outer ring is \( \pi(21^2 - 7^2) = \frac{22}{7} \times 392 \), which works out to exactly 1232 sq. cm.

Exam Tip: Use the algebraic identity \( R^2 - r^2 = (R-r)(R+r) \) to make the multiplication steps much faster and easier.

Question 25. The length of segment APB is if angle is 30 degree and radius of circle is 6 cm
(A) \( \pi \)
(B) \( 2\pi \)
(C) \( 3\pi \)
(D) \( 4\pi \)
Answer: (A) \( \pi \)

In simple words: The central angle is \( 30^\circ \), which is \( \frac{1}{12} \) of the circle. This gives an arc length of \( \frac{1}{12} \times 2\pi \times 6 = \pi\text{ cm} \).

Exam Tip: Note that "segment APB" in the question text refers to the circular arc length APB.

Question 26. An arc makes an angle of 72° at the center of a circle of radius 10 cm . its length will be .
(A) \( 4\pi\text{ cm} \)
(B) \( 8\pi\text{ cm} \)
(C) \( 6\pi \)
(D) \( 7\pi \)
Answer: (A) \( 4\pi\text{ cm} \)
In simple words: The arc length is \( \frac{72}{360} \times 2\pi \times 10 = \frac{1}{5} \times 20\pi = 4\pi\text{ cm} \).

Exam Tip: Simplify the fraction \( \frac{72}{360} \) to \( \frac{1}{5} \) first to make the calculations quick and straightforward.

Question 27. What is the area of a sector of a circle which radius is 5 cm and angle of the sector is 72° ?
(A) \( 25\pi \)
(B) \( 20\pi \)
(C) \( 15\pi \)
(D) \( 5\pi \)
Answer: (D) \( 5\pi \)
In simple words: The sector area is \( \frac{72}{360} \times \pi \times 5^2 = \frac{1}{5} \times 25\pi = 5\pi\text{ cm}^2 \).

Exam Tip: Use the standard formula \( \text{Area} = \frac{\theta}{360^\circ} \times \pi r^2 \) for any sector area calculation.

Question 28. If ABCD is a square which each side is 28 cm, then the area of shaded region is
(A) 42 cm²
(B) 84 cm²
(C) 168 cm²
(D) 336 cm²
Answer: (C) 168 cm²

In simple words: The area of the square is 784 sq. cm. Subtracting the combined area of the four circles (616 sq. cm) leaves 168 sq. cm for the shaded region.

Exam Tip: Each of the four identical inscribed circles has a diameter equal to half of the square's side length.

Question 29. If a wire is bent into the shape of square, the area of square is 196 cm². When the wire is bent into circular shape, the area of circle is
(A) 233 cm²
(B) 626cm²
(C) 154 cm²
(D) none of these
Answer: (D) none of these
In simple words: The wire has a total length of 56 cm. Bending this into a circle yields a radius of \( \frac{98}{11}\text{ cm} \), giving an area of \( \approx 249.45\text{ cm}^2 \), which is not listed.

Exam Tip: The total length of the wire remains constant, meaning the perimeter of the square is equal to the circumference of the circle.

Question 30. Area of circle inscribe in a equilateral triangle is 462 sq cm. The perimeter of the triangle is.
(A) \( 42\sqrt{3}\text{ cm} \)
(B) 126 cm
(C) 72.6cm
(D) 168 cm
Answer: (B) 126 cm
In simple words: The inradius is \( 7\sqrt{3}\text{ cm} \). For an equilateral triangle, this gives a side length of 42 cm, making the total perimeter \( 3 \times 42 = 126\text{ cm} \).

Exam Tip: The inradius \( r \) of an equilateral triangle with side length \( a \) is calculated using the formula \( r = \frac{a}{2\sqrt{3}} \).

