Areas (Circles) Word Problems

Class 10 Previous Year Questions (2014 – 2026)

Topic Overview

Practical applications often combine circular geometry with physics (speed, distance) or daily life (fencing, grazing). Key relationship: $\text{Distance} = \text{Circumference} \times \text{Number of Revolutions}$.

Q1 2026

00:00

The wheels of a bicycle are of diameter 70 cm. How many revolutions will it make to cover a distance of 1.1 km?

Distance = 1.1 km = 110,000 cm.

Circumference = \pi d = (22/7) \times 70 = 220 cm.

Revolutions = 110,000 / 220 = 500.

Ans: 500

Q2 2025

00:00

Find the area of a square that can be inscribed in a circle of radius 8 cm.

Diameter of circle = 16 cm = Diagonal of square.

Area of square = 1/2 \times diagonal^2 = 1/2 \times 16^2 = 128 cm².

Ans: 128 cm²

Q3 2024

00:00

A horse is tied to a peg at one corner of a square grass field of side 15m by means of a 5m long rope. Find the area of that part of the field in which the horse can graze.

Horse grazes in a quadrant of a circle with r = 5m and \theta = 90^{\circ}.

Area = (90/360) \times 3.14 \times 5 \times 5.

Area = 1/4 \times 78.5 = 19.625 m².

Ans: 19.625 m²

Q4 2023

00:00

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/h?

Speed = 66 km/h = 1100 m/min.

Distance in 10 mins = 1100 \times 10 = 11000m = 1,100,000 cm.

Circumference = \pi d = (22/7) \times 80 = 1760/7 cm.

Revolutions = Distance/Circumference = 1,100,000 / (1760/7) = (1,100,000 \times 7) / 1760 = 4375.

Ans: 4375

Q5 2022

00:00

A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters. Find total length of wire used.

Circumference = \pi \times 35 = (22/7) \times 35 = 110 mm.

Length of 5 diameters = 5 \times 35 = 175 mm.

Total length = 110 + 175 = 285 mm.

Ans: 285 mm

Q6 2020

00:00

The cost of fencing a circular field at the rate of ₹24 per meter is ₹5280. The field is to be ploughed at the rate of ₹0.50 per m². Find the cost of ploughing the field.

Circumference = 5280/24 = 220 m.

2 \times (22/7) \times r = 220 \Rightarrow r = 35 m.

Area = (22/7) \times 35 \times 35 = 3850 m².

Cost of ploughing = 3850 \times 0.50 = ₹1925.

Ans: ₹1925

Q5 2024

00:00

In the square field problem (Q1), find the increase in the grazing area if the rope were 10m long instead of 5m. (Use $\pi = 3.14$)

New area $= (90/360) \times 3.14 \times 10^2 = 78.5$ m².

Old area $= 19.625$ m².

Increase in area $= 78.5 - 19.625 = 58.875$ m².

Ans: 58.875 m²

Q6 2022

00:00

A track is in the form of a ring whose inner circumference is 352 m and outer circumference is 396 m. Find the width of the track and its area.

$2\pi r = 352 \Rightarrow 2 \times 22/7 \times r = 352 \Rightarrow r = 56$ m.

$2\pi R = 396 \Rightarrow 2 \times 22/7 \times R = 396 \Rightarrow R = 63$ m.

Width $= R - r = 63 - 56 = 7$ m.

Area $= \pi(R^2 - r^2) = (22/7)(63^2 - 56^2) = (22/7)(63-56)(63+56) = 22 \times 119 = 2618$ m².

Ans: Width = 7m, Area = 2618 m²

00:00

The wheels of a bicycle are of diameter 70 cm. How many revolutions will it make to cover a distance of 1.1 km?

Distance = 1.1 km = 110,000 cm.

Circumference = \pi d = (22/7) \times 70 = 220 cm.

Revolutions = 110,000 / 220 = 500.

Ans: 500

Q2 2025

00:00

Find the area of a square that can be inscribed in a circle of radius 8 cm.

Diameter of circle = 16 cm = Diagonal of square.

Area of square = 1/2 \times diagonal^2 = 1/2 \times 16^2 = 128 cm².

Ans: 128 cm²

Q3 2024

00:00

A horse is tied to a peg at one corner of a square grass field of side 15m by means of a 5m long rope. Find the area of that part of the field in which the horse can graze.

Horse grazes in a quadrant of a circle with r = 5m and \theta = 90^{\circ}.

Area = (90/360) \times 3.14 \times 5 \times 5.

Area = 1/4 \times 78.5 = 19.625 m².

Ans: 19.625 m²

Q4 2023

00:00

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/h?

Speed = 66 km/h = 1100 m/min.

Distance in 10 mins = 1100 \times 10 = 11000m = 1,100,000 cm.

Circumference = \pi d = (22/7) \times 80 = 1760/7 cm.

Revolutions = Distance/Circumference = 1,100,000 / (1760/7) = (1,100,000 \times 7) / 1760 = 4375.

Ans: 4375

Q5 2022

00:00

A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters. Find total length of wire used.

Circumference = \pi \times 35 = (22/7) \times 35 = 110 mm.

Length of 5 diameters = 5 \times 35 = 175 mm.

Total length = 110 + 175 = 285 mm.

Ans: 285 mm

Q6 2020

00:00

The cost of fencing a circular field at the rate of ₹24 per meter is ₹5280. The field is to be ploughed at the rate of ₹0.50 per m². Find the cost of ploughing the field.

Circumference = 5280/24 = 220 m.

2 \times (22/7) \times r = 220 \Rightarrow r = 35 m.

Area = (22/7) \times 35 \times 35 = 3850 m².

Cost of ploughing = 3850 \times 0.50 = ₹1925.

Ans: ₹1925

Q5 2024

00:00

In the square field problem (Q1), find the increase in the grazing area if the rope were 10m long instead of 5m. (Use $\pi = 3.14$)

New area $= (90/360) \times 3.14 \times 10^2 = 78.5$ m².

Old area $= 19.625$ m².

Increase in area $= 78.5 - 19.625 = 58.875$ m².

Ans: 58.875 m²

Q6 2022

00:00

A track is in the form of a ring whose inner circumference is 352 m and outer circumference is 396 m. Find the width of the track and its area.

$2\pi r = 352 \Rightarrow 2 \times 22/7 \times r = 352 \Rightarrow r = 56$ m.

$2\pi R = 396 \Rightarrow 2 \times 22/7 \times R = 396 \Rightarrow R = 63$ m.

Width $= R - r = 63 - 56 = 7$ m.

Area $= \pi(R^2 - r^2) = (22/7)(63^2 - 56^2) = (22/7)(63-56)(63+56) = 22 \times 119 = 2618$ m².

Ans: Width = 7m, Area = 2618 m²