KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.3

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:
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.3 1
PQ = 24 cm, PR = 7 cm,
‘O’ is the centre of circle.
Angle in semicircle,
∠RPQ = 90°
In ⊥∆RPQ, ∠P = 90°
RQ = RP2 + PQ2
= (7)2 + (24)2
= 49 + 576
RQ2 = 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
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.3 2
= 161.3 sq.cm.

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

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°.
MP Board Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 3
Solution:
Radius of the outer circle (R) = 14 cm and θ = 40°
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.3 Q2

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

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:
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.3 5
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 :
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.3 6
∴ 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.
MP Board Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 7
Solution:
Area of the equilateral triangle OAB
MP Board Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 8

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

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.
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.3 9
∴ 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:
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.3 10
= 0.78 sq.cm.

iii) Radius of circle having diameter 2 cm, r = 1 cm.
∴ Area of circle:
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.3 11
= 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.

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

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).
MP Board Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 11
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° = \(\frac{\mathrm{OD}}{\mathrm{OB}}\)
MP Board Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 12

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

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:
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.3 16
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°
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.3 17
∴ 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:
MP Board Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 14
(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.
MP Board Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 15
Similarly, area of another circular path
BCDJKLB = 1100 m2
Area of rectangular path (ABLG) = 106 × 10
= 1060 m2
Similarly, area of another rectangular path
IJDE = 1060 m2
Total area of the track
= (1100 + 1100 + 1060 + 1060) m2
= (2200 + 2120) m2 = 4320 m2.

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

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:
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.3 22
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 = πr2
= \(\frac{22}{7}\) × (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°
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.3 23
∴ 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.

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

Question 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 (see figure). Find the area of the shaded region. (Use π = 3.14 and \(\sqrt{3}\) = 1.73205).
MP Board Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 18
Solution:
Area of ∆ABC = 17320.5 cm2
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.3 Q10
⇒ (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 = \(\frac{\theta}{360^{\circ}}\) × πr2
MP Board Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 20
Now, area of the shaded region = [Area of the equilateral triangle ABC] – [Area of 3 equal sectors]
= 17320.5 cm2 – 15700 cm2 = 1620.5 cm2

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

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:
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.3 27
i) Total area of 9 circular designs each of radius 7 cm
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.3 28

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 = a2
= (42)2
= 1764 sq.cm.
Area of remaining part of handkerchief:
= Area of a square – Area of 9 circles.
= 1764 – 1386
= 378 sq.cm.

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

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
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.3 29
i) Area of quadrant (OACB) of a Circle
r = 3.5 cm, θ = 90°
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.3 29

ii) In ⊥∆BOD, ∠BOD = 90° OB = 3.5 cm; OD = 2 cm
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.3 31
∴ 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)
MP Board Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 24
Solution:
OABC is a square such that its side
OA = 20 cm
∴ OB2 = OA2 + AB2
⇒ OB2 = 202 + 202 = 400 + 400 = 800
⇒ OB = \(\sqrt{800}=20 \sqrt{2}\)
⇒ Radius of the quadrant (r) = \(20 \sqrt{2}\) cm
Now, area of the quadrant OPBQ = \(\frac{1}{4}\) πr2
= \(\frac{1}{4} \times \frac{314}{100}\) × 800 cm2 = 314 × 2 = 628 cm2
Area of the square OABC = 20 × 20 cm2 = 400 cm2
∴ Area of the shaded region
= 628 cm2 – 400 cm2 = 228 cm2

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

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°
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.3 34
= \(\frac{1}{12}\) × 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.

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

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:
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.3 35
i) Area of Part II :
Area of Segment ABC – Area of ∆ABC
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.3 36
= 56 sq.cm.

ii) Area of shaded region Part III :
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.3 37

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.
MP Board Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 29
Solution:
Side of the square = 8 cm
Area of the square ABCD = 8 × 8 cm2 = 64 cm2
Now, radius of the quadrant ADQB = 8 cm
∴ Area of the quadrant ADQB
KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.3 Q16

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