OP Malhotra Class 9 Maths Solutions Chapter 17 Circle Circumference and Area Exercise 17 (A)

S Chand Class 9 ICSE Maths Solutions Chapter 17 Circle: Circumference and Area Ex 17(A)

Unless stated otherwise, use \( \pi = \frac{22}{7} \).

 

Question 1. Find the circumference of the circles whose diameters are :
(i) 49 cm
(ii) 14 cm
(iii) 9.8 cm
(iv) 7 cm
Answer:
(i) Diameter of the circle (d) = 49 cm
Circumference \( = \pi d = \frac{22}{7} \times 49 = 22 \times 7 = 154 \) cm
(ii) Diameter of the circle (d) = 14 cm
Circumference \( = \pi d = \frac{22}{7} \times 14 = 22 \times 2 = 44 \) cm
(iii) Diameter of the circle (d) = 9.8 cm
Circumference \( = \pi d = \frac{22}{7} \times 9.8 = 22 \times 1.4 = 30.8 \) cm
(iv) Diameter of the circle (d) = 7 cm
Circumference \( = \pi d = \frac{22}{7} \times 7 = 22 \) cm
In simple words: To find the distance around a circle (circumference), we multiply its diameter by pi (\( \pi \)). The value of \( \pi \) is approximately \( \frac{22}{7} \). Remember that the diameter is the distance straight across the circle through its center.

🎯 Exam Tip: Always check if the question gives the radius or diameter. Use the correct formula: \( C = \pi d \) (for diameter) or \( C = 2\pi r \) (for radius).

 

Question 2. Find the circumference of the circles whose radii are :
(i) 7 cm
(ii) 28 cm
(iii) 3.5 cm
(iv) 98 m
Answer:
(i) Radius of the circle (r) = 7 cm
Circumference \( = 2\pi r = 2 \times \frac{22}{7} \times 7 = 44 \) cm
(ii) Radius of the circle (r) = 28 cm
Circumference \( = 2\pi r = 2 \times \frac{22}{7} \times 28 = 2 \times 22 \times 4 = 176 \) cm
(iii) Radius of the circle (r) = 3.5 cm
Circumference \( = 2\pi r = 2 \times \frac{22}{7} \times 3.5 = 2 \times 22 \times 0.5 = 22 \) cm
(iv) Radius of the circle (r) = 98 m
Circumference \( = 2\pi r = 2 \times \frac{22}{7} \times 98 = 2 \times 22 \times 14 = 616 \) cm
In simple words: The circumference is the distance around the circle. If you know the radius (distance from the center to the edge), you can find the circumference by multiplying 2, pi (\( \pi \)), and the radius together.

🎯 Exam Tip: Pay attention to the units (cm, m) and make sure your final answer matches the given units.

 

Question 3. Find the circumference of the circles, if
(i) Radius = 4.5 cm
(ii) Diameter = 15 cm (Take \( \pi = 3.14 \) in (i) and (ii) above.)
Answer:
(i) Radius (r) = 4.5 cm
Circumference \( = 2\pi r = 2 \times 3.14 \times 4.5 = 28.26 \) cm \( \approx 28.3 \) cm
(ii) Diameter (d) = 15 cm
Circumference \( = \pi d = 3.14 \times 15 = 47.10 \) cm \( \approx 47.1 \) cm
In simple words: This question asks us to use a different value for pi, which is 3.14 instead of \( \frac{22}{7} \). We use the same formulas, just with the new pi value. Rounding to one decimal place is important for the final answer.

🎯 Exam Tip: Always use the value of \( \pi \) specified in the question. If no value is given, \( \frac{22}{7} \) or 3.14 can be used, but \( \frac{22}{7} \) is often preferred for integer or easily divisible numbers.

