CBSE Class 10 Mathematics Areas Related to Circles MCQs Set 10

Answer: (d) 28.5 cm²

Explanation:
1. Area of the segment = area of the sector − area of the triangle.
2. Sector area = \( \frac{90}{360}\times 3.14\times 10^2 = 78.5 \) cm², and triangle area = \( \frac{1}{2}\times 10\times 10 = 50 \) cm².
3. Segment area = 78.5 − 50 = 28.5 cm².

Teacher's Note:
1. A chord subtending 90° at the centre forms a right triangle with the two radii.
2. Segment = sector − triangle.

Answer: (a) \( \frac{77}{6} \)

Explanation:
1. The shaded region is a sector of the ring between the two circles, with angle 40°.
2. Area = \( \frac{40}{360}\times\frac{22}{7}\left(7^2-3.5^2\right) = \frac{1}{9}\times\frac{22}{7}\times 36.75 \).
3. This equals \( \frac{77}{6} \) (about 12.83), which is option (a).

Teacher's Note:
1. Subtract the inner sector from the outer sector, both with 40°.
2. 7² − 3.5² = 36.75.

Answer: (b) 1253.5 cm²

Explanation:
1. Area of the plate = 40 × 40 = 1600 cm².
2. Area of one hole = \( \frac{22}{7}\times\left(\frac12\right)^2 = \frac{11}{14} \) cm², so 441 holes cover \( 441\times\frac{11}{14} = 346.5 \) cm².
3. Remaining area = 1600 − 346.5 = 1253.5 cm².

Teacher's Note:
1. Radius of each hole is half its diameter.
2. Subtract the total area of the holes from the plate.

Answer: (c) 221.5 cm²

Explanation:
1. Area of the trapezium = \( \frac12\times(7+19)\times 20 = 260 \) cm².
2. Each semicircle has diameter 7 cm, so radius 3.5 cm. The two semicircles together make one circle: \( \frac{22}{7}\times 3.5^2 = 38.5 \) cm².
3. Remaining area = 260 − 38.5 = 221.5 cm².

Teacher's Note:
1. The parallel sides are 15 + 4 = 19 cm and 7 cm.
2. Two identical semicircles add up to one full circle.

Answer: (b) 126 cm²

Explanation:
1. Each semicircle has diameter 42 ÷ 3 = 14 cm, so radius 7 cm.
2. Area of 6 semicircles = \( 6\times\frac{1}{2}\times\frac{22}{7}\times 7^2 = 462 \) cm², and the cloth is 42 × 14 = 588 cm².
3. Coloured area = 588 − 462 = 126 cm².

Teacher's Note:
1. Three semicircles fit along the length, so the diameter is 42 ÷ 3.
2. Coloured = total − uncoloured.

Answer: (b) 259 cm²

Explanation:
1. Area of the rectangle = 28 × 23 = 644 cm². The semicircle has radius 28 ÷ 2 = 14 cm and each quadrant has radius 23 − 16 = 7 cm.
2. Unshaded area = \( \frac12\times\frac{22}{7}\times 14^2 + 2\times\frac14\times\frac{22}{7}\times 7^2 = 308+77 = 385 \) cm².
3. Shaded area = 644 − 385 = 259 cm².

Teacher's Note:
1. Find the radii of the semicircle and the quadrants from the given lengths.
2. Shaded = rectangle − (semicircle + 2 quadrants).

Answer: (a) 14 cm²

Explanation:
1. Perimeter of the sector = \( 2r+\frac{90}{360}\times 2\pi r = 2r+\frac{11}{7}r = \frac{25}{7}r = 25 \), so r = 7 cm.
2. Segment area = sector − triangle = \( \frac14\times\frac{22}{7}\times 49-\frac12\times 49 = 38.5-24.5 \).
3. So the area is 14 cm².

Teacher's Note:
1. Perimeter of a sector = 2r + arc length.
2. The triangle formed by the two radii is right-angled.

Answer: (a) 4 : 1

Explanation:
1. Let the radii be r₁ and r₂. Then r₁ + r₂ = 15 and \( r_1^2+r_2^2 = 153 \).
2. So 2r₁r₂ = 225 − 153 = 72, r₁r₂ = 36, which gives r₁ = 12 and r₂ = 3.
3. The ratio is 12 : 3 = 4 : 1.

