**KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.3** are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Exercise 5.3.

## Karnataka SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Exercise 5.3

{Unless stated otherwise, use \(\pi=\frac{22}{7}\)}

Question 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:

PQ = 24 cm, PR = 7 cm,

‘O’ is the centre of circle.

Angle in semicircle,

∠RPQ = 90°

In ⊥∆RPQ, ∠P = 90°

RQ = RP^{2} + PQ^{2}

= (7)^{2} + (24)^{2}

= 49 + 576

RQ^{2} = 625

∴ RQ = 25 cm.

Diameter, RQ = 25 cm.

∴ Radius, OR = OQ = \(\frac{25}{2}\) cm.

Area of shaded part :

= Area of semicircle – Area of ∆RPQ

= 161.3 sq.cm.

Question 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:

Radius of small circle, r = 7 cm

Radius of big circle, R = 14 cm

∠AOC = θ = 40°

Area of shaded part, ABCD =?

Area of ABCD

= Area of OAC – Area of OBD

Question 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:

Each side of square ABCD = 14 cm.

APD, APC are semicircles.

Area of the shaded region = ?

i) Total area of square

= (Side)^{2}

= (14)^{2}

= 196 sq.cm.

ii) Each radius, r of semicircle

r = \(\frac{14}{2}\) =7cm

Area of two semicircles :

∴ Area of shaded region:

= Area of square – Area of two semicircles

= 196 – 154

= 42 sq. cm

Question 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:

OAB is an equilateral Triangle with side 12 Circle with radius of 6

Each angle of an equilateral triangle is 60°

Angle at the centre in segment, θ = 60°

Area of Major segment + Area of an equilateral triangle OAB =

= 94.28 + 36 × 1.73

= 94.28 + 62.28

= 156.56 sq. cm.

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

Solution:

i) Each side of square ABCD = 4 cm.

∴ Area of Square:

= (Side)^{2}

= (4)^{2}

= 16 sq.cm.

(ii) Area of circle in the corner A is 1 cm.

Radius, r = 1 cm.

θ = 90°

∴ Area of quadrant of circle:

= 0.78 sq.cm.

iii) Radius of circle having diameter 2 cm, r = 1 cm.

∴ Area of circle:

= 3.14 sq. cm.

∴ Area of remaining part of Square :

= Area of Square – Area of 4 quadrants – Area of Circle.

= 16 – 4 × 0.78 – 3.14

= 16 – 3.12 – 3.14

= 16 – 6.26

= 9.74 sq.cm.

Question 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 given figure. Find the area of the design.

Solution:

Radius of the circle r = 32 cm.

∴ Area of the circle = πr^{2}

ii) ∠BAC = 60° ∴ ∠BOC = 120°

(∵ Angle at the centre is double than the angle in the circumference).

Now, in ∆OBC, BC ⊥ OD,

∠BOD = ∠COD = 60°

∴ ∆ABC is an equilateral triangle.

Each side of this triangle,

a= 2 × \(16 \sqrt{3}=32 \sqrt{3} \mathrm{cm}\)

Area of equilateral triangle, ∆ABC :

= 1328.64 sq.cm,

iii) Area of shaded region :

= Area of circle – Area of an equilateral ∆^{le}

= 3218.28 – 1328.64

= 1889.6 sq.cm.

Question 7.

In the Figure given below, 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:

i) Each side of square ABCD =14 cm.

∴ Area of square ABCD

= (Side)^{2}

= (14)^{2}

= 196 sq.cm.

ii) Measure of radii of 4 circles, r = \(\frac{14}{2}=\) = 7 cm.

∵ Distance between tangents which touches circles externally,

d = R + r = 7 + 7 = 14 cm.

Area of segment with centre, A = ?

r = 7 cm, θ = 90°

∴ Total area of 4 sectors

= 4 × 38.5 = 154 sq.cm.

iii) The Area of track :

= (Area of GHIJ – Area of ABCD)

= 196 – 154

= 42 sq.cm.

Question 8.

The Figure given below 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.

Solution:

The distance around the track along its inner edge:

= AB + Arc BEC + CD + Area of Arc DFA

The Area of Track:

= (Area of GHIJ – Area of ABCD) + (Area of semicircle HKI – Area of semicircle BEC) + (Area of semicircle GIJ – Area of semicircle AFD)

= 2120 + 2200

= 4320 sq.m.