 

Question 4. Find the length of the diameter of the circles whose circumference are :
(i) 22 cm
(ii) 8.8 cm
(iii) 44 cm
Answer:
(i) Circumference = 22 cm
Let d be the diameter of the circle.
We know \( C = \pi d \). So, \( d = \frac{\text{Circumference}}{\pi} \)
\( \implies d = \frac{22}{\frac{22}{7}} = \frac{22 \times 7}{22} = 7 \) cm
Diameter = 7 cm
(ii) Circumference = 8.8 cm
\( \implies d = \frac{\text{Circumference}}{\pi} = \frac{8.8}{\frac{22}{7}} = \frac{8.8 \times 7}{22} = 0.4 \times 7 = 2.8 \) cm
Diameter = 2.8 cm
(iii) Circumference = 44 cm
\( \implies d = \frac{\text{Circumference}}{\pi} = \frac{44}{\frac{22}{7}} = \frac{44 \times 7}{22} = 2 \times 7 = 14 \) cm
Diameter = 14 cm
In simple words: If you know the circumference of a circle, you can find its diameter by dividing the circumference by pi (\( \pi \)). This is just rearranging the basic circumference formula to find the diameter instead.

🎯 Exam Tip: Remember the relationship \( C = \pi d \). To find diameter, divide circumference by \( \pi \); to find radius, divide by \( 2\pi \).

 

Question 5. Find the lengths of the radii of the circles whose circumferences are
(i) 11 cm
(ii) 55 cm
(iii) 13.2 m
(iv) 440 m
Answer:
(i) Circumference = 11 cm
Let r be the radius of the circle.
We know \( C = 2\pi r \). So, \( r = \frac{\text{Circumference}}{2\pi} \)
\( \implies 2\pi r = 11 \)
\( \implies 2 \times \frac{22}{7}r = 11 \)
\( \implies r = \frac{11 \times 7}{2 \times 22} = \frac{7}{4} = 1.75 \) cm
Radius = 1.75 cm
(ii) Circumference = 55 cm
\( \implies 2\pi r = 55 \)
\( \implies 2 \times \frac{22}{7}r = 55 \)
\( \implies r = \frac{55 \times 7}{2 \times 22} = \frac{5 \times 7}{2 \times 2} = \frac{35}{4} = 8.75 \) cm
Radius = 8.75 cm
(iii) Circumference = 13.2 m
\( \implies 2\pi r = 13.2 \)
\( \implies 2 \times \frac{22}{7}r = 13.2 \)
\( \implies r = \frac{13.2 \times 7}{2 \times 22} = \frac{0.6 \times 7}{2} = 0.3 \times 7 = 2.1 \) cm
Radius = 2.1 cm
(iv) Circumference = 440 m
\( \implies 2\pi r = 440 \)
\( \implies 2 \times \frac{22}{7}r = 440 \)
\( \implies r = \frac{440 \times 7}{2 \times 22} = \frac{20 \times 7}{2} = 10 \times 7 = 70 \) m
Radius = 70 m
In simple words: This problem asks us to find the radius of a circle when its circumference is given. We use the formula \( C = 2\pi r \) and simply rearrange it to solve for \( r \). This means dividing the circumference by \( 2\pi \).

🎯 Exam Tip: Always keep the formula \( r = \frac{C}{2\pi} \) handy when the circumference is known and the radius needs to be found. Double-check your calculations, especially with fractions.

 

Question 6. Find the radii of the circles if circumference is : (\( \pi = 3.14 \))
(i) 100 m
(ii) 6.4 dm
Answer:
(i) Circumference = 100 m
Let r be the radius of the circle.
\( \implies 2\pi r = 100 \)
\( \implies 2 \times 3.14 \times r = 100 \)
\( \implies r = \frac{100}{2 \times 3.14} = \frac{100}{6.28} \)
\( \implies r = 15.923 \dots \approx 15.9 \) m
Radius = 15.9 m
(ii) Circumference = 6.4 dm
Let r be the radius of the circle.
\( \implies 2\pi r = 6.4 \)
\( \implies 2 \times 3.14 \times r = 6.4 \)
\( \implies 6.28r = 6.4 \)
\( \implies r = \frac{6.4}{6.28} \approx 1.019 \approx 1.02 \) dm
Radius = 1.02 dm
In simple words: Again, we are finding the radius from the circumference, but this time we must use \( \pi = 3.14 \). Make sure to round your final answer correctly to the number of decimal places requested or as appropriate for the problem.

