Area of Trapezium and a Polygon

Q1: Multiple Choice Type:

i. The circumference of a circle is numerically same as its area; then its radius is:

Step 1: Let the radius of the circle be \(r\) units.
Circumference of circle: \[ C = 2 \pi r \] Area of circle: \[ A = \pi r^2 \]Step 2: Given \(C = A\), so: \[ 2 \pi r = \pi r^2 \]Step 3: Divide both sides by \(\pi\) (assuming \(r \neq 0\)): \[ 2r = r^2 \\ r^2 – 2r = 0 \\ r(r – 2) = 0 \]Step 4: Solutions: \[ r = 0 \quad \text{or} \quad r = 2 \] Radius cannot be zero, so radius \(r = 2\) units.
Answer: b. 2 units

ii. The perimeter of a square of side 11 cm is equal to circumference of a circle; then the diameter of the circle is:

Step 1: Perimeter of square: \[ P = 4 \times 11 = 44 \text{ cm} \] Circumference of circle \(C = 2 \pi r\). Given: \[ C = P = 44 \\ 2 \pi r = 44 \\ \Rightarrow r = \frac{44}{2 \pi} = \frac{22}{\pi} \]Step 2: Diameter of circle: \[ d = 2r = 2 \times \frac{22}{\pi} = \frac{44}{\pi} = \frac{44 \times 7}{22} = 14 \text{ cm} \]Answer: c. 14 cm

iii. The circumference of a circle is equal to the sum of the circumferences of two circles with radii 10 cm and 12 cm. The radius of circle formed is:

Step 1: Circumferences of two circles: \[ C_1 = 2 \pi \times 10 = 20 \pi \\ C_2 = 2 \pi \times 12 = 24 \pi \] Sum of circumferences: \[ C = C_1 + C_2 = 20 \pi + 24 \pi = 44 \pi \]Step 2: Let radius of new circle be \(r\): \[ 2 \pi r = 44 \pi \\ \Rightarrow r = 22 \text{ cm} \]Answer: c. 22 cm

iv. The diameter of a circular wheel is 56 cm. The distance moved by it in 100 rounds is:

Step 1: Diameter \(d = 56\) cm, so radius \(r = \frac{56}{2} = 28\) cm.
Circumference of wheel: \[ C = 2 \pi r = 2 \pi \times 28 = 56 \pi \text{ cm} \]Step 2: Distance moved in 100 rounds: \[ D = 100 \times 56 \pi = 5600 \pi = 5600 \times \frac{22}{7} = 17600 \text{ cm} \]Answer: b. 17600 cm

v. A circular wheel of radius 3.5 cm makes 600 rounds in 48 seconds, its speed is:

Step 1: Circumference of wheel: \[ C = 2 \pi r = 2 \pi \times 3.5 = 7 \pi \text{ cm} \]Step 2: Distance moved in 600 rounds: \[ D = 600 \times 7 \pi = 4200 \pi \approx 4200 \times \frac{22}{7} = 13200 \text{ cm} \]Step 3: Speed in cm/s: \[ \text{Speed} = \frac{\text{distance}}{\text{time}} = \frac{13200}{48} = 275 \text{ cm/s} \]Answer: a. 275 cm/s

Q2: Find the radius and area of a circle, whose circumference is:

i. 132 cm

Step 1: Circumference formula: \[ C = 2 \pi r \] Given \(C = 132\) cm, \[ 132 = 2 \pi r \\ \Rightarrow r = \frac{132}{2 \pi} = \frac{66}{\pi} \\ r \approx \frac{66 \times 7}{22} = 21 \text{ cm} \]Step 2: Calculate area: \[ A = \pi r^2 = \pi \times 21^2 = \pi \times 441 = \frac{22}{7} \times 441 = 1386 \text{ cm}^2 \]Answer: Radius = 21 cm, Area ≈ 1386 cm²

ii. 22 m

Step 1: Given circumference \(C = 22\) m, \[ 22 = 2 \pi r \\ \Rightarrow r = \frac{22}{2 \pi} = \frac{11}{\pi} \\ r \approx \frac{11 \times 7}{22} = 3.5 \text{ m} \]Step 2: Calculate area: \[ A = \pi r^2 = \pi \times (3.5)^2 = \pi \times 12.25 = \frac{22}{7} \times 12.25 = 38.50 \text{ m}^2 \]Answer: Radius = 3.5 m, Area = 38.50 m²

