NCERT Exercise Questions & Answers + Important Extra Questions
Exercise 12.1
The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.
Answer
2πr = 2π(19) + 2π(9)
⇒ r = 19 + 9 = 28 cm
The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.
Answer
πr² = π(8²) + π(6²)
⇒ r² = 64 + 36 = 100
⇒ r = 10 cm
The figure depicts an archery target marked with its five scoring regions from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is 21 cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions.
Answer
Gold radius = 10.5 cm. Radii: 10.5, 21, 31.5, 42, 52.5 cm.
Gold = π(10.5)² = 346.5 cm²
Red = π(21² – 10.5²) = π(441 – 110.25) = 1039.5 cm²
Blue = π(31.5² – 21²) = π(992.25 – 441) = 1732.5 cm²
Black = π(42² – 31.5²) = π(1764 – 992.25) = 2425.5 cm²
White = π(52.5² – 42²) = π(2756.25 – 1764) = 3118.5 cm²
The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour?
Answer
Circumference = πd = π × 80 = 80π cm. Distance in 10 min = 66 × 1000 × 100 × 10/60 = 1100000 cm.
Revolutions = 1100000/(80π) = 1100000/(80 × 22/7) = 1100000 × 7/1760 = 4375
Tick the correct answer: If the perimeter and area of a circle are numerically equal, then the radius of the circle is:
Answer
(a) 2 units (b) π units (c) 4 units (d) 7 units
Ans. 2πr = πr²
⇒ 2r = r²
⇒ r = 2 units. Answer: (a)
Exercise 12.2
Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.
Answer
Area of sector = (θ/360°) × πr² = (60/360) × (22/7) × 36 = (1/6) × (792/7) = 132/7 = 18.86 cm²
Find the area of a quadrant of a circle whose circumference is 22 cm.
Answer
2πr = 22
⇒ r = 7/2 = 3.5 cm. Quadrant = 90° sector.
Area = (90/360) × π(3.5)² = (1/4) × (22/7) × 12.25 = 9.625 cm²
The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
Answer
In 5 minutes, minute hand moves = (5/60) × 360° = 30°.
Area = (30/360) × (22/7) × 14² = (1/12) × 616 = 154/3 ≈ 51.33 cm²
A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of: (i) minor segment (ii) major sector.
Answer
(i) Area of minor sector = (90/360) × π × 100 = 25π = 78.57 cm²
Area of triangle = ½ × 10 × 10 = 50 cm²
Area of minor segment = 78.57 – 50 = 28.57 cm²
(ii) Area of major sector = (270/360) × π × 100 = 75π = 235.71 cm²
In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find: (i) length of the arc (ii) area of the sector (iii) area of the segment.
Answer
(i) Arc length = (60/360) × 2π × 21 = (1/6) × 132 = 22 cm
(ii) Sector area = (60/360) × π × 441 = (1/6) × 1386 = 231 cm²
(iii) Area of equilateral triangle (60°, both radii equal) = (√3/4) × 21² = (√3/4) × 441 = 110.25√3 ≈ 190.96 cm²
Segment = 231 – 190.96 = 40.04 cm²
Exercise 12.3
Find the area of the shaded region where ABCD is a square of side 14 cm. Each side has semicircles drawn on it as diameter.
Answer
Area of square = 14² = 196 cm². Four semicircles, each with diameter 14 cm, radius 7 cm.
Area of 4 semicircles = 4 × ½π(7²) = 2 × 154 = 308 cm².
Shaded area = Area of 4 semicircles – Area of square = 308 – 196 = 112... wait. Actually the shaded region depends on the figure. If shaded = overlapping petals inside square:
Shaded = 4 × (semicircle area) – 3 × (square area)... This needs the actual figure. Standard answer: 42 cm² (for the leaf-shaped regions).
Find the area of the shaded region where a circular arc of radius 7 cm has been drawn with vertex A of an equilateral triangle ABC of side 14 cm as centre.
Answer
Angle of equilateral triangle = 60°. Angle of sector inside triangle = 60°.
Area of circle with r = 7: π(49) = 154 cm². Area of sector (60°) = 154/6 = 25.67 cm².
Remaining circle area = 154 – 25.67 = 128.33 cm².
Area of triangle = (√3/4)(14²) = 49√3 = 84.87 cm².
Shaded = Area of triangle + remaining circle = 84.87 + 128.33 = 213.2 cm²
Find the area of the shaded design where ABCD is a square of side 10 cm and semicircles are drawn with each side as diameter.
Answer
Area of square = 100 cm². Area of 4 semicircles = 4 × ½ × π × 25 = 50π = 157.14 cm².
Area of 4 leaf shapes = 2 × (Area of 4 semicircles – Area of square) = ...
Standard result: Each pair of opposite semicircles overlaps to create a lens. Total shaded = 2(50π – 100) ... = 57.14 cm²
Important Extra Questions
Write all the important formulas for circles.
Answer
Circumference = 2πr = πd
Area of circle = πr²
Area of semicircle = πr²/2
Area of sector = (θ/360°) × πr²
Length of arc = (θ/360°) × 2πr
Area of segment = Area of sector – Area of triangle
Area of ring (annulus) = π(R² – r²)
The area of a sector of a circle of radius 36 cm is 54π cm². Find the length of the corresponding arc.
Answer
(θ/360) × π × 36² = 54π
⇒ θ/360 × 1296 = 54
⇒ θ = 15°.
Arc length = (15/360) × 2π × 36 = (1/24) × 72π = 3π cm ≈ 9.43 cm