🎯 Exam Tip: When using \( \pi = 3.14 \), be careful with decimal calculations. It's good practice to keep a few extra decimal places during intermediate steps and round only the final answer.

 

Question 7. The diameter of Venus planet is 12,278 km. Find its circumference.
Answer:
Diameter of Venus planet (d) = 12,278 km
Circumference \( = \pi d = \frac{22}{7} \times 12278 \)
\( = 22 \times 1754 = 38588 \) km
In simple words: We are given the diameter of a large planet, Venus, and asked to find its circumference. We use the basic formula \( C = \pi d \) to calculate the distance around it. This shows how math helps us understand the sizes of celestial bodies.

🎯 Exam Tip: For large numbers, a calculator is usually allowed. Ensure you perform the multiplication and division steps accurately to get the correct final value.

 

Question 8.
(i) Find the perimeter of semi-circular plate of radius 3.85 cm.
(ii) The circumference of a circle exceeds the diameter by 16.8 cm. Find the radius of the circle. (Take \( \pi = \frac{22}{7} \))
Answer:
(i) Radius of the semicircle = 3.85 cm

Perimeter of semicircle \( = \pi r + 2r \)
\( = \frac{22}{7} \times 3.85 + 2 \times 3.85 \)
\( = 22 \times 0.55 + 7.70 \)
\( = 12.10 + 7.70 = 19.80 \) cm
(ii) Let r be the radius of the circle.
Circumference \( = 2\pi r \)
Diameter \( = 2r \)
According to the problem, the circumference exceeds the diameter by 16.8 cm.
\( \implies 2\pi r - 2r = 16.8 \)
\( \implies 2r(\pi - 1) = 16.8 \)
\( \implies 2r\left(\frac{22}{7} - 1\right) = 16.8 \)
\( \implies 2r\left(\frac{22 - 7}{7}\right) = 16.8 \)
\( \implies 2r\left(\frac{15}{7}\right) = 16.8 \)
\( \implies r = \frac{16.8 \times 7}{2 \times 15} \)
\( \implies r = \frac{117.6}{30} = 3.92 \) cm
Radius of the circle = 3.92 cm
In simple words: For part (i), the perimeter of a semi-circle includes its curved edge (half of a full circle's circumference) plus the straight line across its diameter. For part (ii), we set up an equation where the circumference is bigger than the diameter by a certain amount, then we solve for the radius.

🎯 Exam Tip: Remember that the perimeter of a semicircle includes both the curved arc length (\( \pi r \)) and the diameter (or two radii, \( 2r \)). For word problems, clearly define your variables and set up the equation carefully before solving.

 

Question 9. A wire is in form of a circle of radius 42 cm. It is bent into a square. Determine the side of square.
Answer:
Radius of the circular wire (r) = 42 cm
Circumference of the circle \( = 2\pi r = 2 \times \frac{22}{7} \times 42 = 2 \times 22 \times 6 = 264 \) cm
When the wire is bent into a square, the length of the wire remains the same. So, the circumference of the circle becomes the perimeter of the square. This means the total length of the wire is conserved.
Perimeter of square = 264 cm
Side of the square \( = \frac{\text{Perimeter}}{4} = \frac{264}{4} = 66 \) cm
In simple words: First, we find the total length of the wire when it's a circle by calculating its circumference. Then, because the same wire is used to make a square, this length becomes the perimeter of the square. We divide the perimeter by 4 to find the length of one side of the square.

🎯 Exam Tip: Problems involving converting shapes often rely on the principle that the perimeter (or length of wire/material) remains constant. Clearly state this step in your solution.

 

Question 10. In the figure, the radius is 3.5 cm. Find the perimeter of the quarter of the circle correct to one decimal place.
Answer:
Radius of the quarter circle (r) = 3.5 cm

Perimeter of the quarter circle \( = \frac{1}{4} \times 2\pi r + 2r \)
\( = \frac{1}{2}\pi r + 2r \)
\( = \frac{1}{2} \times \frac{22}{7} \times 3.5 + 2 \times 3.5 \)
\( = \frac{1}{2} \times 22 \times 0.5 + 7.0 \)
\( = 11 \times 0.5 + 7.0 \)
\( = 5.5 + 7.0 = 12.5 \) cm
In simple words: The perimeter of a quarter circle includes the length of its curved edge (which is one-fourth of a full circle's circumference) plus the lengths of the two straight sides (which are both radii). We add these three parts to get the total perimeter.

