CBSE Class 10 Mathematics Areas Related to Circles MCQs Set 09

Answer: (b) 33 cm

Explanation:
1. Length of the wire = circumference = \( 2 \times \frac{22}{7} \times 21 = 132 \) cm.
2. Perimeter of the square = 132 cm, so the side = 132 ÷ 4 = 33 cm.

Teacher's Note:
1. The length of the wire does not change when it is bent.
2. Perimeter of a square = 4 × side.

Answer: (a) \( 14.875 \text{ cm}^2 \)

Explanation:
1. Area of trapezium = ½ × (10 + 4) × AB = 24.5, so AB = 3.5 cm.
2. ABE is a quadrant with centre A and radius AB = 3.5 cm: area = \( \frac{1}{4} \times \frac{22}{7} \times 3.5^2 = 9.625 \text{ cm}^2 \).
3. Shaded area = 24.5 − 9.625 = 14.875 cm².

Teacher's Note:
1. First find the height AB from the area of the trapezium.
2. Shaded region = trapezium − quadrant.

Answer: (a) 50°

Explanation:
1. Area of sector ÷ area of circle = x/360.
2. So x/360 = 5/36, which gives x = 360 × 5/36 = 50°.

Teacher's Note:
1. The fraction of the circle taken by a sector equals angle ÷ 360°.

Answer: (a) 77

Explanation:
1. The shaded sectors together make an angle of 60° + 80° + 40° = 180° at the centre.
2. Area = \( \frac{180}{360} \times \frac{22}{7} \times 7 \times 7 = 77 \) cm².

Teacher's Note:
1. Add the angles first, then use one sector formula.
2. 180° is half of the circle.

Answer: (c) \( \left(8\sqrt{3} - \frac{8\pi}{3}\right) \text{ cm}^2 \)

Explanation:
1. Area of the triangle = \( \frac{\sqrt{3}}{4} \times 8^2 = 16\sqrt{3} \), so the right half = \( 8\sqrt{3} \text{ cm}^2 \).
2. Inradius \( r = \frac{8}{2\sqrt{3}} = \frac{4}{\sqrt{3}} \), so \( r^2 = \frac{16}{3} \) and the right half of the circle = \( \frac{1}{2}\pi \times \frac{16}{3} = \frac{8\pi}{3} \).
3. Shaded area = \( \left(8\sqrt{3} - \frac{8\pi}{3}\right) \text{ cm}^2 \).

Teacher's Note:
1. Inradius of an equilateral triangle = side ÷ (2√3).
2. Half triangle minus half circle.

Answer: (b) 16π sq units

Explanation:
1. The shaded sector is a quarter circle drawn in the square of side 8, so its radius is 8 units and its angle is 90°.
2. Area = \( \frac{90}{360} \times \pi \times 8^2 = 16\pi \) sq units.

Teacher's Note:
1. Read the radius from the side of the square that holds the arc.
2. A quarter circle is 1/4 of πr².

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Equating the two formulas: \( \pi r^2 = 2\pi r \), so r = 2 units. Hence A is true.
2. R gives the two formulas, and these are exactly what is equated to obtain r = 2. R is true and explains A.

Teacher's Note:
1. Write the equation first, then solve for r.
2. Check that the formulas quoted in R are the ones used.

Answer: (c) A is true, but R is false.

Explanation:
1. Arc length = \( \frac{40}{360} \times 2 \times 3.14159 \times 8 = 5.585 \approx 5.59 \) cm, so A is true.
2. The expression in R, πr² × θ/360°, is the area of a sector. Arc length = (θ/360°) × 2πr. So R is false.

Teacher's Note:
1. Do not confuse sector area with arc length.
2. Arc length has r to the power 1.

Answer: (d) A is false, but R is true.

Explanation:
1. \( 2\pi r = 221 \) gives \( r = \frac{221 \times 7}{44} = 35.16 \) cm (not 35 cm). So A is false.
2. Circumference = π × diameter is correct, so R is true.