Question 31. The area of a circle inscribe in a equilateral triangle is 154sqcm. Find the perimeter of the triangle.
Answer: Let \( r \) be the radius of the inscribed circle.
The area of the inscribed circle is \( 154\text{ cm}^2 \):
\( \pi r^2 = 154 \)
\( \frac{22}{7} r^2 = 154 \)
\( r^2 = \frac{154 \times 7}{22} = 49 \)
\( r = 7\text{ cm} \)
The inradius of an equilateral triangle with side length \( a \) is:
\( r = \frac{a}{2\sqrt{3}} \)
\( 7 = \frac{a}{2\sqrt{3}} \implies a = 14\sqrt{3}\text{ cm} \)
The perimeter of the equilateral triangle is:
\( \text{Perimeter} = 3a = 3 \times 14\sqrt{3} = 42\sqrt{3}\text{ cm} \approx 72.73\text{ cm} \)
Therefore, the perimeter is \( 42\sqrt{3}\text{ cm} \) (or approximately 72.73 cm).
In simple words: The circle's radius is 7 cm. This means each side of the triangle is \( 14\sqrt{3}\text{ cm} \), giving a total perimeter of \( 42\sqrt{3}\text{ cm} \).

Exam Tip: Expressing side lengths in terms of surds (\( \sqrt{3} \)) is the standard format required in most board examinations.

Question 32. A park is in the form of a rectangle 120m x 100m. At the center of the park there is a circular lawn. The area of park excluding lawn is 8700 square metres. Find the radius of the lawn.
Answer: The dimensions of the rectangular park are \( 120\text{ m} \times 100\text{ m} \).
Total area of the rectangular park \( = 120 \times 100 = 12000\text{ m}^2 \).
The area of the park excluding the circular lawn is \( 8700\text{ m}^2 \).
Therefore, the area of the circular lawn is:
\( \text{Area of lawn} = 12000 - 8700 = 3300\text{ m}^2 \).
Let \( r \) be the radius of the circular lawn:
\( \pi r^2 = 3300 \)
\( \frac{22}{7} r^2 = 3300 \)
\( r^2 = \frac{3300 \times 7}{22} = 150 \times 7 = 1050 \)
\( r = \sqrt{1050} \approx 32.4\text{ m} \)
Therefore, the radius of the lawn is approximately 32.4 m.
In simple words: Subtracting the outer area leaves 3300 sq. m for the lawn. Solving for the radius gives approximately 32.4 meters.

Exam Tip: Be comfortable calculating square roots of non-perfect square numbers using division or estimation methods.

Question 33. The area enclosed between the concentric circle is 770 sqcm. If the radius of the outer circle is 21 cm, find the radius of the inner circle.
Answer: Let \( R \) be the outer radius and \( r \) be the inner radius.
Given \( R = 21\text{ cm} \) and the enclosed area is \( 770\text{ cm}^2 \).
The area between two concentric circles is:
\( \text{Area} = \pi(R^2 - r^2) = 770 \)
\( \frac{22}{7}(21^2 - r^2) = 770 \)
\( 441 - r^2 = \frac{770 \times 7}{22} \)
\( 441 - r^2 = 35 \times 7 = 245 \)
\( r^2 = 441 - 245 = 196 \)
\( r = \sqrt{196} = 14\text{ cm} \)
Therefore, the radius of the inner circle is 14 cm.
In simple words: Solving the concentric area equation shows that the inner circle has a radius of exactly 14 cm.

Exam Tip: Familiarize yourself with square numbers up to 30 (like \( 14^2 = 196 \) and \( 21^2 = 441 \)) to expedite calculation steps.

