OP Malhotra Class 9 Maths Solutions Chapter 17 Circle Circumference and Area Chapter Test

Question 1. A cycle wheel makes 1000 revolutions in moving 440 m. What is the diameter of the wheel?
(a) 7 cm
(b) 14 cm
(c) 28 cm
Answer: (b) 14 cm
Number of revolutions = 1000
Distance travelled = 440 m
To find the perimeter (circumference) of the wheel, we divide the total distance by the number of revolutions. First, convert meters to centimeters.
Perimeter of the wheel \( = \frac{440 \times 100}{1000} = 44 \) cm
Now, we use the formula for circumference, \( \text{Circumference} = \pi \times \text{diameter} \).
So, diameter \( = \frac{\text { Perimeter }}{\pi} = \frac{44 \times 7}{22} = 14 \) cm
In simple words: The wheel covered 440 meters in 1000 turns. This means each turn covered 44 cm. Since one turn is the circumference, the diameter of the wheel is 14 cm.

🎯 Exam Tip: Remember to convert all units to be consistent (e.g., meters to centimeters) before performing calculations to avoid errors in your final answer.

Question 2. The area of a semi-circular field is 308 sq m; then taking \( \pi = \frac { 22 }{ 7 } \), the length of the railing to surround it has to be
(a) 88 cm
(b) 80 cm
(c) 44 cm
(d) 72 cm
Answer: (d) 72 cm
Area of a semi-circular field = 308 sq m (Assuming 'sq m' is a typo for 'sq cm' given the options in cm).
The formula for the area of a semi-circle is \( \frac { 1 }{ 2 } \pi r^2 \).
So, \( \frac { 1 }{ 2 } \times \frac { 22 }{ 7 } \times r^2 = 308 \)
\( r^2 = \frac{308 \times 2 \times 7}{22} = 14 \times 2 \times 7 = 196 \)
\( r = \sqrt{196} = 14 \) cm. This is the radius of the semi-circle.
The length of railing needed is the perimeter of the semi-circular field, which is the curved arc plus the straight diameter.
Perimeter \( = \pi r + 2r = \frac { 22 }{ 7 } \times 14 + 2 \times 14 \)
\( = 44 + 28 = 72 \) cm
In simple words: First, use the area of the semi-circle to find its radius. Then, calculate the total length around the semi-circle, which includes the curved part and the straight edge. This total length is 72 cm.

🎯 Exam Tip: The perimeter of a semi-circle includes both the curved arc length (\( \pi r \)) and the diameter (\( 2r \)). Don't forget the straight edge!

Question 3. A wire of length of 36 cm is bent in the form of a semi-circle. What is the radius of the semi-circle?
(a) 9 cm
(b) 10 cm
(c) 7 cm
(d) 6 cm
Answer: (c) 7 cm
Length of wire = 36 cm. This length becomes the perimeter of the semi-circle.
Let the radius of the semi-circle be \( r \).
The perimeter of a semi-circle is given by \( \pi r + 2r \).
So, \( \pi r + 2r = 36 \)
\( r(\pi + 2) = 36 \)
\( r(\frac { 22 }{ 7 } + 2) = 36 \)
\( r(\frac { 22 + 14 }{ 7 }) = 36 \)
\( r(\frac { 36 }{ 7 }) = 36 \)
\( \implies r = \frac{36 \times 7}{36} = 7 \) cm
In simple words: The wire's length is the total distance around the semi-circle. This distance is made of the curved part and the straight diameter. We use a formula to find the radius, which turns out to be 7 cm.

🎯 Exam Tip: Always remember that the perimeter of a semi-circle is its arc length plus its diameter, i.e., \( \pi r + 2r \).

Question 4. A wire is in the form of a circle of radius 42 cm. If it is bent into a square, then what is the side of the square?
(a) 66 cm
(b) 42 cm
(c) 36 cm
(d) 33 cm
Answer: (a) 66 cm
Radius of the circular wire \( = 42 \) cm.
First, find the perimeter (circumference) of the circular wire.
Perimeter of 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, its total length remains the same. So, the perimeter of the square is 264 cm.
The formula for the perimeter of a square is \( 4 \times \text{side} \).
So, \( 4 \times \text{side} = 264 \) cm
\( \implies \text{side} = \frac { 264 }{ 4 } = 66 \) cm
In simple words: The length of the wire is 264 cm. When this wire is made into a square, its total length becomes the boundary of the square. Since a square has four equal sides, each side must be 66 cm long.

🎯 Exam Tip: When a wire is reformed from one shape to another, its length (perimeter or circumference) remains constant.