Teacher's Note:
1. Use (r₁ + r₂)² = r₁² + r₂² + 2r₁r₂.
2. Check: 144 + 9 = 153.

Answer: (b) 4935 sq. m

Explanation:
1. Inner radius r = \( \frac{437}{2\pi} \) and outer radius R = \( \frac{503}{2\pi} \).
2. Area = \( \pi(R+r)(R-r) = \pi\times\frac{940}{2\pi}\times\frac{66}{2\pi} = \frac{940\times 66}{4\pi} \).
3. With π = 22/7 this is \( \frac{940\times 66\times 7}{4\times 22} = 4935 \) sq. m.

Teacher's Note:
1. Area of a ring = π(R² − r²) = π(R + r)(R − r).
2. Use the given circumferences to find R + r and R − r.

Answer: (b) \( d_1+d_2=d \)

Explanation:
1. \( \pi d_1+\pi d_2=\pi d \), since circumference = π × diameter.
2. Dividing by π gives d₁ + d₂ = d.

Teacher's Note:
1. Circumference is proportional to the diameter.
2. Cancel π from both sides.

Answer: (a) 10.5 cm²

Explanation:
1. Area of the square = 7 × 7 = 49 cm², and the quadrant has area \( \frac14\times\frac{22}{7}\times 49 = 38.5 \) cm².
2. Shaded area = 49 − 38.5 = 10.5 cm².

Teacher's Note:
1. The shaded part is the corner of the square outside the quadrant.
2. Square − quadrant.

Answer: (a) 43.98 sq. cm

Explanation:
1. Area = π(R² − r²) = π(R + r)(R − r) = π × 10 × 1.4.
2. = 3.1416 × 14 = 43.98 sq. cm.

Teacher's Note:
1. Factorise R² − r² to simplify the calculation.
2. 5.7 + 4.3 = 10 and 5.7 − 4.3 = 1.4.

Answer: (b) 81.6 cm

Explanation:
1. Area of the sector = \( \frac{240}{360}\pi(100)^2 = \frac{20000\pi}{3} \).
2. For the new circle, \( \pi r^2 = \frac{20000\pi}{3} \), so \( r=\sqrt{\frac{20000}{3}} \approx 81.6 \) cm.

Teacher's Note:
1. π cancels from both sides.
2. √(20000/3) = √6666.7.

Answer: (d) \( \frac{p}{720^\circ}\times 2\pi R^2 \)

Explanation:
1. Area of a sector = \( \frac{p}{360^\circ}\times\pi R^2 \).
2. Writing \( \frac{p}{360}=\frac{p}{720}\times 2 \) gives \( \frac{p}{720^\circ}\times 2\pi R^2 \), which is option (d).
3. Options (a) and (c) involve 2πR, which is a length, not an area.

Teacher's Note:
1. Sector area = (θ/360) × πr².
2. Check that the unit of the answer is area, not length.

Answer: (b) 4π cm

Explanation:
1. Radius = 5 cm.
2. Arc length = \( \frac{144}{360}\times 2\pi\times 5 = \frac{2}{5}\times 10\pi \) = 4π cm.

Teacher's Note:
1. Arc length = (θ/360) × 2πr.
2. 144/360 = 2/5.

Answer: (c) \( \frac{\pi}{2} \)

Explanation:
1. Let the radius of the outer circle be r. Join B to O and C to O; OQ = BC = r because the diagonals of the square BQCO are equal.
2. So the side of ABCD is r and its perimeter is 4r.
3. Ratio = \( \frac{2\pi r}{4r} = \frac{\pi}{2} \).

Teacher's Note:
1. BQCO is a square, so its diagonals BC and OQ are equal.
2. OQ is a radius of the outer circle.

Answer: (c) 1694 m²

Explanation:
1. Outer radius = 35 + 7 = 42 m.
2. Area of the path = \( \frac{22}{7}\left(42^2-35^2\right) = \frac{22}{7}\times 539 \) = 1694 m².

Teacher's Note:
1. The outer radius includes the width of the path.
2. Path = outer circle − inner circle.

Answer: (c) 25 cm

Explanation:
1. Perimeter = arc length + two radii = \( \frac14\times 2\pi r+2r \).
2. = \( \frac14\times 2\times\frac{22}{7}\times 7+2\times 7 = 11+14 \) = 25 cm.

