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Previous Year Problems on Areas Related to Circles
Question 1. Two circular pieces of equal radii and maximum area, touching each other, are cut out from a rectangular card board of dimensions 14 cm × 7 cm. Find the area of the remaining card board [use π = 22/7]. [2 Marks]
Answer:
The largest circles that fit have diameter 7 cm (the shorter side), so two touching circles of radius 3.5 cm fit along the length 14 cm.
Area of the two circles = \( 2 \times \dfrac{22}{7} \times 3.5^2 = 77 \) cm². Area of the card board = 14 × 7 = 98 cm².
Remaining area = 98 − 77 = 21 cm².
In simple words: The diameter of each circle is the width of the card board.
Teacher's Note:
1. Two circles of diameter 7 cm fit exactly in 14 cm.
2. Subtract the two circles.
Question 2. If the difference between the circumference and the radius of a circle is 37 cm, then using \( \pi = \dfrac{22}{7} \), the circumference (in cm) of the circle is: [1 Mark]
(A) 154 (B) 44 (C) 14 (D) 7
Answer:
Ans. (B) 44
2πr − r = 37, so \( r\left(\dfrac{44}{7} - 1\right) = \dfrac{37r}{7} = 37 \) and r = 7 cm.
Circumference = 2πr = \( 2 \times \dfrac{22}{7} \times 7 = \) 44 cm.
In simple words: Take r common on the left. | Then compute 2πr.
Teacher's Note:
1. First find the radius from the difference.
Answer:
The smaller circle has diameter OB = 7 cm, so its radius is 3.5 cm and its area = \( \dfrac{22}{7} \times 3.5^2 = 38.5 \) cm².
Each of the two lighter parts is a segment: quadrant (\( \dfrac{1}{4} \times \dfrac{22}{7} \times 49 = 38.5 \)) minus △OAC (\( \dfrac{1}{2} \times 7 \times 7 = 24.5 \)) = 14 cm². Two segments = 28 cm².
Shaded area = 38.5 + 28 = 66.5 cm².
In simple words: The shaded region is the small circle and two equal segments.
Teacher's Note:
1. A segment is a quadrant minus a right triangle.
2. Add the three parts.
Question 4. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find (i) the length of the arc (ii) the area of the sector formed by the arc [use π = 22/7]. [3 Marks]
Answer:
(i) Length of the arc = \( \dfrac{60}{360} \times 2 \times \dfrac{22}{7} \times 21 = \) 22 cm.
(ii) Area of the sector = \( \dfrac{60}{360} \times \dfrac{22}{7} \times 21^2 = \) 231 cm².
In simple words: A 60° sector is one sixth of the circle.
Teacher's Note:
1. Use ℓ = (θ/360) × 2πr.
2. Use A = (θ/360) × πr².
Answer:
Area of the quadrant = \( \dfrac{1}{4} \times \dfrac{22}{7} \times 7^2 = 38.5 \) cm². Area of △ODC = \( \dfrac{1}{2} \times OD \times OC = \dfrac{1}{2} \times 4 \times 7 = 14 \) cm².
Area of the shaded region = 38.5 − 14 = 24.5 cm².
In simple words: Cut the triangle ODC out of the quadrant.
Teacher's Note:
1. The two radii OA and OC are 7 cm.
2. Shaded area = quadrant − triangle.
Answer:
In the right triangle AED, \( AD = \sqrt{9^2 + 12^2} = 15 \) cm, so BC = 15 cm and the semicircle has radius 7.5 cm.
Area of the rectangle = 20 × 15 = 300 cm²; area of △AED = ½ × 9 × 12 = 54 cm²; area of the semicircle = \( \dfrac{1}{2} \times 3.14 \times 7.5^2 = 88.31 \) cm².
Shaded area = 300 − 54 + 88.31 = 334.31 cm².
In simple words: Add the semicircle and remove the corner triangle.
Teacher's Note:
1. Find AD with Pythagoras theorem.
2. BC = AD is the diameter of the semicircle.
Answer:
AB = AC = 28 cm and ∠BAC = 90°, so \( BC = 28\sqrt{2} \) cm and the semicircle has radius \( 14\sqrt{2} \) cm.
Area of the semicircle = \( \dfrac{1}{2} \times \dfrac{22}{7} \times (14\sqrt{2})^2 = 616 \) cm².
