CBSE Class 10 Mathematics Areas Related to Circles Assignment Set 12

Board Level Exercise: Very Short Answer Questions (1 Mark Each)

Question 1. If the circumference of a circle and the perimeter of a square are equal, then show that the area of the circle is greater than the area of the square.
Answer:
Let the radius of the circle be r and the side of the square be a. Then 2πr = 4a, so \( a = \dfrac{\pi r}{2} \).
Area of the circle = πr² and area of the square = \( a^2 = \dfrac{\pi^2 r^2}{4} \). The ratio is \( \dfrac{\pi r^2}{\pi^2 r^2 / 4} = \dfrac{4}{\pi} > 1 \).
So the area of the circle is greater than the area of the square.
In simple words: For the same boundary length, the circle encloses more area than a square.

Teacher's Note:
1. Write the radius in terms of the side.
2. Compare the two areas as a ratio.

Question 2. If the perimeter of a circle is equal to that of a square, then find the ratio of their areas.
Answer:
2πr = 4a gives \( r = \dfrac{2a}{\pi} \). Area of the circle : area of the square = \( \pi r^2 : a^2 = \dfrac{4a^2}{\pi} : a^2 = 4 : \pi \).
With \( \pi = \dfrac{22}{7} \): \( 4 : \dfrac{22}{7} = 28 : 22 = \) 14 : 11.
In simple words: The circle has 14 parts of area for every 11 parts of the square.

Teacher's Note:
1. Equate the perimeters.
2. Use π = 22/7.

Question 3. Find the area of the circle that can be inscribed in a square of side 6 cm.
Answer:
The diameter of the inscribed circle equals the side of the square, so r = 3 cm.
Area = πr² = 9π cm².
In simple words: The inscribed circle touches all four sides, so its diameter is the side of the square.

Teacher's Note:
1. Diameter = side.
2. Area = πr².

Question 4. Find the diameter of a circle whose area is equal to the sum of the areas of the two circles of radii 24 cm and 7 cm.
Answer:
\( \pi R^2 = \pi(24^2 + 7^2) = \pi \times 625 \), so R = 25 cm.
The diameter = 50 cm.
In simple words: Add the squares of the two radii (24, 7 and 25 form a Pythagorean triple).

Teacher's Note:
1. The π cancels.
2. Take the square root.

Question 5. Find the area of a sector of a circle of radius 21 cm and central angle 120°.
Answer:
Area = \( \dfrac{120}{360} \times \dfrac{22}{7} \times 21 \times 21 = \) 462 cm².
In simple words: A 120° sector is one third of the circle.

Teacher's Note:
1. Use A = (θ/360) × πr².
2. The sector is a third of the whole circle.

Question 6. The areas of two sectors of two different circles with equal corresponding arc lengths are equal. Is this statement true? Give reasons for your answer.
Answer:
No. The area of a sector is \( A = \dfrac{1}{2}\ell r \). If the arc lengths ℓ are equal but the radii are different, the areas are different (the sector of the larger circle has the larger area).
In simple words: Equal arcs do not mean equal areas when the radii differ.

Teacher's Note:
1. Use A = ℓr/2.
2. The area depends on both ℓ and r.

Short Answer Questions (2 Marks Each)

Teacher's Note:
1. Outer side = diameter.
2. Inner diagonal = diameter.

Teacher's Note:
1. The angle in a semicircle is a right angle.
2. Subtract the triangle from the circle.

Teacher's Note:
1. Break the figure into simple shapes.
2. The semicircle has radius 6 − 4 = 2 m.

Question 10. Find the area of the minor segment of a circle of radius 14 cm, when the angle of the corresponding sector is 60° (use π = 22/7 and \( \sqrt{3} = 1.732 \)).
Answer:
Area of the sector = \( \dfrac{60}{360} \times \dfrac{22}{7} \times 14^2 = 102.67 \) cm².
The triangle formed by the radii is equilateral with side 14 cm: area = \( \dfrac{\sqrt{3}}{4} \times 14^2 = 84.87 \) cm².
Area of the minor segment = 102.67 − 84.87 = 17.80 cm² (approximately).
In simple words: The segment is the sector minus the triangle formed by the two radii and the chord.

Teacher's Note:
1. Radii 14, 14 and angle 60° make an equilateral triangle.
2. Subtract the triangle area.

Teacher's Note:
1. Each arc is centred at a corner with radius half the side.
2. Subtract one full circle from the square.

Question 12. A circular park is surrounded by a road 21 m wide. If the radius of the park is 105 m, find the area of the road.
Answer:
The outer radius = 105 + 21 = 126 m. Area of the road = \( \pi(126^2 - 105^2) = \dfrac{22}{7} \times (126 - 105)(126 + 105) = \dfrac{22}{7} \times 21 \times 231 = \) 15246 m².
In simple words: The road is the ring between the park and the outer circle.

Teacher's Note:
1. Find the outer radius.
2. Use π(R² − r²) = π(R − r)(R + r).

Question 13. A circular pond is 17.5 m in diameter. It is surrounded by a 2 m wide path. Find the cost of constructing the path at the rate of Rs. 25 per m².
Answer:
Radius of the pond = 8.75 m; outer radius = 10.75 m. Area of the path = \( \pi(10.75^2 - 8.75^2) = \pi \times 2 \times 19.5 = 39\pi \) m².
With π = 3.14: area = 122.46 m², cost = 122.46 × 25 = Rs. 3061.50. (With π = 22/7 the area is 122.57 m² and the cost is Rs. 3064.29.)
In simple words: Find the area of the ring-shaped path and multiply by the rate.

Teacher's Note:
1. Outer radius = inner radius + width.
2. Cost = area × rate.