Teacher's Note:
1. Always solve for r instead of trusting the stated value.

Answer: (a) 10 m

Explanation:
1. Radii of the two parks are 8 m and 6 m.
2. \( \pi R^2 = \pi(8^2 + 6^2) = 100\pi \), so R = 10 m.

Teacher's Note:
1. Convert diameters to radii first.
2. Areas add, radii do not.

Answer: (b) 8 cm

Explanation:
1. Area of sector = ½ × arc length × radius.
2. 20π = ½ × 5π × r, so r = 8 cm.

Teacher's Note:
1. Area of sector = ½ l r.

Answer: (c) \( 1694 \text{ m}^2 \)

Explanation:
1. Outer radius = 35 + 7 = 42 m.
2. Area of path = \( \frac{22}{7}(42^2 - 35^2) = \frac{22}{7} \times 77 \times 7 = 1694 \text{ m}^2 \).

Teacher's Note:
1. Path area = outer circle − inner circle.
2. Use a² − b² = (a − b)(a + b).

Answer: (d) 25 cm

Explanation:
1. Arc length = \( \frac{90}{360} \times 2 \times \frac{22}{7} \times 7 = 11 \) cm.
2. Perimeter = arc + 2 radii = 11 + 14 = 25 cm.

Teacher's Note:
1. Do not forget the two bounding radii.

Answer: (c) \( \frac{1}{8}\pi d^2 \)

Explanation:
1. Radius = d/2, so area of the circle = \( \pi \frac{d^2}{4} \).
2. Half of it = \( \frac{1}{8}\pi d^2 \).

Teacher's Note:
1. Convert the diameter to radius before using πr².

Answer: (a) \( \frac{r^2}{4}(\pi - 2) \) sq. units

Explanation:
1. Quadrant OAB has area \( \frac{\pi r^2}{4} \) and triangle OAB has area \( \frac{1}{2}r^2 \).
2. Segment = \( \frac{\pi r^2}{4} - \frac{r^2}{2} = \frac{r^2}{4}(\pi - 2) \).

Teacher's Note:
1. Segment = sector − triangle.

Answer: (c) arc length PQ

Explanation:
1. T lies on the arc, and OT is the diagonal of the unit square, so the radius is already known: \( OT = \sqrt{2} \).
2. Area of a sector = ½ × arc length × radius, so the arc length PQ is exactly the extra data needed.
3. The area of the square and the radius are already known and give no information about the angle POQ.

Teacher's Note:
1. Ask which quantity fixes the angle of the sector.
2. Area of sector = ½ l r.

Answer: (c) \( \frac{81}{4\pi} \text{ cm}^2 \)

Explanation:
1. \( 2\pi r = 18 \), so \( r = \frac{9}{\pi} \) cm.
2. Quadrant area = \( \frac{1}{4}\pi r^2 = \frac{1}{4}\pi \times \frac{81}{\pi^2} = \frac{81}{4\pi} \text{ cm}^2 \).

Teacher's Note:
1. Find r from the circumference first.

Answer: (b) \( 24\pi \) cm

Explanation:
1. \( \frac{1}{4}\pi r^2 = 36\pi \), so \( r^2 = 144 \) and r = 12 cm.
2. Circumference = \( 2\pi \times 12 = 24\pi \text{ cm} \).

Teacher's Note:
1. Simplify 576/16 = 36 first.

Answer: (a) \( \frac{64\pi}{3} \)

Explanation:
1. In 5 minutes the minute hand turns 5 × 6° = 30°.
2. Area = \( \frac{30}{360}\pi \times 16^2 = \frac{256\pi}{12} = \frac{64\pi}{3} \) cm².

Teacher's Note:
1. The minute hand turns 6° every minute.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Wire length = \( 2 \times \frac{22}{7} \times 14 = 88 \) cm, so the side of the square = 22 cm and its area = 484 cm². A is true.
2. The formula in R is correct, so R is true, but it has nothing to do with bending a wire into a square. R is not the explanation of A.

Teacher's Note:
1. Check the units: area is in cm².