Q3: Find the radius and circumference of a circle, whose area is:

i. 154 cm²

Step 1: Write the formula for area of a circle: \[ \text{Area} = \pi r^2 \]Step 2: Substitute given values ( \(\pi = \frac{22}{7}\) ): \[ \frac{22}{7} \times r^2 = 154 \]Step 3: Solve for \(r^2\): \[ r^2 = \frac{154 \times 7}{22} = 49 \]Step 4: Find radius: \[ r = \sqrt{49} = 7 \text{ cm} \]Step 5: Find circumference: \[ \text{Circumference} = 2\pi r = 2 \times \frac{22}{7} \times 7 = 44 \text{ cm} \]Answer: Radius = 7 cm, Circumference = 44 cm

ii. 6.16 m²

Step 1: Use the area formula: \[ \pi r^2 = 6.16 \]Step 2: Substitute value of \(\pi\): \[ \frac{22}{7} \times r^2 = 6.16 \]Step 3: Solve for \(r^2\): \[ r^2 = \frac{6.16 \times 7}{22} = 1.96 \]Step 4: Find radius: \[ r = \sqrt{1.96} = 1.4 \text{ m} \]Step 5: Find circumference: \[ \text{Circumference} = 2\pi r = 2 \times \frac{22}{7} \times 1.4 = 8.8 \text{ m} \]Answer: Radius = 1.4 m, Circumference = 8.8 m

Q4: The circumference of a circular table is 88 m. Find its area.

Step 1: Write the formula for circumference of a circle: \[ \text{Circumference} = 2\pi r \]Step 2: Substitute the given value (take \(\pi = \frac{22}{7}\)): \[ 2 \times \frac{22}{7} \times r = 88 \]Step 3: Solve for radius \(r\): \[ r = \frac{88 \times 7}{44} = 14 \text{ m} \]Step 4: Write the formula for area of a circle: \[ \text{Area} = \pi r^2 \]Step 5: Substitute the value of radius: \[ \text{Area} = \frac{22}{7} \times 14^2 \]Step 6: Simplify: \[ \text{Area} = \frac{22}{7} \times 196 = 616 \text{ m}^2 \]Answer:The area of the circular table is 616 m².

Q5: The area of a circle is 1386 sq. cm, find its circumference.

Step 1: Write the formula for area of a circle: \[ \text{Area} = \pi r^2 \]Step 2: Substitute the given values (take \(\pi = \frac{22}{7}\)): \[ \frac{22}{7} \times r^2 = 1386 \]Step 3: Solve for \(r^2\): \[ r^2 = \frac{1386 \times 7}{22} = 441 \]Step 4: Find the radius: \[ r = \sqrt{441} = 21 \text{ cm} \]Step 5: Write the formula for circumference of a circle: \[ \text{Circumference} = 2\pi r \]Step 6: Substitute the value of radius: \[ \text{Circumference} = 2 \times \frac{22}{7} \times 21 \]Step 7: Simplify: \[ \text{Circumference} = 132 \text{ cm} \]Answer:The circumference of the circle is 132 cm.

Q6: Find the area of a flat circular ring formed by two concentric circles (circles with same centre) whose radii are 9 cm and 5 cm.

Step 1: Write the formula for area of a circular ring: \[ \text{Area of ring} = \pi (R^2 – r^2) \]Step 2: Substitute the given values (take \(\pi = \frac{22}{7}\)): \[ \text{Area} = \frac{22}{7} (9^2 – 5^2) \]Step 3: Simplify the squares: \[ \text{Area} = \frac{22}{7} (81 – 25) \]Step 4: Simplify further: \[ \text{Area} = \frac{22}{7} \times 56 \]Step 5: Final calculation: \[ \text{Area} = 176 \text{ cm}^2 \]Answer:The area of the circular ring is 176 cm².