Area of the segment BDC = quadrant − △ABC = \( \dfrac{1}{4} \times \dfrac{22}{7} \times 28^2 - \dfrac{1}{2} \times 28 \times 28 = 616 - 392 = 224 \) cm².
Shaded area = semicircle − segment = 616 − 224 = 392 cm².
In simple words: The shaded region is the semicircle on the chord minus the segment of the quadrant.
Teacher's Note:
1. Find BC with Pythagoras theorem.
2. The quadrant and the semicircle both have area 616 cm².
Question 8. Find the area of the minor segment of a circle of radius 14 cm, when its central angle is 60°. Also find the area of the corresponding major segment [use π = 22/7 and \( \sqrt{3} = 1.732 \)]. [3 Marks]
Answer:
Area of the sector = \( \dfrac{60}{360} \times \dfrac{22}{7} \times 14^2 = 102.67 \) cm². The triangle with the two radii is equilateral: area = \( \dfrac{\sqrt{3}}{4} \times 14^2 = 84.87 \) cm².
Minor segment = 102.67 − 84.87 = 17.80 cm² (approximately).
Area of the circle = \( \dfrac{22}{7} \times 14^2 = 616 \) cm², so the major segment = 616 − 17.80 = 598.20 cm² (approximately).
In simple words: Segment = sector − triangle, and the major segment is the rest of the circle.
Teacher's Note:
1. The central angle 60° makes an equilateral triangle.
2. Subtract the minor segment from the circle.
Answer:
The diagonal of the lawn is \( 42\sqrt{2} \) m, so OP = OQ = OR = OS = \( 21\sqrt{2} \) m, and the circle through the corners has radius \( 21\sqrt{2} \) m.
Each flower bed is the segment cut off by a side of the square, which subtends a right angle at O: area of the segment = sector (90°) − △OPS = \( \dfrac{1}{4} \times \dfrac{22}{7} \times 882 - \dfrac{1}{2} \times 882 = 693 - 441 = 252 \) m².
Total area of the two flower beds = 2 × 252 = 504 m².
In simple words: Each bed is the part of the circle outside a side of the square.
Teacher's Note:
1. A side of the square subtends 90° at O.
2. Segment = sector − triangle.
Answer:
∠ACB = 90° (angle in a semicircle), so \( BC = \sqrt{13^2 - 12^2} = 5 \) cm.
The shaded region is the part of the circle on the other side of the chords AC and BC from O: area = area of the semicircle (on the side of C) − area of △ABC.
Area of the semicircle = \( \dfrac{1}{2} \times 3.14 \times 6.5^2 = 66.33 \) cm²; area of △ABC = \( \dfrac{1}{2} \times 12 \times 5 = 30 \) cm².
Shaded area = 66.33 − 30 = 36.33 cm².
In simple words: The two shaded pieces are the circle minus the triangle and minus the other semicircle.
Teacher's Note:
1. The angle in a semicircle is 90°.
2. Radius = 13/2 cm.
Answer:
Area of the ring = \( \pi(14^2 - 7^2) = 462 \) cm². Area of the part ABDC of the ring inside the angle = \( \dfrac{40}{360} \times \dfrac{22}{7} \times (14^2 - 7^2) = \dfrac{154}{3} \) cm².
Shaded area = \( 462 - \dfrac{154}{3} = \dfrac{1232}{3} = \) 410.67 cm².
In simple words: Remove the 40° piece of the ring from the whole ring.
Teacher's Note:
1. The ring is π(R² − r²).
2. The piece is 40/360 of the ring.
Answer:
In the right triangle OAB, \( \tan\theta = \dfrac{AB}{OA} \), so AB = r tan θ, and \( \sec\theta = \dfrac{OB}{OA} \), so OB = r sec θ. Hence PB = OB − OP = r sec θ − r.
The arc AP has length \( \dfrac{\pi r \theta}{180} \).
Perimeter of the shaded region = AB + BP + arc AP = \( r\tan\theta + r\sec\theta - r + \dfrac{\pi r\theta}{180} = \) \( r\left[\tan\theta + \sec\theta + \dfrac{\pi\theta}{180} - 1\right] \). Hence proved.
(Area of the shaded region = △OAB − sector OAP = \( \dfrac{1}{2}r \cdot r\tan\theta - \dfrac{\pi r^2\theta}{360} = \dfrac{r^2}{2}\left(\tan\theta - \dfrac{\pi\theta}{180}\right) \).)