Answer: (c) A is true, but R is false.

Explanation:
1. Area = \( \frac{30}{360} \times \pi \times 16 = 4.19 \) cm². A is true.
2. \( \frac{\theta}{360} \times 2\pi r \) is the arc length formula, so R is false.

Teacher's Note:
1. Area has r², arc length has r.

Answer: (b) \( 1211.05 \text{ cm}^2 \)

Explanation:
1. Each blade sweeps a sector of radius 34 cm and angle 120°.
2. Area = \( \frac{120}{360} \times \frac{22}{7} \times 34 \times 34 = 1211.05 \) cm².

Teacher's Note:
1. A wiper blade sweeps a sector.

Answer: (b) \( 81.85 \text{ km}^2 \)

Explanation:
1. The area is a sector of radius 12.5 km and angle 60°.
2. Area = \( \frac{60}{360} \times \frac{22}{7} \times 12.5^2 = 81.85 \text{ km}^2 \).

Teacher's Note:
1. The light covers a sector.

Answer: (b) 4375

Explanation:
1. Distance in 10 minutes = 66000 m ÷ 6 = 11000 m = 1100000 cm.
2. Circumference of a wheel = \( \frac{22}{7} \times 80 = \frac{1760}{7} \) cm.
3. Revolutions = \( 1100000 \times \frac{7}{1760} = 4375 \).

Teacher's Note:
1. Convert km/h to cm per 10 minutes.

Answer: (a) \( \frac{11}{3} \) cm

Explanation:
1. Triangle formed by the two radii and the chord has two equal sides and a 60° angle, so it is equilateral and r = 3.5 cm.
2. Arc length = \( \frac{60}{360} \times 2 \times \frac{22}{7} \times 3.5 = \frac{11}{3} \) cm.

Teacher's Note:
1. Equal radii with 60° between them make an equilateral triangle.

Answer: (b) \( 75.56 \text{ m}^2 \)

Explanation:
1. The calf grazes a quadrant. New radius = 11.5 m.
2. Increase = \( \frac{1}{4} \times 3.14 \times (11.5^2 - 6^2) = \frac{1}{4} \times 3.14 \times 96.25 = 75.56 \text{ m}^2 \).

Teacher's Note:
1. Both regions are quadrants since a corner of a square is 90°.

Answer: (b) \( \frac{22}{7} \text{ cm}^2 \)

Explanation:
1. In 15 minutes the minute hand turns 90°.
2. Area = \( \frac{90}{360} \times \pi \times 2^2 = \pi = \frac{22}{7} \text{ cm}^2 \).

Teacher's Note:
1. 15 minutes = a quarter of a revolution.

Answer: (a) Rs. 1925

Explanation:
1. Circumference = 5280 ÷ 24 = 220 m, so \( 2\pi r = 220 \) and r = 35 m.
2. Area = \( \frac{22}{7} \times 35^2 = 3850 \text{ m}^2 \).
3. Cost = 3850 × 0.50 = Rs. 1925.

Teacher's Note:
1. Fencing cost gives the circumference.

Answer: (b) \( 21 \text{ cm}^2 \)

Explanation:
1. Each circle has diameter 7 cm (radius 3.5 cm).
2. Area of cardboard = 98 cm²; two circles = \( 2 \times \frac{22}{7} \times 3.5^2 = 77 \) cm².
3. Remaining = 98 − 77 = 21 cm².

Teacher's Note:
1. The width 7 cm fixes the diameter.

Answer: (b) \( 154 \text{ m}^2 \)

Explanation:
1. The angle at a corner of a rectangle is 90°, so the cow grazes a quadrant of radius 14 m.
2. Area = \( \frac{1}{4} \times \frac{22}{7} \times 14^2 = 154 \text{ m}^2 \).

Teacher's Note:
1. Both sides (20 m and 16 m) are longer than 14 m, so the full quadrant fits.