Q7: The radii of the inner and outer circumferences of a circular running track are 63 m and 70 m respectively. Find:

i. the area of the track

Step 1: Write the formula for area of a circular track (ring): \[ \text{Area} = \pi (R^2 – r^2) \]Step 2: Substitute the given values (take \(\pi = \frac{22}{7}\)): \[ \text{Area} = \frac{22}{7} (70^2 – 63^2) \]Step 3: Evaluate the squares: \[ \text{Area} = \frac{22}{7} (4900 – 3969) \]Step 4: Simplify: \[ \text{Area} = \frac{22}{7} \times 931 \]Step 5: Final calculation: \[ \text{Area} = 2926 \text{ m}^2 \]Answer: Area of the track = 2926 m²

ii. the difference between the lengths of the two circumferences of the track

Step 1: Write the formula for circumference of a circle: \[ \text{Circumference} = 2\pi r \]Step 2: Find the outer circumference: \[ C_1 = 2 \times \frac{22}{7} \times 70 = 440 \text{ m} \]Step 3: Find the inner circumference: \[ C_2 = 2 \times \frac{22}{7} \times 63 = 396 \text{ m} \]Step 4: Find the difference: \[ \text{Difference} = 440 – 396 = 44 \text{ m} \]Answer: Difference between the two circumferences = 44 m

Q8: A circular field of radius 105 m has a circular path of uniform width of 5 m along and inside its boundary. Find the area of the path.

Step 1: Understand the figure:
The circular path lies inside the field.
Outer radius \(R = 105\) m
Inner radius \(r = 105 – 5 = 100\) m
Step 2: Write the formula for area of a circular path (ring): \[ \text{Area of path} = \pi (R^2 – r^2) \]Step 3: Substitute the values (take \(\pi = \frac{22}{7}\)): \[ \text{Area} = \frac{22}{7} (105^2 – 100^2) \]Step 4: Evaluate the squares: \[ \text{Area} = \frac{22}{7} (11025 – 10000) \]Step 5: Simplify: \[ \text{Area} = \frac{22}{7} \times 1025 \]Step 6: Final calculation: \[ \text{Area} = 3221 \frac{3}{7} \text{ m}^2 \]Answer: The area of the circular path is \(3221 \frac{3}{7}\) m².

Q9: A wire, when bent in the form of a square, encloses an area of 484 cm². Find:

i. One side of the square

Step 1: Write the formula for area of a square: \[ \text{Area} = (\text{side})^2 \]Step 2: Substitute the given area: \[ (\text{side})^2 = 484 \]Step 3: Find the side: \[ \text{side} = \sqrt{484} = 22 \text{ cm} \]Answer: Side of the square = 22 cm

ii. Length of the wire

Step 1: The wire forms the perimeter of the square: \[ \text{Perimeter of square} = 4 \times \text{side} \]Step 2: Substitute the value of side: \[ \text{Length of wire} = 4 \times 22 = 88 \text{ cm} \]Answer: Length of the wire = 88 cm

iii. The largest area enclosed when the same wire is bent to form a circle

Step 1: The length of the wire becomes the circumference of the circle: \[ 2\pi r = 88 \]Step 2: Substitute \(\pi = \frac{22}{7}\): \[ 2 \times \frac{22}{7} \times r = 88 \]Step 3: Solve for radius \(r\): \[ r = \frac{88 \times 7}{44} = 14 \text{ cm} \]Step 4: Write the formula for area of a circle: \[ \text{Area} = \pi r^2 \]Step 5: Substitute the value of radius: \[ \text{Area} = \frac{22}{7} \times 14^2 = 616 \text{ cm}^2 \]Answer:Largest area enclosed = 616 cm²

Q10: A wire, when bent in the form of a square, encloses an area of 196 cm². If the same wire is bent to form a circle, find the area of the circle.

Step 1: Find the side and perimeter of the square. Given: Area of the square = \( 196\text{ cm}^{2} \)
Formula: \( \text{Area} = \text{side}^{2} \) \[ \text{side} = \sqrt{196} \\ \text{side} = 14\text{ cm} \] Length of wire = Perimeter of square
\[ \text{Length} = 4 \times \text{side} \\ \text{Length} = 4 \times 14 = 56\text{ cm} \]Step 2: Find the radius of the circle. The length of the wire is now the circumference of the circle.
Circumference (\( C \)) = \( 56\text{ cm} \)
Formula: \( C = 2\pi r \) \[ 2 \times \frac{22}{7} \times r = 56 \\ \frac{44}{7} \times r = 56 \\ r = \frac{56 \times 7}{44} \\ r = \frac{14 \times 7}{11} = \frac{98}{11}\text{ cm} \]Step 3: Calculate the area of the circle. Formula: \[ \text{Area} = \pi r^{2} \\ \text{Area} = \frac{22}{7} \times \frac{98}{11} \times \frac{98}{11} \\ \text{Area} = \frac{2 \times 14 \times 98}{11} \\ \text{Area} = \frac{2744}{11} \\ \text{Area} \approx 249.45\text{ cm}^{2} \]Answer: The area of the circle is \( 249.45\text{ cm}^{2} \).