Teacher's Note:
1. A quadrant has two straight edges of length r.
2. Arc length = one quarter of the circumference.

Answer: (d) quadrupled

Explanation:
1. 2πr = 4π gives r = 2 and the area is 4π. 2πr = 8π gives r = 4 and the area is 16π.
2. 16π is 4 times 4π, so the area is quadrupled.

Teacher's Note:
1. Doubling the radius multiplies the area by 4.
2. Compare the two areas directly.

Answer: (b) 19%

Explanation:
1. New radius = 0.9r, so the new area = \( \pi(0.9r)^2 = 0.81\pi r^2 \).
2. Decrease = (1 − 0.81) × 100 = 19%.

Teacher's Note:
1. Area depends on the square of the radius.
2. 0.9² = 0.81.

Answer: (b) 14 cm

Explanation:
1. Perimeter = πr + 2r = \( \left(\frac{22}{7}+2\right)r = \frac{36}{7}r = 36 \).
2. So r = 7 cm and the diameter is 14 cm.

Teacher's Note:
1. A semicircle's perimeter includes the straight diameter.
2. πr + 2r = r(π + 2).

Answer: (a) \( \pi d(2r+d) \)

Explanation:
1. The outer radius is r + d, so the area of the path = \( \pi\left[(r+d)^2-r^2\right] \).
2. = \( \pi\left(2rd+d^2\right) = \pi d(2r+d) \).

Teacher's Note:
1. Outer circle − inner circle.
2. Expand (r + d)² carefully.

Answer: (a) \( 14\sqrt{3} \) m

Explanation:
1. Semi-perimeter s = (35 + 53 + 66) ÷ 2 = 77 m. By Heron's formula, area = \( \sqrt{77\times 42\times 24\times 11} = 924 \) m².
2. The circle has area 2 × 924 = 1848 m², so \( \frac{22}{7}r^2 = 1848 \) and r² = 588.
3. So r = \( \sqrt{588} = 14\sqrt{3} \) m.

Teacher's Note:
1. Use Heron's formula for the triangle.
2. 588 = 196 × 3.

Answer: (a) 5 cm²

Explanation:
1. The inscribed square has side √2 cm, with its base centred on the diameter. Half of the base is √2/2.
2. The radius satisfies \( r^2=\left(\sqrt{2}\right)^2+\left(\frac{\sqrt2}{2}\right)^2 = \frac52 \).
3. A square inscribed in a full circle has diagonal 2r, so its area is \( \frac12(2r)^2 = 2r^2 = 5 \) cm².

Teacher's Note:
1. Use Pythagoras theorem for the radius to the top corner.
2. Area of a square = ½ × diagonal².

Answer: (c) 0.7 and 0.75

Explanation:
1. Let the centres be A and B and the circles meet at P and Q. In the triangle APC, with C the mid-point of AB, AC = √3 and AP = 2, so cos θ = √3/2 and θ = ∠PAC = 30°. The chord PQ subtends 60° at A.
2. The common region is made of two equal segments, each cut by the chord PQ: area = \( 2\left(\frac{60}{360}\pi\times 2^2-\frac12\times 2^2\sin 60^\circ\right) \).
3. = \( 2\left(\frac{2\pi}{3}-\sqrt{3}\right) = 2(2.09-1.73) = 0.72 \), which lies between 0.7 and 0.75.

Teacher's Note:
1. Each common segment is a sector of 60° minus an equilateral triangle of side 2.
2. Use the given values of π and √3.

Answer: (a) 4 : 7

Explanation:
1. Let the radius of the incircle be r. Circumference = 2πr.
2. Joining the centre to the vertices splits the triangle into three triangles of height r, so area = \( \frac12 r\times(\text{perimeter}) = \frac12 r\times 7\pi \).
3. Ratio = \( \frac{2\pi r}{\frac12 r\times 7\pi} = \frac47 \).

Teacher's Note:
1. Area of a triangle = ½ × inradius × perimeter.
2. r and π cancel in the ratio.

Answer: (d) 49 cm²

Explanation:
1. Area = \( \pi r^2 = \pi\times\frac{49}{\pi} \) = 49 cm².