In simple words: The tangent and secant of θ give the two straight sides of the shaded region.
Teacher's Note:
1. Use the right triangle OAB.
2. Add the arc length to the two straight sides.
Answer:
∠ACB = 90° (angle in a semicircle), so ∠CBA = 90° − 30° = 60°.
By the alternate segment theorem, the angle between the tangent CP and the chord CA equals the angle in the alternate segment: ∠PCA = ∠CBA = 60°.
In simple words: The tangent-chord angle equals the angle in the alternate segment.
Teacher's Note:
1. Find ∠CBA first.
2. Apply the alternate segment theorem.
Answer:
BC = 14 cm, so the semicircle has radius 7 cm. Area of the rectangle = 21 × 14 = 294 cm²; area of the semicircle = \( \dfrac{1}{2} \times \dfrac{22}{7} \times 7^2 = 77 \) cm².
Area of the shaded region = 294 − 77 = 217 cm².
Perimeter of the shaded region = AB + AD + DC + arc of the semicircle = 21 + 14 + 21 + \( \dfrac{22}{7} \times 7 \) = 21 + 14 + 21 + 22 = 78 cm.
In simple words: The semicircle is cut out of the rectangle, so its arc replaces the side BC in the perimeter.
Teacher's Note:
1. Diameter 14 cm gives radius 7 cm.
2. The side BC is not part of the shaded boundary.
Answer:
Area of the big semicircle (radius 4.5 cm) = \( \dfrac{1}{2} \times \dfrac{22}{7} \times 4.5^2 = 31.82 \) cm². Area of the circle (radius 2.25 cm) = \( \dfrac{22}{7} \times 2.25^2 = 15.91 \) cm². Area of each small semicircle (radius 1.5 cm) = \( \dfrac{1}{2} \times \dfrac{22}{7} \times 1.5^2 = 3.54 \) cm².
Two small semicircles lie inside the big semicircle and are not shaded, and the third one below the base is shaded: area = 31.82 − 15.91 − 2 × 3.54 + 3.54 = 31.82 − 15.91 − 3.54.
Shaded area = \( \dfrac{99}{8} = \) 12.375 cm² (about 12.38 cm²).
In simple words: Subtract what is not shaded and add the shaded small semicircle that lies outside.
Teacher's Note:
1. Three small semicircles have the same area.
2. Net effect: big semicircle − circle − one small semicircle.
Answer:
Area of the ring = \( \dfrac{22}{7}(42^2 - 21^2) = \dfrac{22}{7} \times 1323 = 4158 \) cm². Area of the ring part inside ∠AOB = \( \dfrac{60}{360} \times 4158 = 693 \) cm².
Area of the shaded region = 4158 − 693 = 3465 cm².
In simple words: The unshaded sixth of the ring is removed from the whole ring.
Teacher's Note:
1. The ring area is π(R² − r²).
2. A 60° piece is one sixth of it.
Free study material for Mathematics
Chapter 11 Areas Related to Circles Printable Assignments & Solutions for Class 10 Mathematics
Class 10 Mathematics Chapter 11 Areas Related to Circles Printable Assignments
Review targeted chapter assignments for Class 10 Mathematics Chapter 11 Areas Related to Circles. Built according to official CBSE guidelines, these downloadable problem sets help students build accuracy and prepare effectively for school tests.
Why Practice Class 10 Mathematics Assignments?
- Curriculum Standards: Assignments match modern CBSE sample formats to ensure relevant preparation.
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How to Approach Mathematics Chapter 11 Areas Related to Circles Assignments
- Concept Foundation: Review the NCERT book for Class 10 Mathematics thoroughly before diving into assignment tasks.
- Self-Evaluation: Solve exercises independently before inspecting professional answer guides.
- Progress Monitoring: Note down complex formulas or concepts, clearing them up using available online practice aids.
FAQs
You can download free PDF assignments for Class 10 Mathematics Chapter 11 Areas Related to Circles from StudiesToday.com. These practice sheets have been updated for the 2026-27 session covering all concepts from latest NCERT textbook.
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Yes. These assignments are designed as per the latest CBSE syllabus for 2026. We have included huge variety of question formats such as MCQs, Case-study based questions and important diagram-based problems found in Chapter 11 Areas Related to Circles.
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