Q11: The radius of a circular wheel is 42 cm. Find the distance travelled by it in:

i. 1 revolution

Step 1: Distance travelled in one revolution = Circumference of the wheel \[ \text{Circumference} = 2\pi r \]Step 2: Substitute the values: \[ 2 \times \frac{22}{7} \times 42 \]Step 3: Simplify: \[ = 264 \text{ cm} \]Answer: Distance in 1 revolution = 264 cm

ii. 50 revolutions

Step 1: Distance in 1 revolution = 264 cm
Step 2: Distance in 50 revolutions: \[ 264 \times 50 \]Step 3: Calculate: \[ = 13200 \text{ cm} \]Answer: Distance in 50 revolutions = 13,200 cm

iii. 200 revolutions

Step 1: Distance in 1 revolution = 264 cm
Step 2: Distance in 200 revolutions: \[ 264 \times 200 \]Step 3: Calculate: \[ = 52800 \text{ cm} \]Answer: Distance in 200 revolutions = 52,800 cm

Q12: The diameter of a wheel is 0.70 m. Find the distance covered by it in 500 revolutions. If the wheel takes 5 minutes to make 500 revolutions, find its speed in:

Distance covered in 500 revolutions

Step 1:Diameter of the wheel \(= 0.70\,m\)
Step 2:Radius of the wheel \[ r = \frac{\text{Diameter}}{2} \\ r = \frac{0.70}{2} = 0.35\,m \]Step 3:Circumference of the wheel \[ C = 2\pi r \\ C = 2 \times \frac{22}{7} \times 0.35 \\ C = 2.2\,m \]Step 4:Distance covered in \(500\) revolutions \[ \text{Distance} = 500 \times C \\ \text{Distance} = 500 \times 2.2 = 1100\,m \]Answer: Distance covered \(= 1100\,m\)

i. Speed of the wheel in m/s

Step 1:Time taken \[ 5\,\text{minutes} = 5 \times 60 = 300\,\text{seconds} \]Step 2:Speed \[ \text{Speed} = \frac{\text{Distance}}{\text{Time}} \\ \text{Speed} = \frac{1100}{300} = \frac{11}{3}\,m/s \\ \text{Speed} = 3 \frac{2}{3}\,m/s \]Answer: Speed \(= 3 \frac{2}{3}\,m/s\)

ii. Speed of the wheel in km/hr

Step 1:Conversion formula \[ 1\,m/s = 3.6\,km/hr \]Step 2:\[ \text{Speed in km/hr} = 3.67 \times 3.6 \\ \text{Speed} = 13.2\,km/hr \]Answer: Speed \(= 13.2\,km/hr\)

Q13: A bicycle wheel, diameter 56 cm, is making 45 revolutions in every 10 seconds. At what speed, in kilometre per hour, is the bicycle travelling?

Step 1: Find the circumference of the wheel. \[ \text{Diameter} = 56 \text{ cm} \\ \Rightarrow r = \frac{56}{2} = 28 \text{ cm} \\ \text{Circumference} = 2\pi r = 2 \times \frac{22}{7} \times 28 = 2 \times 22 \times 4 = 176 \text{ cm} \]Step 2: Find the distance travelled in 10 seconds. \[ \text{Distance} = \text{Circumference} \times \text{Number of revolutions} = 176 \times 45 = 7920 \text{ cm} \]Step 3: Convert distance to meters. \[ 7920 \text{ cm} = \frac{7920}{100} = 79.2 \text{ m} \]Step 4: Find speed in meters per second. \[ \text{Speed} = \frac{79.2 \text{ m}}{10 \text{ s}} = 7.92 \text{ m/s} \]Step 5: Convert speed to km/h. \[ 7.92 \times \frac{18}{5} = 28.512 \text{ km/h} \]Answer: The speed of the bicycle is 28.512 km/h.