Teacher's Note:
1. π cancels with the π in the denominator.
2. Square the radius first.

Answer: (c) \( \frac{5}{4} \)

Explanation:
1. \( \frac{\pi r_1^2}{\pi r_2^2}=\frac{25}{16} \) gives \( \frac{r_1}{r_2}=\frac54 \).
2. Circumferences are in the same ratio as radii: 5 : 4, that is \( \frac54 \).

Teacher's Note:
1. Areas are in the ratio of the squares of the radii.
2. Circumferences are in the ratio of the radii.

Answer: (b) \( \frac{49\pi}{12} \) cm²

Explanation:
1. Radius = 7 cm.
2. Area = \( \frac{30}{360}\times\pi\times 7^2 = \frac{49\pi}{12} \) cm².

Teacher's Note:
1. Sector area = (θ/360) × πr².
2. 30/360 = 1/12.

Answer: (a) The area of the sector of a circle of radius 6 cm and central angle 60° is \( \frac{132}{14} \) cm².

Explanation:
1. Statement (a): area = (60/360) × 3.14 × 36 = 18.84 cm², which is not 132/14 (about 9.43). So (a) is not correct.
2. Statement (b): minor sector = (60/360) × 3.14 × 196 = 102.57, and the major sector = 615.44 − 102.57 = 512.87 cm². Correct.
3. Statement (c): 2π(5) : π(5²) = 10 : 25 = 2 : 5. Correct. Statement (d): area = ½ × arc × radius = ½ × 22 × 6 = 66 cm². Correct.
4. The statement that is not correct is (a).

Teacher's Note:
1. Check each statement with the sector formulae.
2. Sector area = ½ × arc length × radius.

Answer: (d) A line segment joining any two points of a circle is called an arc.

Explanation:
1. Statements (a), (b) and (c) are correct definitions.
2. A line segment joining two points of a circle is a chord, not an arc. An arc is a part of the circle itself.
3. So the incorrect statement is (d).

Teacher's Note:
1. A chord is a line segment; an arc is a curve.
2. Concentric circles share a centre.

Answer: (a) 2 units

Explanation:
1. Perimeter = 2πr and area = πr², and these are numerically equal: 2πr = πr².
2. Dividing by πr gives r = 2 units.

Teacher's Note:
1. Divide both sides by πr (r is not zero).
2. Check: 2π(2) = 4π and π(2)² = 4π.

Answer: (a) \( \frac{\pi r^2\theta}{360} \)

Explanation:
1. The sector is the fraction \( \frac{\theta}{360} \) of the whole circle.
2. Area = \( \frac{\theta}{360}\times\pi r^2 = \frac{\pi r^2\theta}{360} \).

Teacher's Note:
1. Fraction of the circle × area of the circle.
2. Options (c) and (d) have r to the first power, so they are lengths.

Answer: (b) \( R_1^2+R_2^2=R^2 \)

Explanation:
1. \( \pi R_1^2+\pi R_2^2=\pi R^2 \).
2. Dividing by π, \( R_1^2+R_2^2=R^2 \), which is option (b).

Teacher's Note:
1. Areas add up; radii do not.
2. Divide the equation by π.

Answer: (a) 10 m

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

Teacher's Note:
1. Use radius = half of the diameter.
2. 8² + 6² = 10².

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

Explanation:
1. 2 × (22/7) × r = 176 gives r = 28 cm. The Assertion is true.
2. The Reason is the formula used, so it explains the Assertion: option (a).

Teacher's Note:
1. Use circumference = 2πr with π = 22/7.
2. Divide 176 by 44/7.

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

Explanation:
1. R = 5 m and r = 3 m, so area = π(25 − 9) = 16π m². The Assertion is true.
2. The Reason is the formula for the area of a ring, so it explains the Assertion: option (a).

Teacher's Note:
1. Convert diameters to radii first.
2. Ring area = π(R² − r²).

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

Explanation:
1. The circumference is 22 cm, so 2πr = 22 and r = 3.5 cm (π = 22/7). Area = (22/7) × 3.5² = 38.5 cm², not 40 cm². The Assertion is false.
2. The Reason is true: the wire forms the boundary of the circle.
3. A false, R true: option (d).

Teacher's Note:
1. Find the radius from the circumference first.
2. A wire bent into a circle keeps its length as the perimeter.