Maharashtra Board Class 7 Chapter 11 Set 42 Circle Solutions

Question 1. Complete the table below: Solution:
(i) Radius (r) = 7 cm
Diameter (d) = \(2r\)
= \(2 \times 7 = 14\) cm
Circumference (c) = \( \pi d\)
= \( \frac{22}{7} \times 14 \)
= \( 44 \) cm
(ii) Diameter (d) = 28 cm
Radius (r) = \( \frac{d}{2} = \frac{28}{2} = 14 \) cm
Circumference (c) = \( \pi d\)
= \( \frac{22}{7} \times 28 \)
= \( 88 \) cm
(iii) Circumference (c) = 616 cm
\( \therefore \pi d = 616 \)
\( \implies \frac{22}{7} \times d = 616 \)
\( \implies d = 616 \times \frac{7}{22} \)
\( \implies d = 196 \) cm
\( \therefore \) Diameter (d) = 196 cm
Radius (r) = \( \frac{d}{2} = \frac{196}{2} = 98 \) cm
(iv) Circumference (c) = 72.6 cm
\( \therefore \pi d = 72.6 \)
\( \implies \frac{22}{7} \times d = 72.6 \)
\( \implies d = 72.6 \times \frac{7}{22} = \frac{726}{10} \times \frac{7}{22} = \frac{33 \times 7}{10} \)
\( \implies d = 23.1 \) cm
\( \therefore \) Diameter (d) = 23.1 cm
Radius (r) = \( \frac{d}{2} = \frac{23.1}{2} \)
= 11.55 cm
Answer:

In simple words: This question requires calculating the missing values (radius, diameter, or circumference) for a circle using the standard formulas \(d = 2r\) and \(c = \pi d\) or \(c = 2 \pi r\). Each part provides one known value to find the others.

🎯 Exam Tip: Remember the basic formulas for circle properties: diameter is twice the radius (\(d=2r\)) and circumference is pi times the diameter (\(c=\pi d\)). Ensure accurate calculation, especially when working with decimals or fractions for pi (\(22/7\)).

Question 2. If the circumference of a circle is 176 cm, find its radius. Solution: Circumference (c) = 176 cm
\( \therefore 2 \pi r = 176 \)
\( \implies 2 \times \frac{22}{7} \times r = 176 \)
\( \implies \frac{44}{7} \times r = 176 \)
\( \implies r = 176 \times \frac{7}{44} = 28 \) cm
\( \therefore \) The radius of the circle is 28 cm.
Answer: The radius of the circle is 28 cm. In simple words: Given the circumference of a circle, we use the formula \(c = 2\pi r\) to find the radius by rearranging the equation and solving for \(r\).

🎯 Exam Tip: When given circumference, remember to use the formula \(C = 2 \pi r\). Accurately substitute the value of \( \pi \) (usually \(22/7\) or \(3.14\)) and perform algebraic rearrangement to find the radius. Double-check your multiplication and division.

Question 3. The radius of a circular garden is 56 m. What would it cost to put a 4-round fence around this garden at a rate of 40 rupees per metre? Solution: Radius of the circular garden (r) = 56 m
\( \therefore \) Circumference of the circular garden (c) = \(2 \pi r\)
= \( 2 \times \frac{22}{7} \times 56 \)
= \( 352 \) m
\( \therefore \) Length of the wire required to put 1-round fence = Circumference
\( \therefore \) Length of wire required to put a 4-round fence = \( 4 \times \) Circumference
= \( 4 \times 352 \)
= \( 1408 \) m
\( \therefore \) Cost of wire per meter = Rs 40
\( \therefore \) Total cost = length of wire required \( \times \) cost of the wire
= \( 1408 \times 40 \)
= Rs 56320
\( \therefore \) The cost to put a 4-round fence around the garden is Rs 56320.
Answer: The cost to put a 4-round fence around the garden is Rs 56320. In simple words: First, calculate the circumference of the circular garden. Then, multiply the circumference by 4 to get the total length of wire needed for four rounds of fencing. Finally, multiply this total length by the cost per meter to find the total cost.

🎯 Exam Tip: This problem involves multiple steps. Break it down: calculate circumference, then total length for multiple rounds, and finally, the total cost. Pay attention to units (meters for length, rupees for cost).

Question 4. The wheel of a bullock cart has a diameter of 1.4 m. How many rotations will the wheel complete as the cart travels 1.1 km? Solution: Diameter of the wheel of the bullock cart (d) = 1.4 m
Circumference of the wheel of the bullock cart (c) = \( \pi d\)
= \( \frac{22}{7} \times 1.4 = \frac{22}{7} \times \frac{14}{10} = \frac{44}{10} = 4.4 \) m
Distance covered in 1 rotation = Circumference of the wheel
= 4.4 m
Number of rotations = \( \frac{\text{Distance covered}}{\text{Circumference}} \)
= \( \frac{1.1 \text{ km}}{4.4 \text{ m}} \)
= \( \frac{1.1 \times 1000 \text{ m}}{4.4 \text{ m}} \) ...[1 km = 1000 m]
= \( \frac{11 \times 1000}{44} \)
= 250
\( \therefore \) The wheel of the bullock cart will complete 250 rotations as the cart travels 1.1 km.
Answer: The wheel will complete 250 rotations. In simple words: Calculate the circumference of the wheel, which represents the distance covered in one rotation. Convert the total travel distance to the same unit (meters), then divide the total distance by the circumference to find the number of rotations.

🎯 Exam Tip: Unit consistency is crucial! Ensure both total distance and circumference are in the same units (e.g., meters) before performing division. A common mistake is to mix kilometers and meters.

Questions And Activities

Question 1. Identify the radii, chords and diameters in the circle alongside and write their names in the table below: (Textbook pg. no. 75)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक वृत्त दिखाता है जिसका केंद्र 'O' है। वृत्त पर विभिन्न बिंदु A, B, C, D, E, F स्थित हैं। रेखाखंड केंद्र से परिधि तक खींचे गए हैं (जैसे OA, OB), साथ ही वृत्त के भीतर विभिन्न जीवाएँ और व्यास भी दर्शाए गए हैं। Solution:
Answer:

In simple words: This question asks to identify and list the specific line segments in the given circle that represent radii (lines from center to circumference), chords (lines connecting two points on the circumference), and diameters (chords passing through the center).

🎯 Exam Tip: Clearly distinguish between radius, chord, and diameter. A radius always connects the center to the circumference. A chord connects any two points on the circumference. A diameter is a special type of chord that passes through the center.