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NCERT Solutions Class 10 Maths Chapter 12 – Areas Related to Circles Exercise 12.3

NCERT Solutions for Class 10 Maths Chapter 12 – Areas Related to Circles Exercise 12.3 are provided to help students understand and apply the concept of areas related to circles in a variety of problems. This exercise focuses on the combination of areas of different circular sections such as sectors, segments, and composite shapes.

Exercise 12.3 covers:

NCERT Solutions Class 10 Maths Chapter 12 – Areas Related to Circles Exercise 12.3

Area Related to Circles

Difficult

The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.

Area Related to Circles

Medium

From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in the following figure. Find the area of the remaining portion of the square.

Area Related to Circles

Medium

In the following figure, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.

Area Related to Circles

Medium

AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see the following figure). If ∠AOB = 30°, find the area of the shaded region

Area Related to Circles

Medium

In the following figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use π = 3.14)

Area Related to Circles

Medium

On a square handkerchief, nine circular designs each of radius 7 cm are made (see the following figure). Find the area of the remaining portion of the handkerchief.

Area Related to Circles

Medium

\begin{array}{l} {The area of an equilateral triangle ABC is 17320.5 cm}^{2} . \\ With each vertex of the triangle as centre, a circle is \\ drawn with radius equal to half the length of the side \\ of the triangle (see the following figure) . Find the \\ area of the shaded region. (Use π = 3.14 and \sqrt{3} = 1.73205) \end{array}

Area Related to Circles

Medium

In the following figure, AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.

Area Related to Circles

Medium

The following figure depicts a racing track whose left and right ends are semicircular.

The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find:

the distance around the track along its inner edge
the area of the track.

Area Related to Circles

Medium

In the following figure, ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touches externally two of the remaining three circles. Find the area of the shaded region.

Area Related to Circles

Medium

In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in the following figure. Find the area of the design (shaded region).

Area Related to Circles

Medium

Find the area of the shaded region in the following figure, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre.

Area Related to Circles

Medium

\begin{array}{l} A chord of a circle of radius 12 cm subtends an \\ angle of 120° at the centre. Find the area of the \\ corresponding segment of the circle. \\ (Use π = 3.14 and \sqrt{3} = 1.73) \end{array}

Area Related to Circles

Medium

Find the area of the shaded region in the following figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles.

Area Related to Circles

Medium

\begin{array}{l} Find the area of the shaded region in the following \\ figure, if radii of the two concentric circles with \\ centre O are 7 cm and 14 cm respectively and \\ ∠ AOC = 40°. \end{array}

Area Related to Circles

Medium

Find the area of the shaded region in the following figure, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.

Area Related to Circles

Medium

\begin{array}{l} Tick the correct answer in the following: \\ Area of a sector of angle p (in degrees) of a circle \\ with radius R is \\ (A) \frac{P}{180} \times 2 π R                               (B) \frac{P}{180} \times π R^{2} \\ (C) \frac{P}{360} \times 2 π R                              (D) \frac{P}{720} \times 2 π R^{2} \end{array}

Area Related to Circles

Medium

\begin{array}{l} A round table cover has six equal designs as shown \\ in the following figure. If the radius of the cover is \\ 28 cm, find the cost of making the designs at the \\ {rate of ₹ 0.35 per cm}^{2} . (Use \sqrt{3} = 1.7) \end{array}

Area Related to Circles

Medium

To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use π = 3.14)

Area Related to Circles

Medium

A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.

Area Related to Circles

Medium

An umbrella has 8 ribs which are equally spaced (see the following figure). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

Area Related to Circles

Medium

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 which divide the circle into 10 equal sectors as shown in the following figure. Find:
(i) the total length of the silver wire required.
(ii) the area of each sector of the brooch.

Area Related to Circles

Medium

\begin{array}{l} A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m \\ long rope (see the following figure) . Find \\ (i) the area of that part of the field in which the horse can graze. \\ (ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use π = 3.14) \end{array}

Area Related to Circles

Medium

Calculate the area of the designed region in the following figure common between the two quadrants of circles of radius 8 cm each.

NCERT Solutions Class 10 Maths Chapter 12 – Areas Related to Circles Exercise 12.3

Area of sectors and segments involving different geometric shapes.
Solving problems related to composite figures formed by combining parts of a circle with other shapes.
Using trigonometric identities and circle properties to calculate areas in various geometrical contexts.

