# NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.3 – Area related to Circles

Download NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.3 – Area related to Circles. This Exercise contains 16 questions, for which detailed answers have been provided in this note. In case you are looking at studying the remaining Exercise for Class 10 for Maths NCERT solutions for other Chapters, you can click the link at the end of this Note.

### NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.3 – Area related to Circles

NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.3 – Area related to Circles

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

Solution:

Using Pythagoras theorem in △PQR (Angle of semi-circle is right angle)
QR2 = PQ2 + PR2
QR2 = 242 + 72 = 576 + 49 = 625 = 252
QR = 25 cm

Radius of the circle = ½ QR = $\cfrac { 25 }{ 2 }$ cm

Area of shaded region = Area of semicircle – Area of △PQR
= $\cfrac { 1 }{ 2 }$ x Π x ${ \left( \cfrac { 25 }{ 2 } \right) }^{ 2 }$$\cfrac { 1 }{ 2 }$ x 24 x 7
= $\cfrac { 1 }{ 2 }$ x $\cfrac { 22 }{ 7 }$ x $\cfrac { 625 }{ 4 }$ – 12 x 7
= $\cfrac { 11\times 625 }{ 28 }$ – 84
= $\cfrac { 6875 - 2352 }{ 28 }$
= $\cfrac { 4523 }{ 28 }$ cm2

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

Solution:

Shaded area = Area of sector of larger circle – Area of sector of smaller circle
= $\cfrac { { 40 }^{ \circ } }{ { 360 }^{ \circ } }$ x Π x 14² – $\cfrac { { 40 }^{ \circ } }{ { 360 }^{ \circ } }$ x Π x 7²
= $\cfrac { 1 }{ 9 }$ x $\cfrac { 22 }{ 7 }$ x 196 – $\cfrac { 1 }{ 9 }$ x $\cfrac { 22 }{ 7 }$ x 49
= $\cfrac { 22\times 28 }{ 9 }$$\cfrac { 22\times 7 }{ 9 }$
= $\cfrac { 616-154 }{ 9 }$
= $\cfrac { 462 }{ 9 }$
= $\cfrac { 154 }{ 3 }$ cm2

3. 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.

Solution:

Side of square = 14 cm
Radius of circles = $\cfrac { 1 }{ 2 }$ x 14 = 7 cm

Area of shaded region = Area of square – Area of two semicircles
= 14 x 14 – 2 x ($\cfrac { 1 }{ 2 }$ x Π x 7²)
= 196 – $\cfrac { 22 }{ 7 }$ x 49
= 196 – 154
= 42 cm2

4. 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 centre.

Solution:

∠ AOB = 60° (OAB is equilateral triangle)
Reflex ∠ AOB = 360° – 60° = 300°

Shaded area = Area of triangle + Area of major sector of the circle
= $\cfrac { \sqrt { 3 } }{ 4 }$ x (12)² x $\cfrac { { 300 }^{ \circ } }{ { 360 }^{ \circ } }$ x Π x 6²
= $\cfrac { \sqrt { 3 } }{ 4 }$ x 144 + $\cfrac { 5 }{ 6 }$ x $\cfrac { 22 }{ 7 }$ x 36
= (36√3 + $\cfrac { 660 }{ 7 }$) cm2

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

Solution:

Side of the square = 4 cm
Radius of inner circle = $\cfrac { 1 }{ 2 }$ x 2 = 1 cm
Remaining area = Area of square – (Area of circle + Area of 4 quadrants)
= 4 x 4 – (Π  x (1)² + 4 x ($\cfrac { 1 }{ 4 }$ x Π x (1)²))
= 16 – (Π + Π)
= 16 – 2 x $\cfrac { 22 }{ 7 }$
= $\cfrac { 112-44 }{ 7 }$
= $\cfrac { 68 }{ 7 }$ cm2

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 the figure. Find the area of the design.

Solution:

Let O be the centre of the circle and the side of the triangle be s
Draw an altitude AOD from A on BC

∠ ABD = 60°
In △ABD
sin 60° = $\cfrac { AD }{ AB }$
$\cfrac { \sqrt { 3 } }{ 2 }$ = $\cfrac { AD }{ s }$
AD = $\cfrac { \sqrt { 3 }s }{ 2 }$
AO = Radius = 32 cm
OD = AD – AO = $\cfrac { \sqrt { 3 }s }{ 2 }$ – 32
Since, centroid O divides the median into the ratio 2:1
Therefore,
$\cfrac { AO }{ OD }$ = $\cfrac { 2 }{ 1 }$
$\cfrac { 32 }{ \cfrac { \sqrt { 3 } s }{ 2 } -\quad 32 } =\cfrac { 2 }{ 1 }$
32 = √3s- 64
96 = s√3
S = 32√3 cm
Area of design =  Area of circle – Area of triangle
= Π x 32² – $\cfrac { \sqrt { 3 } }{ 4 }$ x (32√3)²
= $\cfrac { 22 }{ 7 }$ x 1024 – $\cfrac { \sqrt { 3 } }{ 4 }$ x 3072
= ($\cfrac { 22528 }{ 7 }$ – 768√3) cm2