Question 34. A chord AB of a circle, of radius 14cm makes an angle of 60 degree at the centre of the circle. Find the area of the minor segment of the circle.
Answer: Let the radius be \( r = 14\text{ cm} \) and the angle be \( \theta = 60^\circ \).
The area of the minor segment is:
\( \text{Area of segment} = \text{Area of sector OAPB} - \text{Area of equilateral triangle OAB} \)
\( \text{Area of sector} = \frac{60^\circ}{360^\circ} \times \pi r^2 = \frac{1}{6} \times \frac{22}{7} \times 14 \times 14 = \frac{308}{3} \approx 102.67\text{ cm}^2 \).
\( \text{Area of triangle} = \frac{\sqrt{3}}{4} r^2 = \frac{\sqrt{3}}{4} \times 14 \times 14 = 49\sqrt{3} \approx 49 \times 1.732 = 84.87\text{ cm}^2 \).
\( \text{Area of segment} = 102.67 - 84.87 = 17.80\text{ cm}^2 \).
Therefore, the area of the minor segment is approximately 17.80 sq. cm.
In simple words: The sector has an area of 102.67 sq. cm, and the triangle inside has an area of 84.87 sq. cm. The remaining segment is 17.80 sq. cm.

Exam Tip: Minor segment area is always calculated as \( \text{Area of Sector} - \text{Area of Triangle} \). Use \( \sqrt{3} \approx 1.732 \) unless specified otherwise.

Question 35. Area of quadrant of a circle which circumference is 44cm is equal to
(A) 154 cm²
(B) 38.5cm²
(C) 77cm²
(D) 22 cm²
Answer: (B) 38.5cm²
In simple words: The circumference of 44 cm means the radius is 7 cm. A quadrant (\( \frac{1}{4} \)) of this circle has an area of exactly 38.5 sq. cm.

Exam Tip: A quadrant represents a central angle of exactly \( 90^\circ \), which constitutes \( \frac{1}{4} \) of the circle's total area.

Question 36. What is the area of shaded region? If ABCD is a square of side 14 cm and A × D and B × C are semicircle?
(A) 21 cm²
(B) 42 cm²
(C) 63 cm²
(D) 84 cm²
Answer: (B) 42 cm²

In simple words: The square's total area is 196 sq. cm. Subtracting the area of the two semicircles combined (154 sq. cm) leaves exactly 42 sq. cm for the remaining shaded parts.

Exam Tip: The two semicircles constructed on opposite sides with diameter equal to the square's side length combine to form one complete circle.

Question 37. A copper wire is bent in the form of square, enclosing an area of 625 sq.cm. If the same wire is bent in the form of a circle what will be its area?
Answer: Area of the square \( = 625\text{ cm}^2 \).
Side of the square \( s = \sqrt{625} = 25\text{ cm} \).
Total length of the wire \( = \text{Perimeter of square} = 4s = 4 \times 25 = 100\text{ cm} \).
When bent into a circle, the circumference of the circle is equal to the length of the wire:
\( 2\pi r = 100 \implies r = \frac{50}{\pi}\text{ cm} \).
The area of the circle is:
\( \text{Area} = \pi r^2 = \pi \left(\frac{50}{\pi}\right)^2 = \frac{2500}{\pi} \approx \frac{2500}{3.1416} \approx 795.77\text{ cm}^2 \).
Therefore, the area of the circle is approximately 795.77 sq. cm.
In simple words: The wire is 100 cm long. Bending it into a circle gives a radius of about 15.9 cm, which results in a circular area of approximately 795.77 sq. cm.

Exam Tip: Always equate the perimeter of the first shape to the perimeter of the second shape when a wire is reshaped.

Question 38. A road which is 7m wide surrounded a circular park whose circumference is 352 m. Find the area of the road.
Answer: Let \( r \) be the radius of the inner circular park.
Circumference of the park \( = 2\pi r = 352\text{ m} \).
\( 2 \times \frac{22}{7} \times r = 352 \)
\( r = \frac{352 \times 7}{44} = 8 \times 7 = 56\text{ m} \).
The outer radius \( R \) (including the 7 m wide road) is:
\( R = r + 7 = 56 + 7 = 63\text{ m} \).
The area of the road is the area of the circular ring:
\( \text{Area of road} = \pi(R^2 - r^2) = \frac{22}{7} \times (63^2 - 56^2) \)
\( \text{Area of road} = \frac{22}{7} \times (63 - 56)(63 + 56) \)
\( \text{Area of road} = \frac{22}{7} \times 7 \times 119 = 22 \times 119 = 2618\text{ m}^2 \).
Therefore, the area of the road is 2618 sq. m.
In simple words: The inner radius is 56 m and the outer is 63 m. The surface area of the road around the park is exactly 2618 sq. m.