Q14: A roller has a diameter of 1.4 m. Find:

i. its circumference

Step 1: Given diameter \(d = 1.4\, m\), radius \(r = \frac{d}{2} = \frac{1.4}{2} = 0.7\, m\).
Step 2: Calculate the circumference \(C\): \[ C = 2\pi r = 2 \times \frac{22}{7} \times 0.7 = 2 \times \frac{22}{7} \times \frac{7}{10} = 2 \times \frac{22}{10} = \frac{44}{10} = 4.4\, m \]Answer: The circumference of the roller is 4.4 m.

ii. the number of revolutions it makes while travelling 61.6 m.

Step 1: Total distance travelled = 61.6 m.
Step 2: Number of revolutions \(n = \frac{\text{distance}}{\text{circumference}} = \frac{61.6}{4.4} = 14\).
Answer: The roller makes 14 revolutions while travelling 61.6 m.

Q15: Find the area of the circle, length of whose circumference is equal to the sum of the lengths of the circumferences with radii 15 cm and 13 cm.

Step 1: Calculate the circumference of the two circles. \[ C_1 = 2\pi \times 15 = 2 \times \frac{22}{7} \times 15 = \frac{660}{7} \text{ cm} \\ C_2 = 2\pi \times 13 = 2 \times \frac{22}{7} \times 13 = \frac{572}{7} \text{ cm} \]Step 2: Find the sum of the circumferences. \[ C = C_1 + C_2 = \frac{660}{7} + \frac{572}{7} = \frac{1232}{7} \text{ cm} \]Step 3: Let the radius of the required circle be \(r\). Its circumference is equal to \(C\), so: \[ 2\pi r = \frac{1232}{7} \\ 2 \times \frac{22}{7} \times r = \frac{1232}{7} \\ \frac{44}{7} r = \frac{1232}{7} \\ r = \frac{1232}{7} \times \frac{7}{44} = \frac{1232}{44} = 28 \text{ cm} \]Step 4: Find the area of the circle: \[ \text{Area} = \pi r^2 = \frac{22}{7} \times 28^2 = \frac{22}{7} \times 784 = 22 \times 112 = 2464 \text{ cm}^2 \]Answer: The area of the circle is 2464 cm².

Q16: A piece of wire of length 108 cm is bent to form a semicircular arc bounded by its diameter. Find its radius and area enclosed.

Step 1: Let the radius of the semicircle be \(r\).
Step 2: The length of the wire forms the perimeter of the semicircle, which consists of the semicircular arc plus the diameter: \[ \text{Perimeter} = \pi r + 2r = r(\pi + 2) \] Given, \[ r(\pi + 2) = 108 \] Substitute \(\pi = \frac{22}{7}\): \[ r \left( \frac{22}{7} + 2 \right) = 108 \\ r \left( \frac{22}{7} + \frac{14}{7} \right) = 108 \\ r \times \frac{36}{7} = 108 \\ r = \frac{108 \times 7}{36} = \frac{756}{36} = 21 \text{ cm} \]Step 3: Calculate the area enclosed by the semicircle: \[ \text{Area} = \frac{1}{2} \pi r^2 = \frac{1}{2} \times \frac{22}{7} \times 21^2 = \frac{1}{2} \times \frac{22}{7} \times 441 \\ = \frac{1}{2} \times 22 \times 63 = 11 \times 63 = 693 \text{ cm}^2 \]Answer: Radius of the semicircle is 21 cm and the area enclosed is 693 cm².

Q17: In the following figure, a rectangle ABCD encloses three circles. If BC = 14 cm, find the area of the shaded portion.

Step 1: Length of rectangle = diameter of 3 circles + diameter of one semi-circle \[ l = 14 + 14 + 14 + \frac{14}{2} \\ l = 14 + 14 + 14 + 7 = 49 \text{ cm} \]Breadth of rectangle = BC = 14 cm
Area of rectangle: \[ Area = l \times b = 49 \times 14 = 686 \text{ cm}^2 \]Step 2: There are 3 identical circles and one semi-circle placed side by side
Area of one circle: \[ = \pi r^2 = \pi \times 7^2 = 49\pi \]Area of three and half circles: \[ = 3.5 \times 49\pi = 171.5\pi \]Using \(\pi = \frac{22}{7}\): \[ 171.5\pi = 171.5 \times \frac{22}{7} = 539\text{ cm}^2 \]Step 3: Area of shaded portion: \[ = \text{Area of rectangle} – \text{Area of circles} \\ = 686 – 539 \\ = 147\text{ cm}^2 \]Answer: Area of the shaded portion = 147 cm²