# 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:

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 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 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.
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 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.

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.
= $$\frac{7}{2}$$ = 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°

∴ 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 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).

Solution:
Area of ∆ABC = 17320.5 cm2

⇒ (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

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

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 = 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.

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
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 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
∴ 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

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 cm2 = 64 cm2