🎯 Exam Tip: When dealing with parts of a circle, always include the straight line segments (radii or diameter) in your perimeter calculation, not just the curved arc length.

 

Question 11.
(i) The inner circumference of a circular track is 440 m. The track is 14 m wide. Find the diameter of the outer circle of the track. (Take \( \pi = \frac{22}{7} \))
(ii) The inner edge of a circular track is 440 m long. If the track is 10 m wide, find the length of the outer edge of the track.
Answer:
(i) Inner circumference of the circular track \( = 440 \) m
Let the inner radius be r.
\( \implies 2\pi r = 440 \)
\( \implies 2 \times \frac{22}{7} \times r = 440 \)
\( \implies r = \frac{440 \times 7}{2 \times 22} = \frac{20 \times 7}{2} = 10 \times 7 = 70 \) m

Width of track = 14 m
Radius of the outer circle (R) \( = r + \text{width of track} = 70 + 14 = 84 \) m
Diameter of the outer circle \( = 2R = 2 \times 84 = 168 \) m
(ii) Inner edge (circumference) of the circular track \( = 440 \) m
Let the inner radius be r.
\( \implies 2\pi r = 440 \)
\( \implies r = \frac{440 \times 7}{2 \times 22} = 70 \) m
Width of track = 10 m
Outer radius (R) \( = r + \text{width of track} = 70 + 10 = 80 \) m
Length of outer edge (circumference) \( = 2\pi R = 2 \times \frac{22}{7} \times 80 = \frac{3520}{7} \approx 502.86 \) m
In simple words: This question involves a circular track with an inner and outer boundary. For part (i), we use the inner circumference to find the inner radius. Then, we add the track's width to get the outer radius, and finally, double it to find the outer diameter. For part (ii), we follow similar steps to find the outer radius and then calculate the circumference of the outer edge.

🎯 Exam Tip: Always draw a simple diagram for track problems. Clearly label inner radius (r) and outer radius (R), where \( R = r + \text{width} \). This helps avoid errors in calculation.

 

Question 12. A garden roller has a circumference of 3 metres. How many revolutions does it make in moving a distance of 21 metres?
Answer:
Circumference of roller = 3 m
Distance travelled = 21 m
The number of revolutions is found by dividing the total distance covered by the distance covered in one revolution (which is the circumference).
Number of revolutions \( = \frac{\text{Distance travelled}}{\text{Circumference}} = \frac{21}{3} = 7 \)
The roller makes 7 revolutions.
In simple words: Imagine the roller turns one full time, it moves forward by a distance equal to its circumference. To find how many times it turns for a longer distance, we just divide the total distance by the distance of one turn.

🎯 Exam Tip: For problems involving revolutions, understand that one revolution covers a distance equal to the circumference of the wheel or roller. Ensure consistent units for distance and circumference.

 

Question 13. A bicycle wheel, diameter 70 cm, is making 25 revolutions in 10 sec. At what speed in km/hr is the bicycle travelling?
Answer:
Diameter of the cycle wheel (d) = 70 cm
Circumference \( = \pi d = \frac{22}{7} \times 70 = 22 \times 10 = 220 \) cm
Distance travelled in 25 revolutions \( = 220 \times 25 = 5500 \) cm
Convert cm to m: \( 5500 \text{ cm} = 55 \) m
Time taken = 10 sec
Speed \( = \frac{\text{Distance}}{\text{Time}} = \frac{55 \text{ m}}{10 \text{ sec}} = 5.5 \text{ m/sec} \)
To convert m/sec to km/hr, multiply by \( \frac{18}{5} \).
Speed in km/hr \( = 5.5 \times \frac{18}{5} = \frac{55}{10} \times \frac{18}{5} = \frac{11}{2} \times \frac{18}{5} = \frac{11 \times 9}{5} = \frac{99}{5} = 19.8 \) km/hr
In simple words: First, we find how far the wheel travels in one turn using its diameter. Then, we find the total distance traveled in 25 turns. After that, we calculate the speed using distance and time, and finally, convert this speed from meters per second to kilometers per hour.

