CBSE Class 10 Mathematics Areas Related to Circles MCQs Set 18

Answer: (a) R₁ + R₂ = R

Explanation:
1. 2πR₁ + 2πR₂ = 2πR.
2. Dividing by 2π gives R₁ + R₂ = R.

Teacher's Note:
1. Circumference is proportional to radius, so radii add just as circumferences do.
2. Areas, by contrast, do not add this way.

Answer: (a) 10 m

Explanation:
1. The radii of the two parks are 8 m and 6 m.
2. πR² = π(8² + 6²) = 100π, so R² = 100.
3. R = 10 m.

Teacher's Note:
1. Equate areas, not circumferences or diameters.
2. The 6, 8, 10 triple makes the calculation quick.

Answer: (a) 22 cm

Explanation:
1. Circumference − diameter = πd − d = d(π − 1) = 15.
2. With π = 22/7, d × 15/7 = 15, so d = 7 cm.
3. Circumference = (22/7) × 7 = 22 cm.

Teacher's Note:
1. Take d common from πd − d.
2. Check: 22 − 7 = 15.

Answer: (d) becomes four times

Explanation:
1. Circumference 4π means radius 2, and circumference 8π means radius 4.
2. Area changes from 4π to 16π.
3. 16π ÷ 4π = 4, so the area becomes four times.

Teacher's Note:
1. Doubling the radius multiplies the area by 4.
2. Area is proportional to the square of the radius.

Answer: (b) 110 m

Explanation:
1. Circumferences are in the ratio of the radii, so R : r = 23 : 22. Let R = 23k and r = 22k.
2. Width of path = R − r = k = 5 m.
3. Inner radius = 22 × 5 = 110 m.

Teacher's Note:
1. The ratio of circumferences equals the ratio of radii.
2. Width of the path is the difference of the radii.

Answer: (c) 77 cm²

Explanation:
1. Side of square = 9 cm, so the length of the wire = 36 cm.
2. Perimeter of semicircle = πr + 2r = r(π + 2) = r × 36/7 = 36, so r = 7 cm.
3. Area = ½πr² = ½ × (22/7) × 49 = 77 cm².

Teacher's Note:
1. Remember that the perimeter of a semicircle includes the diameter.
2. The wire length is the same in both shapes.

Answer: (b) 21 m

Explanation:
1. Let the width be w. Outer radius − inner radius = w.
2. 2π(R − r) = 132, so 2 × (22/7) × w = 132.
3. w = 132 × 7 ÷ 44 = 21 m.

Teacher's Note:
1. The difference of circumferences is 2π times the width.
2. Check: 2 × 22/7 × 21 = 132.

Answer: (b) 1320 m

Explanation:
1. Circumference = πd = (22/7) × 140 = 440 m.
2. Wire for three rounds = 3 × 440 = 1320 m.

Teacher's Note:
1. Do not stop at one round: read the question for the number of times.
2. Use the diameter directly in πd.

Answer: (c) 78.5 m²

Explanation:
1. At a corner the cow can reach a quarter of a circle of radius 10 m, since the rope is shorter than the side.
2. Area = ¼ × 3.14 × 10 × 10 = 78.5 m².

Teacher's Note:
1. The angle at the corner of a square is 90°, so the sector is a quadrant.
2. Check that the rope length does not exceed the sides.

Answer: (d) 5 cm

Explanation:
1. Area of the rectangle = 5 × 15 = 75 cm².
2. Each square has area 75 ÷ 3 = 25 cm², so its side is 5 cm.
3. Three squares of side 5 cm placed side by side fill the 5 × 15 rectangle exactly.

Teacher's Note:
1. Divide the total area by 3 first.
2. The shape check confirms the cut is possible.

Answer: (c) 288 cm²

Explanation:
1. The diagonal of the square equals the diameter, 24 cm.
2. Area of a square = ½ × (diagonal)² = ½ × 576 = 288 cm².

Teacher's Note:
1. Area = d²/2 is a handy formula when the diagonal d is known.
2. Side = 24/√2 = 12√2 cm, which gives the same area.

Answer: (c) 168 cm²

Explanation:
1. The largest circle has diameter 28 cm, so its radius is 14 cm.
2. Area of circle = (22/7) × 14 × 14 = 616 cm².
3. Remaining area = 784 − 616 = 168 cm².