These solutions are explained step-by-step, helping students grasp these concepts easily and apply them to real-life problems in the Class 10 exams.

Q1. Find the area of the shaded region in the figure, if PQ = 24 cm, PR = 7 cm, and O is the center of the circle.

Difficulty Level: Medium

Known/Given:

PQ = 24 cm, PR = 7 cm, O is the center of the circle.
Use π = 22/7

Unknown:

Area of the shaded region in the figure.

Reasoning:

Visually, it’s clear that:

$\text{Area of the shaded region} = \text{Area of semicircle RPQ} - \text{Area of ΔRPQ}$ $Area of the shaded region = Area of semicircle RPQ - Area of ΔRPQ$

Since we don’t know RQ (diameter), OR, and PQ (radii of the circle), we are unable to directly find the areas of the semicircle or the triangle RPQ using Heron’s formula.
We need to find RQ:

$\text{Angle subtended by arc at any point on the circle is half the angle subtended at the center.}$ $Angle subtended by arc at any point on the circle is half the angle subtended at the center.$

$\angle ROQ = \frac{180^\circ}{2} = 90^\circ$ $∠ ROQ = 2180 ^{\circ} = 9 0^{\circ}$

Therefore, ΔRPQ is a right-angled triangle. Using Pythagoras’ theorem:

$RQ^2 = PR^2 + PQ^2 = 7^2 + 24^2 = 49 + 576 = 625$ $R Q^{2} = P R^{2} + P Q^{2} = 7^{2} + 2 4^{2} = 49 + 576 = 625$

$RQ = \sqrt{625} = 25 \text{ cm}$ $RQ = 625 = 25 cm$
Area of the shaded region:

$\text{Area of shaded region} = \frac{1}{2} \times \pi \times RQ^2 - \text{Area of triangle RPQ}$ $Area of shaded region = 21 \times π \times R Q^{2} - Area of triangle RPQ$

Using the formula:

$\text{Area of semicircle RPQ} = \frac{1}{2} \times \pi \times RQ^2 = \frac{1}{2} \times \frac{22}{7} \times 25^2 = 161.54 \, \text{cm}^2$ $Area of semicircle RPQ = 21 \times π \times R Q^{2} = 21 \times 722 \times 2 5^{2} = 161.54 cm^{2}$

$\text{Area of triangle RPQ} = \frac{1}{2} \times PQ \times PR = \frac{1}{2} \times 24 \times 7 = 84 \, \text{cm}^2$ $Area of triangle RPQ = 21 \times PQ \times PR = 21 \times 24 \times 7 = 84 cm^{2}$

Therefore, the area of the shaded region is:

$161.54 - 84 = 77.54 \, \text{cm}^2$ $161.54 - 84 = 77.54 cm^{2}$

Q2. Find the area of the shaded region in the given figure, if radii of the two concentric circles with center O are 7 cm and 14 cm respectively and ∠AOC = 40°.

Difficulty Level: Medium

Known/Given:

Radii of two concentric circles are 7 cm and 14 cm.
∠AOC = 40°.

Unknown:

Area of the shaded region.

Reasoning:
The area of the shaded region can be calculated by subtracting the area of the sector of the smaller circle from the area of the sector of the larger circle.

$\text{Area of shaded region} = \text{Area of sector AOC} - \text{Area of sector BDO}$

$Area of shaded region = Area of sector AOC - Area of sector BDO$

Using the sector area formula:

$\text{Area of sector} = \frac{\theta}{360} \times \pi r^2$

$Area of sector = 360 θ \times π r^{2}$

For the larger circle (radius = 14 cm):

$\text{Area of sector AOC} = \frac{40}{360} \times \pi \times 14^2 = \frac{40}{360} \times \pi \times 196 = 21.77 \, \text{cm}^2$

$Area of sector AOC = 36040 \times π \times 1 4^{2} = 36040 \times π \times 196 = 21.77 cm^{2}$

For the smaller circle (radius = 7 cm):

$\text{Area of sector BDO} = \frac{40}{360} \times \pi \times 7^2 = \frac{40}{360} \times \pi \times 49 = 5.43 \, \text{cm}^2$

$Area of sector BDO = 36040 \times π \times 7^{2} = 36040 \times π \times 49 = 5.43 cm^{2}$

Thus, the area of the shaded region is:

$\text{Shaded area} = 21.77 - 5.43 = 16.34 \, \text{cm}^2$

$Shaded area = 21.77 - 5.43 = 16.34 cm^{2}$

Q3. Find the area of the shaded region in the given figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles.