Answer: (b) Rs. 3061.50

Explanation:
1. Inner radius = 8.75 m, outer radius = 10.75 m.
2. Area of path = 3.14 × (10.75² − 8.75²) = 3.14 × 2 × 19.5 = 122.46 m².
3. Cost = 122.46 × 25 = Rs. 3061.50.

Teacher's Note:
1. Path area = outer circle − inner circle.

Answer: (a) \( 24\sqrt{21} - 77 \text{ m}^2 \)

Explanation:
1. Semi-perimeter s = 24, so the area of the triangle = \( \sqrt{24 \times 9 \times 8 \times 7} = 24\sqrt{21} \text{ m}^2 \).
2. The three sectors of radius 7 m have angles A, B, C adding to 180°, so their total area = \( \frac{180}{360} \times \frac{22}{7} \times 49 = 77 \text{ m}^2 \).
3. Ungrazed area = 24√21 − 77 m².

Teacher's Note:
1. Use Heron's formula.
2. The three angles of a triangle add to 180°.

Answer: (b) 1 : 3 : 5

Explanation:
1. Let the radii be x, 2x, 3x (ratio of radii = ratio of diameters).
2. Areas: inner circle = πx²; middle ring = π(4x² − x²) = 3πx²; outer ring = π(9x² − 4x²) = 5πx².
3. Ratio = 1 : 3 : 5.

Teacher's Note:
1. Areas of rings are differences of circle areas.

Answer: (b) \( 32.14 \text{ cm}^2 \)

Explanation:
1. Right-angled isosceles triangle: \( 2r^2 = 25 \), so \( r^2 = 12.5 \).
2. Minor segment = \( \frac{\pi r^2}{4} - \frac{r^2}{2} = 9.82 - 6.25 = 3.57 \) cm²; major segment = πr² − 3.57 = 39.29 − 3.57 = 35.71 cm².
3. Difference = 35.71 − 3.57 = 32.14 cm².

Teacher's Note:
1. Segment = sector − triangle.

Answer: (a) \( 66 \text{ cm}^2 \)

Explanation:
1. Time = 35 minutes, so the angle swept = 35 × 6° = 210°.
2. Area = \( \frac{210}{360} \times \frac{22}{7} \times 36 = 66 \text{ cm}^2 \).

Teacher's Note:
1. Minute hand: 6° per minute.

Answer: (b) \( 308 \text{ cm}^2 \)

Explanation:
1. The three sectors have the same radius 14 cm and angles A, B, C with A + B + C = 180°.
2. Total area removed = \( \frac{180}{360} \times \frac{22}{7} \times 14^2 = 308 \text{ cm}^2 \).

Teacher's Note:
1. Equal radii let you add the angles.

Answer: (a) \( 28.5 \text{ cm}^2 \)

Explanation:
1. Sector OAB = ¼ × 3.14 × 100 = 78.5 cm².
2. Triangle OAB = ½ × 10 × 10 = 50 cm².
3. Minor segment = 78.5 − 50 = 28.5 cm².

Teacher's Note:
1. Segment = sector − triangle.

Answer: (b) \( 15246 \text{ m}^2 \)

Explanation:
1. Outer radius = 105 + 21 = 126 m.
2. Area = \( \frac{22}{7}(126^2 - 105^2) = \frac{22}{7} \times 21 \times 231 = 15246 \text{ m}^2 \).

Teacher's Note:
1. Road area = outer − inner circle.

Answer: (c) 285 mm

Explanation:
1. Circumference = \( \frac{22}{7} \times 35 = 110 \) mm.
2. Five diameters = 5 × 35 = 175 mm.
3. Total = 110 + 175 = 285 mm.

Teacher's Note:
1. Add the rim and all five diameters.

Answer: (b) \( 96.25 \text{ mm}^2 \)

Explanation:
1. Each sector has angle 360° ÷ 10 = 36° and radius 17.5 mm.
2. Area = \( \frac{36}{360} \times \frac{22}{7} \times 17.5^2 = 96.25 \text{ mm}^2 \).

Teacher's Note:
1. Ten equal sectors: one tenth of the circle each.