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

Solution:

Let the radius of each circle be r
r + r = Side of the square = 14 cm
2r = 14 cm
r = 7 cm

Shaded area = Area of square – Area of four quadrants
= 14 x 14 – 4 x ($\cfrac { 1 }{ 4 }$ x Π x 7²)
= 196 – $\cfrac { 22 }{ 7 }$ x 49
= 196 – 154
= 42 cm2

8. The given figure depicts a racing track whose left and right ends are semi-circular. 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

According to the question, the lengths and radii of different parts of the track are as shown in the figure

Solution:

The distance around the track along its inner edge
= AB + arc BCD + DE + arc EFA
= 106 + $\cfrac { 1 }{ 2 }$ x 2Π($\cfrac { 60 }{ 2 }$) + 106 + $\cfrac { 1 }{ 2 }$ x 2Π($\cfrac { 60 }{ 2 }$)
= 212 + 2 x $\cfrac { 22 }{ 7 }$ x 30
= 212 + $\cfrac { 1320 }{ 7 }$
= $\cfrac { 2804 }{ 7 }$ m

Area of the track
= (ar(PQST) + Area of semicircles QRS and TUP) – (ar(ABDE) + Area of semicircles BCD and EFA)
= (106 × 80 + $\cfrac { 1 }{ 2 }$ x Π x ${ \left( \cfrac { 80 }{ 2 } \right) }^{ 2 }$+$\cfrac { 1 }{ 2 }$ x Π x ${ \left( \cfrac { 80 }{ 2 } \right) }^{ 2 }$) – (106 × 60 + $\cfrac { 1 }{ 2 }$ x Π x ${ \left( \cfrac { 60 }{ 2 } \right) }^{ 2 }$+$\cfrac { 1 }{ 2 }$ x Π x ${ \left( \cfrac { 60 }{ 2 } \right) }^{ 2 }$)
= (106 × 80 + $\cfrac { 22 }{ 7 }$ x 40²) – (106 × 60 + $\cfrac { 22 }{ 7 }$ x 30²)
= (8480 + $\cfrac { 35200 }{ 7 }$) – (6360 + $\cfrac { 19800 }{ 7 }$)
= 8480 – 6360 + $\cfrac { 35200-19800 }{ 7 }$
= 2120 + $\cfrac { 15400 }{ 7 }$
= 2120 + 2200
= 4320 m2

9. In the given 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.

Solution:

Radius of smaller circle = $\cfrac { 7 }{ 2 }$ cm
Shaded area = Area of semicircle ACB + Area of smaller circle – Area of △ABC
=$\cfrac { 1 }{ 2 } \times \pi \times 7^{ 2 }+\pi \times { \left( \cfrac { 7 }{ 2 } \right) }^{ 2 }-\cfrac { 1 }{ 2 } \times AB\times OC$
=$\cfrac { 1 }{ 2 }$ x$\cfrac { 22 }{ 7 }$ x 49 + $\cfrac { 22 }{ 7 }$ x $\cfrac { 49 }{ 4 }$$\cfrac { 1 }{ 2 }$ x 14 x 7
= 77 + $\cfrac { 77 }{ 2 }$ – 49
= 28 + $\cfrac { 77 }{ 2 }$
= 28 + 38.5
= 66.5 cm2

10. The area of an equilateral triangle ABC is 17320.5 cm2. 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. Find the area of the shaded region.(Use π = 3.14 and √3 = 1.73205)

Solution:

Let the side of the triangle be s
Area = $\cfrac { \sqrt { 3 } }{ 4 }$ x s²
17320.5 = $\cfrac { 1.73205 }{ 4 }$ x s²
s2 = 4 x $\cfrac { 17320.5 }{ 1.73205 }$
s2 = 40000
s = 200 cm
Therefore, radius of each circle = $\cfrac { 200 }{ 2 }$ = 100 cm
Area of sectors of circle = 60° (ABC is equilateral triangle)
Shaded area = Area of triangle – Area of 3 sectors
= 17320.5 – 3 x ($\cfrac { { 60 }^{ \circ } }{ { 360 }^{ \circ } }$ x Π x 100² )
= 17320.5 – 3 x $\cfrac { 1 }{ 6 }$ x 3.14 x 10000
= 17320.5 – 15700
= 1620.5 cm2