Difficulty Level: Medium

Known/Given:

ABCD is a square with side 14 cm.
APD and BPC are semicircles.

Unknown:

Area of the shaded region.

Reasoning:

The diameter of each semicircle is equal to the side of the square:

$\text{Diameter of each semicircle} = 14 \, \text{cm}$ $Diameter of each semicircle = 14 cm$

Therefore, the radius of each semicircle is:

$\text{Radius} = \frac{14}{2} = 7 \, \text{cm}$ $Radius = 214 = 7 cm$
The area of the shaded region is the area of the square minus the areas of the two semicircles:

$\text{Area of shaded region} = \text{Area of square} - 2 \times \text{Area of semicircle}$ $Area of shaded region = Area of square - 2 \times Area of semicircle$

The area of the square is:

$\text{Area of square} = 14^2 = 196 \, \text{cm}^2$ $Area of square = 1 4^{2} = 196 cm^{2}$

The area of one semicircle is:

$\text{Area of semicircle} = \frac{1}{2} \times \pi \times 7^2 = \frac{1}{2} \times \pi \times 49 = 76.96 \, \text{cm}^2$ $Area of semicircle = 21 \times π \times 7^{2} = 21 \times π \times 49 = 76.96 cm^{2}$

Therefore, the area of the shaded region is:

$\text{Shaded area} = 196 - 2 \times 76.96 = 196 - 153.92 = 42.08 \, \text{cm}^2$ $Shaded area = 196 - 2 \times 76.96 = 196 - 153.92 = 42.08 cm^{2}$

Q4. Find the area of the shaded region in the given figure, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as center.

Difficulty Level: Medium

Known/Given:

Radius of the circular arc = 6 cm
Side of equilateral triangle OAB = 12 cm

Unknown:

Area of the shaded region.

Reasoning:

The angle of the sector is 60° since each angle of an equilateral triangle is 60°.
The area of the shaded region is the area of the circle sector minus the area of triangle OAB.
Area of the circle sector:

$\text{Area of sector} = \frac{60}{360} \times \pi \times 6^2 = \frac{1}{6} \times \pi \times 36 = 18.85 \, \text{cm}^2$ $Area of sector = 36060 \times π \times 6^{2} = 61 \times π \times 36 = 18.85 cm^{2}$
Area of the equilateral triangle OAB:
Using the formula for the area of an equilateral triangle:

$\text{Area of triangle} = \frac{\sqrt{3}}{4} \times 12^2 = \frac{\sqrt{3}}{4} \times 144 = 62.35 \, \text{cm}^2$ $Area of triangle = 4 3 \times 1 2^{2} = 4 3 \times 144 = 62.35 cm^{2}$
Area of the shaded region:

$\text{Shaded area} = \text{Area of sector} - \text{Area of triangle OAB} = 18.85 - 62.35 = 55.5 \, \text{cm}^2$ $Shaded area = Area of sector - Area of triangle OAB = 18.85 - 62.35 = 55.5 cm^{2}$

Q5. From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in the given figure. Find the area of the remaining portion of the square.

Difficulty Level: Medium

Known/Given:

Side of square = 4 cm
Radius of each quadrant = 1 cm
Diameter of the cut circle = 2 cm, so radius = 1 cm

Unknown:

Area of the remaining portion of the square.

