Chapter 11: Areas Related to Circles — Online MCQ Test
MATHS · Grade 10 · CBSE(NCERT)
Practice Chapter 11: Areas Related to Circles with a free chapter-wise online MCQ test.
This chapter covers: Circle Circumference Sector Segment Arc Length Area of sector Area of segment Composite figures Perimeter.
AI-generated questions from basic to board-exam level, with instant results and explanations.
Chapter 11: Areas Related to Circles — Important Questions & Answers
What is the circumference of a circle of radius r?
- A. πr
- B. 2πr
- C. πr²
- D. 2πr²
Answer: B. 2πr
The circumference of a circle is given by C = 2πr, where r is the radius.
The circumference of a circle is given by C = 2πr, where r is the radius.
The area of a circle with diameter d is:
- A. πd²
- B. 2πd²
- C. πd²/4
- D. πd/2
Answer: C. πd²/4
Radius is d/2, so area = π(d/2)² = πd²/4.
Radius is d/2, so area = π(d/2)² = πd²/4.
The length of an arc of a circle of radius r and central angle θ is:
- A. θ/360 × 2πr
- B. θ/360 × πr²
- C. 360/θ × 2πr
- D. θ/180 × πr²
Answer: A. θ/360 × 2πr
Arc length is the same fraction of the circumference as θ is of 360°.
Arc length is the same fraction of the circumference as θ is of 360°.
The area of a sector of angle θ in a circle of radius r is:
- A. θ/360 × 2πr
- B. θ/180 × πr²
- C. θ/360 × πr²
- D. 360/θ × πr²
Answer: C. θ/360 × πr²
A sector with angle θ occupies θ/360 part of the circle, so its area is θ/360 × πr².
A sector with angle θ occupies θ/360 part of the circle, so its area is θ/360 × πr².
Using π = 22/7, the circumference of a circle of radius 7 cm is:
- A. 22 cm
- B. 44 cm
- C. 49 cm
- D. 154 cm
Answer: B. 44 cm
Circumference = 2πr = 2 × 22/7 × 7 = 44 cm.
Circumference = 2πr = 2 × 22/7 × 7 = 44 cm.
Using π = 22/7, find the area of a sector of radius 21 cm and angle 60°.
- A. 231 cm²
- B. 462 cm²
- C. 693 cm²
- D. 1386 cm²
Answer: A. 231 cm²
Area of sector = 60/360 × 22/7 × 21 × 21 = 231 cm².
Area of sector = 60/360 × 22/7 × 21 × 21 = 231 cm².
A racing track consists of a rectangle of length 100 m and breadth 50 m with semicircular ends on the breadths. What is its perimeter?
- A. 200 + 25π m
- B. 200 + 50π m
- C. 300 + 50π m
- D. 100 + 100π m
Answer: B. 200 + 50π m
The two semicircular ends together form one full circle of diameter 50 m. Perimeter = 2 × 100 + π × 50 = 200 + 50π m.
The two semicircular ends together form one full circle of diameter 50 m. Perimeter = 2 × 100 + π × 50 = 200 + 50π m.
Find the area of the minor segment of a circle of radius 14 cm when the central angle is 60°, using π = 22/7.
- A. 308/3 - 49√3 cm²
- B. 308/3 + 49√3 cm²
- C. 154 - 49√3 cm²
- D. 49√3 - 308/3 cm²
Answer: A. 308/3 - 49√3 cm²
Area of sector = 60/360 × 22/7 × 14² = 308/3 cm². The triangle is equilateral with area 49√3 cm², so segment area = 308/3 - 49√3 cm².
Area of sector = 60/360 × 22/7 × 14² = 308/3 cm². The triangle is equilateral with area 49√3 cm², so segment area = 308/3 - 49√3 cm².
Three goats are tied at the three vertices of an equilateral triangular field of side 14 m, each with a rope of length 7 m. Using π = 22/7, find the area of the field not grazed by the goats.
- A. 49√3 - 77 m²
- B. 77 - 49√3 m²
- C. 49√3 - 154 m²
- D. 98√3 - 77 m²
Answer: A. 49√3 - 77 m²
Area of triangle = 49√3 m². The grazed area is three 60° sectors of radius 7 m, equal to 77 m²; hence ungrazed area = 49√3 - 77 m².
Area of triangle = 49√3 m². The grazed area is three 60° sectors of radius 7 m, equal to 77 m²; hence ungrazed area = 49√3 - 77 m².
Two equal circles of radius 14 cm have their centres 14 cm apart. Using π = 22/7, find the area common to both circles.
- A. 616/3 - 98√3 cm²
- B. 308/3 - 49√3 cm²
- C. 616/3 + 98√3 cm²
- D. 308/3 + 49√3 cm²
Answer: A. 616/3 - 98√3 cm²
The common region is made of two identical 60° segments. Each segment has area 308/3 - 49√3, so total common area = 616/3 - 98√3 cm².
The common region is made of two identical 60° segments. Each segment has area 308/3 - 49√3, so total common area = 616/3 - 98√3 cm².