NCERT Solutions for Class 10 Maths Exercise 12.3

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NCERT Solutions for Class 10 Maths Exercise 12.3 Class 10 Maths book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 10 Maths chapter wise NCERT solution for Maths Book for all the chapters can be downloaded from our website and myCBSEguide mobile app for free.

NCERT solutions for Maths Area Related to Circles Download as PDF

NCERT Solutions for Class 10 Maths Exercise 12.3

NCERT Solutions for Class 10 Maths Area Related to Circles

Unless stated otherwise, take
1. Find the area of the shaded region in figure, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.

Ans. RPQ = [Angle in semi-circle is ]

= = 49 + 576 = 625

RQ = 25 cm

Diameter of the circle = 25 cm

Radius of the circle = cm

Area of the semicircle =

= =

Area of right triangle RPQ =

= =

Area of shaded region = Area of semicircle – Area of right triangle RPQ

= =


2. Find the area of the shaded region in figure, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and AOC =

Ans. Area of shaded region = Area of sector OAC – Area of sector OBD

=

=

=

=

=


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

Ans. Area of shaded region

= Area of square ABCD – (Area of semicircle APD + Area of semicircle BPC)

=

=

= 196 – 154 =


NCERT Solutions for Class 10 Maths Exercise 12.3

4. Find the area of the shaded region in 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.

Ans. Area of shaded region

= Area of circle + Area of equilateral triangle OAB – Area common to the circle and the triangle

=

=

= =

=


NCERT Solutions for Class 10 Maths Exercise 12.3

5. 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 figure. Find the area of the remaining portion of the figure.

Ans. Area of remaining portion of the square

= Area of square – (4 Area of a quadrant + Area of a circle)

=

= =


NCERT Solutions for Class 10 Maths Exercise 12.3

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

Ans. Area of design = Area of circular table cover – Area of the equilateral triangle ABC

= ………(i)

G is the centroid of the equilateral triangle.

radius of the circumscribed circle = cm

According to the question,

= 48 cm

Now,

= 3072

Required area = [From eq. (i)]

=

=


NCERT Solutions for Class 10 Maths Exercise 12.3

7. In figure ABCD is a square of side 14 cm. With centers 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.

Ans. Area of shaded region = Area of square – 4 Area of sector

=

= = 196 – 154 =


NCERT Solutions for Class 10 Maths Exercise 12.3

8. 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:

(i) the distance around the track along its inner edge.

(ii) the area of the track.

Ans. (i)Distance around the track along its inner edge

=

= = = m

(ii)Area of track =

=

= =


NCERT Solutions for Class 10 Maths Exercise 12.3

9. In 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.

Ans. Area of shaded region = Area of circle + Area of semicircle ACB – Area of ACB

=

= = =


NCERT Solutions for Class 10 Maths Exercise 12.3

10. The area of an equilateral triangle ABC is. 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 figure). Find the area of the shaded region.

Ans. Area of equilateral triangle = = 17320.5

cm

Area of shaded region = Area of

ABC –

= 17320.5 – 15700 =


NCERT Solutions for Class 10 Maths Exercise 12.3

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

Ans. Area of remaining portion of handkerchief = Area of square ABCD – Area of 9 circular designs

=

= 1764 – 1386 =


NCERT Solutions for Class 10 Maths Exercise 12.3

12. In figure, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the:

(i) quadrant OACB

(ii) shaded region

Ans. (i)Area of quadrant OACB =

= =

(ii) Area of shaded region = Area of quadrant OACB – Area of OBD

=

=

= = cm2


NCERT Solutions for Class 10 Maths Exercise 12.3

13. In figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region.

Ans. OB =

=

= OA = cm

Area of shaded region = Area of quadrant OPBQ – Area of square OABC

=

=

=

=


NCERT Solutions for Class 10 Maths Exercise 12.3

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

Ans. Area of shaded region = Area of sector OAB – Area of sector OCD

=

=

= =

= =


NCERT Solutions for Class 10 Maths Exercise 12.3

15. In 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.

Ans. In right triangle BAC, [Pythagoras theorem]

=

BC = cm

Radius of the semicircle = cm

Required area = Area BPCQB

= Area BCQB – Area BCPB

= Area BCQB – (Area BACPB – Area BAC)

=

=

= 154 – (154 – 98) =


NCERT Solutions for Class 10 Maths Exercise 12.3

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

Ans. In right triangle ADC, [Pythagoras theorem]

=

AC = = cm

Draw BMAC.

Then AM = MC = AC = = cm

In right triangle AMB,

[Pythagoras theorem]

= 64 – 32 = 32

BM = cm

Area of ABC =

= =

Half Area of shaded region

=

= =

Area of designed region

= =

NCERT Solutions for Class 10 Maths Exercise 12.3

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