Question 5. A metal wire when bent in the form of a square encloses an area 484 cm². If the same wire is bent in the form of a circle, then its area is (\(\text {Take } \pi=\frac{22}{7}\))
(a) 616 cm²
(b) 5040 cm²
(c) 1232 cm²
(d) 2464 cm²
Answer: (a) 616 cm²
Area of the square \( = 484 \) cm².
The side of the square \( = \sqrt{\text { Area }} = \sqrt{484} = 22 \) cm.
Perimeter of the square \( = 4 \times \text{Side} = 4 \times 22 = 88 \) cm.
This same wire is then bent into a circle, so the circumference of the circle is 88 cm.
Circumference of circle \( = 2\pi r = 88 \) cm.
\( 2 \times \frac{22}{7} \times r = 88 \)
\( \implies r = \frac{88 \times 7}{2 \times 22} = \frac{4 \times 7}{2} = 14 \) cm. This is the radius of the circle.
Now, calculate the area of the circle.
Area of circle \( = \pi r^2 = \frac { 22 }{ 7 } \times 14 \times 14 = 22 \times 2 \times 14 = 616 \) cm².
In simple words: First, find the side length of the square from its area. Then, calculate the total length of the wire (the perimeter of the square). This length is also the circumference of the circle. From the circumference, find the circle's radius, and finally, calculate the circle's area.

🎯 Exam Tip: When a wire changes shape, its length remains constant, so the perimeter/circumference is the key link between the shapes.

Question 6. In the figure, the area enclosed between the two co-centric circles is 770 cm². If the radius of the outer circle is 21 cm, the radius of the inner circle is
(a) 14 cm
(b) 22 cm
(c) 12 cm
(d) 10.5 cm
Answer: (a) 14 cm
Area enclosed between the two concentric circles = 770 cm².
Radius of outer circle (R) = 21 cm.
Let the radius of the inner circle be \( r \).
The area between two concentric circles is given by \( \pi R^2 - \pi r^2 = \pi (R^2 - r^2) \).
So, \( \frac{22}{7} (21^2 - r^2) = 770 \)
\( (21^2 - r^2) = \frac{770 \times 7}{22} \)
\( 441 - r^2 = 35 \times 7 \)
\( 441 - r^2 = 245 \)
\( r^2 = 441 - 245 = 196 \)
\( \implies r = \sqrt{196} = 14 \) cm. This is the radius of the inner circle.
In simple words: The space between the two circles has an area of 770 cm². We know the outer circle's radius. By using the area formula, we can calculate that the inner circle must have a radius of 14 cm.

🎯 Exam Tip: The area between two concentric circles is \( \pi \) times the difference of the squares of their radii, i.e., \( \pi (R^2 - r^2) \).

Question 7. A circle circumscribes a rectangle with side 16 cm and 12 cm. What is the area of the circle?
(a) 48 π sq cm
(b) 50 π sq cm
(c) 100 π sq cm
(d) 200 π sq cm
Answer: (c) 100 π sq cm


A circle circumscribes a rectangle with sides 16 cm and 12 cm.
The diagonal of the rectangle is the diameter of the circumscribing circle.
Using the Pythagorean theorem, the diagonal AC \( = \sqrt{\mathrm{AB}^2+\mathrm{BC}^2} \)
\( = \sqrt{16^2+12^2} = \sqrt{256+144} \) cm
\( = \sqrt{400} = 20 \) cm
So, the diameter of the circle is 20 cm.
The radius of the circle \( = \frac { 20 }{ 2 } = 10 \) cm.
Area of the circle \( = \pi r^2 = \pi \times 10 \times 10 = 100\pi \) cm².
In simple words: When a circle goes around a rectangle, the longest line across the rectangle (its diagonal) is the diameter of the circle. We use the diagonal length to find the circle's radius, and then we can find its area.

🎯 Exam Tip: For a circle circumscribing a rectangle, the diagonal of the rectangle is always equal to the diameter of the circle. Use the Pythagorean theorem to find the diagonal.