🎯 Exam Tip: Be careful with unit conversions. Ensure all distances are in meters or kilometers and all times are in seconds or hours, as required by the final speed unit.

 

Question 14. A small cart is being driven at 10 km per hour. If each wheel is 91 cm in diameter, find the number of revolutions made by each wheel per minute.
Answer:
Diameter of wheel (d) = 91 cm
Circumference \( = \pi d = \frac{22}{7} \times 91 = 22 \times 13 = 286 \) cm
Speed of cart = 10 km/hr
Distance travelled in 1 hour = 10 km
Convert speed to cm per minute for consistency:
Distance travelled in 1 minute \( = \frac{10 \text{ km}}{60 \text{ min}} \times 1 \text{ min} = \frac{10}{60} \text{ km} = \frac{1}{6} \text{ km} \)
\( \frac{1}{6} \text{ km} = \frac{1}{6} \times 1000 \text{ m} = \frac{1000}{6} \text{ m} = \frac{500}{3} \text{ m} \)
\( \frac{500}{3} \text{ m} = \frac{500}{3} \times 100 \text{ cm} = \frac{50000}{3} \text{ cm} \)
Number of revolutions in one minute \( = \frac{\text{Distance travelled in 1 minute}}{\text{Circumference}} \)
\( = \frac{\frac{50000}{3}}{286} = \frac{50000}{3 \times 286} = \frac{50000}{858} \)
\( \approx 58.275 \approx 58 \frac{118}{429} \) revolutions
In simple words: We first calculate how far the wheel travels in one turn. Then, we find out how much distance the cart covers in one minute. Finally, by dividing the distance covered in one minute by the distance of one wheel turn, we find how many times the wheel rotates in a minute.

🎯 Exam Tip: Be very careful with unit conversions for speed and distance. It's often easiest to convert everything to the smallest common unit (like cm and seconds) first, or ensure consistency throughout the calculation.

 

Question 15. How long will a boy take to go four times round a circular field whose radius is 35 m, walking at 5 km per hour ?
Answer:
Radius of circular field (r) = 35 m
Circumference of the field \( = 2\pi r = 2 \times \frac{22}{7} \times 35 = 2 \times 22 \times 5 = 220 \) m
Distance travelled in 4 rounds \( = 220 \times 4 = 880 \) m
Speed of walking = 5 km/hour
Convert speed to m/hr: \( 5 \text{ km/hr} = 5 \times 1000 \text{ m/hr} = 5000 \text{ m/hr} \)
Time taken \( = \frac{\text{Distance}}{\text{Speed}} = \frac{880 \text{ m}}{5000 \text{ m/hr}} = \frac{880}{5000} \) hours
\( = \frac{88}{500} = \frac{22}{125} \) hours
To convert to minutes: \( \frac{22}{125} \times 60 = \frac{1320}{125} = \frac{264}{25} = 10.56 \) minutes
\( 10.56 \) minutes \( = 10 \) minutes and \( 0.56 \times 60 \) seconds \( = 10 \) minutes and \( 33.6 \) seconds
Approximately 10 minutes 34 seconds.
In simple words: First, we find the distance around the circular field. Then, we calculate the total distance the boy walks by multiplying this by four. We convert the boy's speed to meters per hour and then divide the total distance by his speed to find the time taken in hours, which is then converted to minutes and seconds.

🎯 Exam Tip: Ensure consistent units throughout your calculation. Converting all values to the same unit (e.g., meters and hours, or meters and minutes) at the start can prevent errors.

