Chapter Exam

Chapter 11: Areas Related to Circles — Online MCQ Test

Q: What is the circumference of a circle of radius r?

Answer: B. 2πr — The circumference of a circle is given by C = 2πr, where r is the radius.

Q: The area of a circle with diameter d is:

Answer: C. πd²/4 — Radius is d/2, so area = π(d/2)² = πd²/4.

Q: The length of an arc of a circle of radius r and central angle θ is:

Answer: A. θ/360 × 2πr — Arc length is the same fraction of the circumference as θ is of 360°.

Q: The area of a sector of angle θ in a circle of radius r is:

Answer: C. θ/360 × πr² — A sector with angle θ occupies θ/360 part of the circle, so its area is θ/360 × πr².

Q: Using π = 22/7, the circumference of a circle of radius 7 cm is:

Answer: B. 44 cm — Circumference = 2πr = 2 × 22/7 × 7 = 44 cm.

Q: Using π = 22/7, find the area of a sector of radius 21 cm and angle 60°.

Answer: A. 231 cm² — Area of sector = 60/360 × 22/7 × 21 × 21 = 231 cm².

Q: 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?

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.

Q: Find the area of the minor segment of a circle of radius 14 cm when the central angle is 60°, using π = 22/7.

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.

Q: 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.

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.

Q: Two equal circles of radius 14 cm have their centres 14 cm apart. Using π = 22/7, find the area common to both circles.

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.

MATHS · Grade 10 · CBSE(NCERT)

Practice Chapter 11: Areas Related to Circles with a free chapter-wise online MCQ test for CBSE(NCERT) Grade 10 MATHS. 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.

Questions

20m

Time Limit

Attempts Left

10 random questions from this chapter (mixed difficulty)
Questions you've seen before won't repeat until the pool resets
You have 20 minutes — exam auto-submits when time is up
Maximum 3 attempts per chapter
Results and explanations shown immediately after submission

New here? Register free — includes 3 free chapter exams.

Chapter 11: Areas Related to Circles — Important Questions & Answers (FAQ)

Frequently asked questions from CBSE(NCERT) Grade 10 MATHS — Chapter 11: Areas Related to Circles, with answers and explanations. These are sample questions; the exam has its own separate question set.

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 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.

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°.

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.

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.

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.

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.

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.

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.

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.

← Previous Chapter

Chapter 10: Circles

Next Chapter →

Chapter 12: Surface Areas and Volumes