# NCERT Solutions for Class 10 Maths Exercise 12.3

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NCERT solutions for Maths Area Related to Circles

## NCERT Solutions for Class 10 Maths Area Related to Circles

###### 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.

= 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.

= 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:

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 =

= =

=

= =

Area of designed region

= =

## NCERT Solutions for Class 10 Maths Exercise 12.3

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