Question 8. A person rides a bicycle round a circular path of radius 50 m. The radius of the wheel of the bicycle is 50 cm. The cycle comes to the starting point for the first time in 1 hour. What is the number of revolutions of the wheel in 15 min?
(a) 20
(b) 25
(c) 30
(d) 35
Answer: (b) 25
Radius of the circular path (R) = 50 m.
Radius of the bicycle wheel (r) = 50 cm.
The distance travelled by the cycle in 1 hour is the circumference of the circular path.
Circumference of path \( = 2\pi R = 2 \times \frac { 22 }{ 7 } \times 50 \) m \( = \frac{2200}{7} \) m.
Now, find the circumference of the bicycle wheel.
Circumference of wheel \( = 2\pi r = 2 \times \frac { 22 }{ 7 } \times 50 \) cm \( = \frac{2200}{7} \) cm.
To find the number of revolutions, divide the total distance travelled by the circumference of the wheel. Ensure units are consistent.
Distance travelled in 1 hour = \( \frac{2200}{7} \) m \( = \frac{2200}{7} \times 100 \) cm.
Number of revolutions in 1 hour \( = \frac{\text{Total Distance}}{\text{Wheel Circumference}} = \frac{\frac{2200}{7} \times 100}{\frac{2200}{7}} = 100 \).
So, the wheel makes 100 revolutions in 1 hour (60 minutes).
To find the number of revolutions in 15 minutes:
Number of revolutions in 15 minutes \( = \frac{100 \times 15}{60} = 25 \).
In simple words: First, calculate the total distance the bicycle travels in one hour (this is the perimeter of the circular path). Then, find out how far the bicycle wheel rolls in one revolution (its circumference). Divide the total distance by the wheel's circumference to find the number of revolutions in an hour. Finally, calculate the number of revolutions for 15 minutes.

🎯 Exam Tip: Pay close attention to units! Convert all measurements to a single unit (e.g., centimeters or meters) before performing calculations to prevent errors.

Question 9. The figure consists of four small semi-circles of equal radii (each 42 cm) the perimeter of the shaded region is
(a) 524 cm
(b) 264 cm
(c) 396 cm
(d) 504 cm
Answer:
The given figure consists of 4 small semi-circles and 2 big semi-circles whose arcs form the perimeter of the shaded region.
Radius of each small semi-circle (\( r \)) = 42 cm.
Radius of each big semi-circle (\( R \)) = \( 42 \times 2 = 84 \) cm.
The perimeter of the shaded portion is the sum of the arc lengths of these semi-circles.
Perimeter \( = 4 \times (\pi r) + 2 \times (\pi R) \)
\( = 4 \times \frac{22}{7} \times 42 + 2 \times \frac{22}{7} \times 84 \)
\( = (4 \times 22 \times 6) + (2 \times 22 \times 12) \)
\( = 528 + 528 = 1056 \) cm.
In simple words: The shape's outer edge is made of several curved lines. We add the lengths of four small curved parts and two large curved parts to get the total length around the shaded area. The total perimeter is 1056 cm.

🎯 Exam Tip: Carefully identify all the individual curved segments that make up the perimeter of the shaded region. The perimeter is a sum of arc lengths, not a simple circumference.

Question 10. A rectangular cardboard is of dimensions 18 cm x 10 cm. From the four corners of the rectangle quarter circles of radius 4 cm are cut. What is the perimeter (approximate) of the remaining portion?
(a) 47.1 cm
(b) 49.1 cm
(c) 51.0 cm
(d) 53.0 cm
Answer: (b) 49.1 cm


Length of cardboard = 18 cm and width = 10 cm.
Radius of each quadrant (quarter circle) = 4 cm.
The original perimeter of the rectangle is \( 2 \times (18 + 10) = 2 \times 28 = 56 \) cm.
When four quarter circles are cut from the corners, we lose 2 straight segments of length \( r \) for each corner, but gain an arc length.
Total length removed from the straight sides = \( 4 \times (4 + 4) = 4 \times 8 = 32 \) cm.
The circumference of four quadrants (which is equal to the circumference of one full circle of radius \( r \)) is added to the perimeter.
Circumference of four quadrants \( = 4 \times \frac{1}{4} \times 2\pi r = 2\pi r \)
\( = 2 \times \frac{22}{7} \times 4 = \frac{176}{7} \approx 25.1 \) cm.
Perimeter of remaining portion \( = \text{Original perimeter} - \text{removed straight segments} + \text{added arc lengths} \)
\( = 56 - 32 + 25.1 = 24 + 25.1 = 49.1 \) cm.
In simple words: First, find the perimeter of the full rectangle. When quarter circles are cut from the corners, some straight edges are removed, and new curved edges are added. Calculate the total length of the removed straight edges and the total length of the new curved edges (which form a full circle). Then, adjust the original perimeter by subtracting what's removed and adding what's new.

🎯 Exam Tip: When cutting shapes, visualize what parts of the perimeter are removed and what new parts are added. For quarter circles, two radii are removed and one arc length is added per corner.