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

Solution:

Side of the square = 3 × diameter = 6 × radius = 42 cm

Remaining area = Area of square – Area of nine circles
= 42 × 42 – 9 × π × 72
= 1764 – 9 x $\cfrac { 22 }{ 7 }$ x 49
= 1764 – 1386
= 378 cm2

12. In the given 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

Solution:

Radius of the quadrant of the circle = 3.5 cm

Area of quadrant = $\cfrac { 1 }{ 4 }$ x Π x 3.5² = $\cfrac { 1 }{ 4 }$ x Π x ${ \left( \cfrac { 7 }{ 2 } \right) }^{ 2 }$
= $\cfrac { 1 }{ 4 }$ x $\cfrac { 22 }{ 7 }$ x $\cfrac { 49 }{ 4 }$
= $\cfrac { 77 }{ 8 }$ cm2

Area of shaded region = Area of quadrant – Area of △BOD
= $\cfrac { 77 }{ 8 }$$\cfrac { 1 }{ 2 }$ x 3.5 x 2
= $\cfrac { 77 }{ 8 }$$\cfrac { 7 }{ 2 }$
= $\cfrac { 77-28 }{ 8 }$
= $\cfrac { 49 }{ 8 }$ cm2

13. In the given 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)

Solution:

Using Pythagoras theorem
OB2 = OC2 + OA2
OB2 = 202 + 202
OB2 = 2 × 202
OB = 20√2 cm

Radius of quadrant = OB = 20√2 cm

Shaded area = Area of quadrant – Area of square
= $\cfrac { 1 }{ 4 }$ x Π x (20√2)² – 20 x 20
= $\cfrac { 1 }{ 4 }$ x 3.14 x 800 – 400
= 628 – 400
= 228 cm2

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

Solution:

Radius of smaller sector = 7 cm
Radius of larger sector = 21 cm
Angle of sectors = 30°

Shaded area = Area of larger sector – Area of smaller sector
= $\cfrac { { 30 }^{ \circ } }{ { 360 }^{ \circ } }$ x Π x 21² – $\cfrac { { 30 }^{ \circ } }{ { 360 }^{ \circ } }$ x Π x 7²
= $\cfrac { 1 }{ 12 }$ x $\cfrac { 22 }{ 7 }$ x 441 – $\cfrac { 1 }{ 12 }$ x $\cfrac { 22 }{ 7 }$ x 49
= $\cfrac { 11\times 63 }{ 6 }$$\cfrac { 11\times 7 }{ 6 }$
= $\cfrac { 693 }{ 6 }$$\cfrac { 77 }{ 6 }$
= $\cfrac { 616 }{ 6 }$
= $\cfrac { 308 }{ 3 }$ cm2

15. In the given 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.

Solution:

Using Pythagoras theorem
BC2 = AB2 + AC2
BC2 = 142 + 142
BC2 = 2 × 142
BC = 14√2 cm

Radius of the semicircle = $\cfrac { BC }{ 2 }$ = 7√2 cm

Shaded area = Area of semicircle – Area of segment of quadrant
= Area of semicircle – (Area of quadrant – Area of △ABC)
=$\cfrac { 1 }{ 2 } \times \pi \times (7\sqrt { 2 } )^{ 2 }-(\cfrac { 1 }{ 4 } \times \pi \times 14^{ 2 }-\cfrac { 1 }{ 2 } \times 14\times 14)$
= $\cfrac { 1 }{ 2 }$ x $\cfrac { 22 }{ 7 }$ x 98 – ($\cfrac { 1 }{ 4 }$ x $\cfrac { 22 }{ 7 }$ x 196 – $\cfrac { 196 }{ 2 }$)
=154-(154-98)
=  98 cm2

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

Solution:

The design is a combination of two segments

Area of one segment = Area of quadrant – Area of triangle
= $\cfrac { 1 }{ 4 } \times \pi \times 8^{ 2 }-\cfrac { 1 }{ 2 } \times 8\times 8$
= $\cfrac { 1 }{ 4 }$ x $\cfrac { 22 }{ 7 }$ x 64 – $\cfrac { 64 }{ 2 }$
= $\cfrac { 352 }{ 7 }$ – 32
= $\cfrac { 352-224 }{ 7 }$
= $\cfrac { 128 }{ 7 }$ cm2

Area of design = Area of two segments = 2 × $\cfrac { 128 }{ 7 }$
= $\cfrac { 256 }{ 7 }$ cm2

NCERT Solutions for Class 10 Maths Chapter 12 Exercise 12.3 – Area related to Circles, has been designed by the NCERT to test the knowledge of the student on the topic – Areas of Combinations of Plane Figures

Download NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.3 – Area related to Circles