Exam Tip: Calculate the inner radius first using the given inner circumference, then add the road width to find the outer radius.

Question 39. The diagram shows a sector of a circle of radius r cm containing an angle \( \theta^\circ \). The area of sector is A sq.cm. and perimeter of the sector is 50cm.Prove that
i. \( \theta^\circ = \frac{360}{\pi} \left( \frac{25}{r} - 1 \right) \)
ii. \( A = 25r - r^2 \)
Answer: Let us write down the given information:
Radius \( = r \), central angle \( = \theta^\circ \), area of sector \( = A \), and perimeter \( = 50\text{ cm} \).
The perimeter of a sector consists of two radii plus the arc length:
\( \text{Perimeter} = 2r + \frac{\theta^\circ}{360^\circ} \times 2\pi r = 50 \)
\( \frac{\theta^\circ}{360^\circ} \times 2\pi r = 50 - 2r \)
\( \theta^\circ = \frac{360^\circ(50 - 2r)}{2\pi r} = \frac{360^\circ \times 2(25 - r)}{2\pi r} = \frac{360^\circ}{\pi} \left(\frac{25 - r}{r}\right) = \frac{360^\circ}{\pi} \left(\frac{25}{r} - 1\right) \).
This proves part (i).

Now, for part (ii):
The area of a sector \( A \) is:
\( A = \frac{\theta^\circ}{360^\circ} \pi r^2 \)
Substitute \( \theta^\circ = \frac{360^\circ}{\pi} \left(\frac{25 - r}{r}\right) \) into the area formula:
\( A = \left[ \frac{\frac{360^\circ}{\pi} \left(\frac{25 - r}{r}\right)}{360^\circ} \right] \pi r^2 \)
\( A = \frac{1}{\pi} \left(\frac{25 - r}{r}\right) \pi r^2 \)
\( A = (25 - r) \cdot r = 25r - r^2 \).
This proves part (ii).

In simple words: We express the perimeter to get an equation for \( \theta \), then substitute this expression directly into the area formula to arrive at the simplified quadratic formula for \( A \).

Exam Tip: Note that the perimeter of a sector always includes the two bounding straight radii \( 2r \) in addition to the curved outer arc length.

Question 40. Four equal circles, each of radius 7cm touch each other as shown in figure. Find the area included between them.
Answer: Let us join the centers of the four circles \( A, B, C, \) and \( D \).
This forms a square \( ABCD \) with side length \( = 2r = 2 \times 7 = 14\text{ cm} \).
Area of this square \( ABCD = 14 \times 14 = 196\text{ cm}^2 \).
Inside the square, there are 4 sectors (each representing a quadrant of a circle of radius 7 cm).
The area of these 4 quadrants combined is equal to the area of one full circle of radius 7 cm:
\( \text{Area of quadrants} = \pi r^2 = \frac{22}{7} \times 7 \times 7 = 154\text{ cm}^2 \).
The area included between the circles (the shaded region) is:
\( \text{Area of shaded region} = \text{Area of square} - \text{Area of quadrants} = 196 - 154 = 42\text{ cm}^2 \).
Therefore, the enclosed area between them is 42 sq. cm.

In simple words: The central area is calculated by taking the square formed by the circles' centers (196 sq. cm) and subtracting one full circle's area (154 sq. cm), leaving exactly 42 sq. cm.

Exam Tip: Joining the centers of four mutually touching circles of equal radius always yields a perfect square of side length \( 2r \).