Teacher's Note:
1. The circle touches all four sides of the square.
2. Subtract the circle from the square.

Answer: (d) 4 : π

Explanation:
1. Let radius = r and side = s. Then 2πr = 4s, so s = πr/2.
2. Area of circle = πr² and area of square = π²r²/4.
3. Ratio = πr² ÷ (π²r²/4) = 4 : π.

Teacher's Note:
1. For equal perimeters the circle always has the larger area.
2. The ratio 4 : π is greater than 1, which confirms this.

Answer: (c) 11 cm

Explanation:
1. Length of the wire = 2πr = 2 × (22/7) × 7 = 44 cm.
2. Side of the square = 44 ÷ 4 = 11 cm.

Teacher's Note:
1. The wire length stays the same, so perimeters are equal.
2. Divide by 4 for a square.

Answer: (d) 72 units

Explanation:
1. πr² = 48π, so r² = 48 and r = 4√3.
2. For an equilateral triangle of side a, inradius r = a/(2√3), so a = 2√3 × 4√3 = 24.
3. Perimeter = 3 × 24 = 72 units.

Teacher's Note:
1. The inradius of an equilateral triangle is one third of its altitude.
2. Altitude = 3r = 12√3 and a = altitude × 2/√3 = 24.

Answer: (d) 49π m²

Explanation:
1. Side of the square = 11 m, so its perimeter is 44 m.
2. 2πr = 44, so with π = 22/7 we get r = 7 m.
3. Area of circle = πr² = 49π m².

Teacher's Note:
1. Equal perimeters mean 2πr equals 4 × side.
2. Use π = 22/7 to find r.

Answer: (a) \( \frac{a^2}{6} \) cm²

Explanation:
1. Inradius of the triangle r = \( \frac{a}{2\sqrt{3}} \).
2. The diagonal of the inscribed square is 2r, so its area = ½(2r)² = 2r².
3. Area = \( 2 × \frac{a^2}{12} = \frac{a^2}{6} \).

Teacher's Note:
1. Area of a square with diagonal d is d²/2.
2. Square r² = a²/12.

Answer: (b) 42 cm²

Explanation:
1. The centres form a square of side 14 cm, with area 196 cm².
2. Each circle contributes a quarter inside the square, so the four quarters make one full circle of area (22/7) × 49 = 154 cm².
3. Enclosed space = 196 − 154 = 42 cm².

Teacher's Note:
1. Each corner of the square contains a 90° sector.
2. The four 90° sectors together give a whole circle.

Answer: (b) 1 : 10

Explanation:
1. Area of minor sector = ¼ × (22/7) × 49 = 38.5 cm², and area of the triangle = ½ × 7 × 7 = 24.5 cm².
2. Smaller segment = 38.5 − 24.5 = 14 cm².
3. Whole circle = 154 cm², so the larger segment = 154 − 14 = 140 cm².
4. Ratio = 14 : 140 = 1 : 10.

Teacher's Note:
1. Segment area = sector area − triangle area.
2. The larger segment is the rest of the circle.

Answer: (d) 2 cm

Explanation:
1. AC = √(36 + 64) = 10 cm.
2. For a right triangle, r = (AB + BC − AC)/2 = (8 + 6 − 10)/2 = 2 cm.
3. Check: area = 24 = r × s = r × 12, so r = 2.

Teacher's Note:
1. The incircle radius equals area divided by semi-perimeter.
2. In a right triangle it also equals the sum of the legs minus the hypotenuse, halved.

Answer: (b) 7 cm

Explanation:
1. The centres form an equilateral triangle of side 2r.
2. Area = (√3/4)(2r)² = √3 r² = 49√3, so r² = 49.
3. r = 7 cm.

Teacher's Note:
1. Circles touching externally have centre distance equal to the sum of the radii, here 2r.
2. The area of an equilateral triangle of side s is (√3/4)s².

Answer: (c) 30π cm

Explanation:
1. πR² = π(81 + 144) = 225π, so R = 15 cm.
2. Circumference = 2πR = 30π cm.

Teacher's Note:
1. Add areas, not radii.
2. The 9, 12, 15 triple gives the radius quickly.