**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 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 the outer circle (R) = 14 cm and θ = 40°

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 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:

Area of the equilateral triangle OAB

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 (he radius 32 cm, a design Is formed leaving an equilìtcral triangle ABC in the middle as shown in the figure. Find the area of the design (shaded region).

Solution:

Let O be the centre of a circular table and ABC be the equilateral triangle.

Then we draw OD ⊥ BC

In ∆OBD, we have: cos 60° = \(\frac{\mathrm{OD}}{\mathrm{OB}}\)

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 given 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.

Solution:

The distance around the track along the inner edge = Perimeter of GHI + Perimeter JKL + GL + IJ

= (π × 30 + π × 30 + 106 + 106)

= (60π + 212) m

= (60 × \(\frac{22}{7}\) + 212)m = \(\frac{2804}{7}\) m.

Similarly, area of another circular path

BCDJKLB = 1100 m^{2}

Area of rectangular path (ABLG) = 106 × 10

= 1060 m^{2}

Similarly, area of another rectangular path

IJDE = 1060 m^{2}

Total area of the track

= (1100 + 1100 + 1060 + 1060) m^{2}

= (2200 + 2120) m^{2} = 4320 m^{2}.

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 figure). Find the area of the shaded region. (Use π = 3.14 and \(\sqrt{3}\) = 1.73205).

Solution:

Area of ∆ABC = 17320.5 cm^{2}

⇒ (side)^{2} = 40000

⇒ (side)^{2} = (200)^{2}

⇒ side = 200 cm

∴ Radius of each circle = \(\frac{200}{2}\) cm = 100 cm

Since, each angle of an equilateral triangle is 60°,

∠A = ∠B = ∠C = 60°

Area of a sector having angle of sector (θ), as 60° and radius (r) = 100 cm = \(\frac{\theta}{360^{\circ}}\) × πr^{2}

Now, area of the shaded region = [Area of the equilateral triangle ABC] – [Area of 3 equal sectors]

= 17320.5 cm^{2} – 15700 cm^{2} = 1620.5 cm^{2}

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

Solution:

OABC is a square such that its side

OA = 20 cm

∴ OB^{2} = OA^{2} + AB^{2}

⇒ OB^{2} = 20^{2} + 20^{2} = 400 + 400 = 800

⇒ OB = \(\sqrt{800}=20 \sqrt{2}\)

⇒ Radius of the quadrant (r) = \(20 \sqrt{2}\) cm

Now, area of the quadrant OPBQ = \(\frac{1}{4}\) πr^{2}

= \(\frac{1}{4} \times \frac{314}{100}\) × 800 cm^{2} = 314 × 2 = 628 cm^{2}

Area of the square OABC = 20 × 20 cm^{2} = 400 cm^{2}

∴ Area of the shaded region

= 628 cm^{2} – 400 cm^{2} = 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:

Side of the square = 8 cm

Area of the square ABCD = 8 × 8 cm^{2} = 64 cm^{2}

Now, radius of the quadrant ADQB = 8 cm

∴ Area of the quadrant ADQB

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