∴ Total area of Track = 4320 sq.m.

Question 9.

In the figure given below, 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:

i) Radius of Big circle is the diameter of a small circle.

∴ Radius of small circle

= \(\frac{7}{2}\) = 3.5 cm

∴ Area of smalle circle = πr^{2}

= \(\frac{22}{7}\) × (3.5)^{2}

= \(\frac{22}{7}\) × 12.25

= 3.14 × 12.25

= 38.47 sq.cm.

ii) Area of segment OCB = ?

Radius, r = 7 cm, θ = 90°

∴ Area of sector OCB – Area of ∆OBC

= 38.47 – 24.50

= 13.97 sq.cm.

Similarly, Area of sector OAC – Area of ∆OAC

= 38.47 – 24.50

= 13.97 sq.cm.

∴ Total area of shaded region :

= Area of small circle + Area of Minor segment OBC + Area of Minor segment OAC

= 38.47 + 13.97 + 13.97

= 66.41 sq.cm.

Question 10.

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 Figure). Find the area of the shaded region.

(Use π = 314 and \(\sqrt{3}\) = 1.73205).

Solution:

i) Area of an equilateral ∆ABC :

= 17320.5 cm^{2}

\(\frac{\sqrt{3} a^{2}}{4}\) = 17320.5

ii) Area formed by centre A:

r = 100 cm, θ = 60°

Similarly Area of sectors formed by centres B and C is 5233 sq.cm.

∴ Total Area of THREE sectors :

= 3 × 5233 = 15699 sq.cm.

∴ Area of the shaded region:

= Area of an equilateral Triangle – Area of three sectors.

= 17320.5 – 15699

= 1621.5 sq. cm.

Question 11.

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

Solution:

i) Total area of 9 circular designs each of radius 7 cm

ii) All the circles touches externally.

∴ Sum of the diameter of 3 circles in first Row =

14 + 14 + 14 = 42 cm.

∴ Length of each side of square ABCD,

a = 42 cm.

∴ Area of square ABCD = a^{2}

= (42)^{2}

= 1764 sq.cm.

Area of remaining part of handkerchief:

= Area of a square – Area of 9 circles.

= 1764 – 1386

= 378 sq.cm.

Question 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

i) quadrant OACB,

ii) shaded region.

Solution:

Radius of Circle, OA = OB = 3.5 cm.

OD = 2 cm.

i) Area of a quadrant of a circle

ii) Area of the shaded region

i) Area of quadrant (OACB) of a Circle

r = 3.5 cm, θ = 90°

ii) In ⊥∆BOD, ∠BOD = 90° OB = 3.5 cm; OD = 2 cm

∴ Area of shaded region,

= Area of quadrant OACB – Area of ABOD

= 9.61 – 3.5

= 6.11 sq.cm.

Question 13.

In Figure (i) given below, 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:

i) Each side of square OABC = 20 cm.

∴ OA = 20 cm, AB = 20 cm.

∴ Area of square PABC = (Side)^{2}

= (20)^{2}

= 400 sq.cm.

ii) Diagonal OB is drawn in OABC square.

In ⊥∆OAB, OA = 20 cm.

AB = 20 cm.

OB =?

OB^{2} = OA^{2} + AB^{2}

= (20)^{2} + (20)^{2}

= 400 + 400

= 800

∴ Area of shaded region = 628 – 400 = 228 cm^{2}.

Question 14.

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

Solution:

i) Area of segment OAPB = ?

r = 21 cm, θ = 30°

= \(\frac{1}{12}\) × 3.14 × 49

= 12.82 sq.cm.

∴ Area of the shaded region = Area of Sector OAPB – Area of segment OCQD

= 115.4 – 12.82

= 102.58 sq.cm.

Question 15.

In the given figure (i), 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:

i) Area of Part II :

Area of Segment ABC – Area of ∆ABC

= 56 sq.cm.

ii) Area of shaded region Part III :

Question 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:

i) Area of Square, ABCD = a^{2} = (8)^{2} = 64 cm^{2}.

ii) Sum of Areas of Part II and Part III = Area of the segment with centre D and radius of 8 cm.

iii) Area of Part I = (Sum of Part I, II and III) – (Sum of the area of Part II, III)

= Area of Square ABCD – (Sum of the area of part II and III)

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