Reasoning:

The area of the square is:

$\text{Area of square} = 4^2 = 16 \, \text{cm}^2$ $Area of square = 4^{2} = 16 cm^{2}$
The area of the cut circle:

$\text{Area of circle} = \pi \times 1^2 = \pi \times 1 = 3.14 \, \text{cm}^2$ $Area of circle = π \times 1^{2} = π \times 1 = 3.14 cm^{2}$
The area of each quadrant is:

$\text{Area of each quadrant} = \frac{1}{4} \times \pi \times 1^2 = \frac{1}{4} \times 3.14 = 0.785 \, \text{cm}^2$ $Area of each quadrant = 41 \times π \times 1^{2} = 41 \times 3.14 = 0.785 cm^{2}$

The total area of the quadrants cut from all four corners is:

$\text{Total area of quadrants} = 4 \times 0.785 = 3.14 \, \text{cm}^2$ $Total area of quadrants = 4 \times 0.785 = 3.14 cm^{2}$
The area of the remaining portion of the square is:

$\text{Remaining area} = \text{Area of square} - \text{Area of circles} - \text{Area of quadrants} = 16 - 3.14 - 3.14 = 9.72 \, \text{cm}^2$ $Remaining area = Area of square - Area of circles - Area of quadrants = 16 - 3.14 - 3.14 = 9.72 cm^{2}$

FAQs: Class 10 Maths Chapter 12 – Areas Related to Circles Exercise 12.3

Q1. What is the focus of Exercise 12.3?
Answer:
Exercise 12.3 focuses on solving problems involving composite figures where the areas of sectors, segments, and other shapes are combined to find the total area.

Q2. How do I calculate the area of a sector in composite figures?
Answer:
To calculate the area of a sector in composite figures, use the formula for the area of a sector:

$A_{\text{sector}} = \frac{\theta}{360^\circ} \times \pi r^2$

$A_{sector} = 360 ^{\circ} θ \times π r^{2}$

where

$\theta$

$θ$ is the central angle of the sector and

$r$

$r$ is the radius of the circle. Add or subtract other areas as required by the problem.

Q3. How is the area of a segment calculated in composite problems?
Answer:
The area of a segment is calculated by subtracting the area of the triangle formed by the chord and radius from the area of the sector:

$A_{\text{segment}} = A_{\text{sector}} - A_{\text{triangle}}$

$A_{segment} = A_{sector} - A_{triangle}$

Q4. How do I handle composite figures in area problems?
Answer:
For composite figures:

Break down the figure into simpler parts (like sectors, triangles, or rectangles).
Calculate the area of each part.
Add or subtract the areas of the parts to get the total area, depending on the problem.

Q5. How do NCERT Solutions help with exam preparation?
Answer:
NCERT Solutions provide clear explanations and step-by-step solutions to problems involving composite areas. By practicing these solutions, students can easily apply the formulae and solve complex geometry problems efficiently, helping them excel in Class 10 exams.

NCERT Solutions Class 10 Maths Chapter 12 – Areas Related to Circles Exercise 12.3

NCERT Solutions Class 10 Maths Chapter 12 – Areas Related to Circles Exercise 12.3

NCERT Solutions Class 10 Maths Chapter 12 – Areas Related to Circles Exercise 12.3

Q1. Find the area of the shaded region in the figure, if PQ = 24 cm, PR = 7 cm, and O is the center of the circle.

Q2. Find the area of the shaded region in the given figure, if radii of the two concentric circles with center O are 7 cm and 14 cm respectively and ∠AOC = 40°.

Q3. Find the area of the shaded region in the given figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles.

Q4. Find the area of the shaded region in the given figure, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as center.

Q5. From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in the given figure. Find the area of the remaining portion of the square.

FAQs: Class 10 Maths Chapter 12 – Areas Related to Circles Exercise 12.3

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NCERT Solutions Class 10 Maths Chapter 12 – Areas Related to Circles Exercise 12.3

NCERT Solutions Class 10 Maths Chapter 12 – Areas Related to Circles Exercise 12.3

NCERT Solutions Class 10 Maths Chapter 12 – Areas Related to Circles Exercise 12.3

Q1. Find the area of the shaded region in the figure, if PQ = 24 cm, PR = 7 cm, and O is the center of the circle.

Q2. Find the area of the shaded region in the given figure, if radii of the two concentric circles with center O are 7 cm and 14 cm respectively and ∠AOC = 40°.

Q3. Find the area of the shaded region in the given figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles.

Q4. Find the area of the shaded region in the given figure, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as center.

Q5. From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in the given figure. Find the area of the remaining portion of the square.

FAQs: Class 10 Maths Chapter 12 – Areas Related to Circles Exercise 12.3

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