 

Question 16. A wheel of a cart is making 4 revolutions per second. If the diameter of the wheel is 84 cm, find its speed.
Answer:
Diameter of the wheel (d) = 84 cm
Circumference \( = \pi d = \frac{22}{7} \times 84 = 22 \times 12 = 264 \) cm
Distance covered in 1 second (4 revolutions) \( = 4 \times 264 = 1056 \) cm
This is the speed in cm/sec.
Speed \( = 1056 \) cm/sec
To find speed in km/hr:
Distance covered in 1 hour \( = 1056 \text{ cm/sec} \times 60 \text{ sec/min} \times 60 \text{ min/hr} \)
\( = 1056 \times 3600 \) cm
\( = 3801600 \) cm
Convert cm to km: \( 3801600 \text{ cm} = \frac{3801600}{100 \times 1000} \text{ km} = \frac{3801600}{100000} \text{ km} = 38.016 \) km
Speed of the cart \( = 38.016 \) km/hr
In simple words: We first find the distance covered in one turn of the wheel. Then, we calculate the total distance covered in one second by multiplying this by the number of revolutions per second. Finally, we convert this speed from centimeters per second to kilometers per hour.

🎯 Exam Tip: When converting speed units, remember to multiply by the appropriate conversion factors for both time and distance. For example, to go from cm/s to km/hr, you'd multiply by 3600 (seconds in an hour) and divide by 100,000 (cm in a km).

 

Question 17. The figure shows a running track in which AF = CD = 100 m and FED and ABC are semicircles whose diameters FD and AC are equal. How much will an athlete run in taking one complete round of the track?
Answer:
From the figure, the straight parts of the track are AF and CD, both 100 m.
The curved parts are two semicircles, FED and ABC, with diameters FD and AC respectively.
The diagram indicates the diameter of these semicircles is 70 m.
Circumference of two semicircular parts \( = \) Circumference of one full circle with diameter 70 m
\( = \pi d = \frac{22}{7} \times 70 = 22 \times 10 = 220 \) m

Length of one complete round \( = \text{Length of straight sections} + \text{Circumference of two semicircles} \)
\( = (\text{AF} + \text{CD}) + (\text{Circumference of semicircles FED and ABC}) \)
\( = (100 + 100) + 220 \) m
\( = 200 + 220 = 420 \) m
The athlete will run 420 m in one complete round.
In simple words: The running track has two straight parts and two curved parts. The curved parts are semicircles that, when put together, form a full circle. We add the lengths of the two straight parts and the circumference of this combined full circle to get the total distance for one round.

🎯 Exam Tip: For composite shapes like a running track, break the shape down into simpler geometric figures (rectangles and circles/semicircles). The perimeter is the sum of the lengths of all outer boundaries.

 

Question 18. A circular field has a perimeter of 660 m. A plot in the shape of a square having its vertices on the circumference is marked in the field, calculate the area of the square plot. \( \left(\pi=\frac{22}{7}\right) \).
Answer:
Perimeter (circumference) of the circular field = 660 m
Let r be the radius and d be the diameter of the circular field.
Circumference \( = \pi d = 660 \)
\( \implies \frac{22}{7} \times d = 660 \)
\( \implies d = \frac{660 \times 7}{22} = 30 \times 7 = 210 \) m
So, the diameter of the circle is 210 m.
Since the vertices of the square lie on the circumference of the circle, the diagonal of the square is equal to the diameter of the circle. This is a key geometric property.
Diagonal of the square \( = \text{diameter of the circle} = 210 \) m
Area of a square can be calculated using its diagonal (D) by the formula: Area \( = \frac{D^2}{2} \)
Area of square plot \( = \frac{(210)^2}{2} = \frac{210 \times 210}{2} = 210 \times 105 = 22050 \) m\(^2\)
The area of the square plot is 22050 m\(^2\).
In simple words: First, we use the given perimeter of the circular field to find its diameter. Because the square is inside the circle with its corners touching the edge, the diagonal of the square is the same length as the circle's diameter. Finally, we use this diagonal length to calculate the area of the square.

🎯 Exam Tip: Remember the relationship between an inscribed square and its circle: the diagonal of the square is equal to the diameter of the circle. Also, know the formula for the area of a square in terms of its diagonal (\( \text{Area} = \frac{\text{diagonal}^2}{2} \)).