Question 14. Find the difference of the areas of a sector of angle 120° and its corresponding major sector of a circle of radius 21 cm.
Answer:
The major sector has angle 360° − 120° = 240°. The difference = \( \dfrac{240 - 120}{360} \times \dfrac{22}{7} \times 21^2 = \dfrac{1}{3} \times \dfrac{22}{7} \times 441 = \) 462 cm².
In simple words: The major sector is bigger by one third of the circle.

Teacher's Note:
1. The corresponding major sector has 240°.
2. Subtract the two sector areas.

Long Answer Questions (3 Marks Each)

Teacher's Note:
1. The two semicircular ends make one full circle of radius 2 m.
2. The straight part of the slot is 20 − 4 = 16 m long.

Teacher's Note:
1. Each angle of an equilateral triangle is 60°.
2. Add the three sectors.

Question 17. The diameters of the front and rear wheels of a tractor are 80 cm and 2 m respectively. Find the number of revolutions that the rear wheel will make in covering a distance in which the front wheel makes 1400 revolutions.
Answer:
Distance covered by the front wheel = \( 1400 \times \pi \times 80 \) cm. The rear wheel has diameter 200 cm, so its circumference is \( \pi \times 200 \).
Number of revolutions of the rear wheel = \( \dfrac{1400 \times 80\pi}{200\pi} = \) 560.
In simple words: The larger wheel turns fewer times: the number of turns is inversely proportional to the diameter.

Teacher's Note:
1. Distance = revolutions × circumference.
2. Convert 2 m to 200 cm.

Teacher's Note:
1. The sum of the angles of any quadrilateral is 360°.
2. Subtract the circle from the trapezium.

Question 19. Three circles each of radius 3.5 cm are drawn in such a way that each of them touches the other two. Find the area enclosed between these circles.
Answer:
The centres form an equilateral triangle of side 7 cm. Its area is \( \dfrac{\sqrt{3}}{4} \times 7^2 = 21.22 \) cm² (with \( \sqrt{3} = 1.732 \)).
The three sectors of 60° inside the triangle together form a semicircle: \( \dfrac{1}{2} \times \dfrac{22}{7} \times 3.5^2 = 19.25 \) cm².
Area enclosed = 21.22 − 19.25 = 1.97 cm² (approximately).
In simple words: The curved triangle between the circles is an equilateral triangle minus three sectors.

Teacher's Note:
1. The sides of the triangle are 2 × radius = 7 cm.
2. The three sectors make half a circle.

Question 20. On a square cardboard sheet of area 784 cm², four congruent circular plates of maximum size are placed such that each circular plate touches the other two plates and each side of the square sheet is tangent to two circular plates. Find the area of the square sheet not covered by the circular plates.
Answer:
Side of the sheet = 28 cm. Each side carries two plates touching it, so the diameter of a plate = 14 cm and the radius is 7 cm.
Area covered by the four plates = \( 4 \times \dfrac{22}{7} \times 49 = 616 \) cm².
Uncovered area = 784 − 616 = 168 cm².
In simple words: Two plates fit along each side of the sheet, so each diameter is half the side.

Teacher's Note:
1. Side = √784 = 28 cm.
2. Subtract the four circles from the square.

Very Long Answer Questions (4 Marks Each)

Teacher's Note:
1. Convert 50 cm to 0.5 m.
2. Subtract the circular tiles from the floor.

Teacher's Note:
1. Squares of 1, 2, 3 are 1, 4, 9.
2. Take successive differences.

Question 23. The area of a sector of central angle 200° of a circle is 770 cm². Find the length of the corresponding arc of this sector.
Answer:
\( \dfrac{200}{360} \times \dfrac{22}{7} \times r^2 = 770 \), so \( r^2 = \dfrac{770 \times 360 \times 7}{200 \times 22} = 441 \) and r = 21 cm.
Arc length = \( \dfrac{200}{360} \times 2 \times \dfrac{22}{7} \times 21 = \dfrac{220}{3} = \) \( 73\dfrac{1}{3} \) cm.
In simple words: Find r from the area first and then use the arc formula.

Teacher's Note:
1. Area of the sector = (θ/360) πr².
2. Arc = (θ/360) × 2πr.

Question 24. Sides of a triangular field are 15 m, 16 m and 17 m. With the three corners of the field a cow, a buffalo and a horse are tied separately with ropes of length 7 m each to graze in the field. Find the area of the field which cannot be grazed by the three animals.
Answer:
s = 24 m: area of the field = \( \sqrt{24 \times 9 \times 8 \times 7} = 24\sqrt{21} \) m².
The three sectors of radius 7 m have angles A, B, C which add up to 180°: area grazed = \( \dfrac{1}{2} \times \dfrac{22}{7} \times 7^2 = 77 \) m². (The ropes of length 7 m do not overlap since the shortest side is 15 m > 14 m.)
Area that cannot be grazed = (24√21 − 77) m² (about 33 m²).
In simple words: The grazed regions are three sectors that together make half a circle.

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

Question 25. All the vertices of a rhombus lie on a circle. Find the area of the rhombus, if the area of the circle is 1256 cm² (use π = 3.14).
Answer:
A rhombus whose vertices lie on a circle is a cyclic rhombus, so it is a square, and its diagonals are diameters of the circle.
πr² = 1256 gives r² = 400, r = 20 cm and the diagonal = 40 cm. Area of the square = ½ × 40 × 40 = 800 cm².
In simple words: A cyclic rhombus must be a square.

Teacher's Note:
1. Opposite angles of a cyclic quadrilateral add up to 180°.
2. Area of a square = ½ × diagonal².