RBSE Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3

Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3 Textbook Exercise Questions and Answers.

RBSE Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3

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

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

Solution:
According to the question,
PQ = 24 cm and PR = 7 cm.
RQ is a diameter of the circle.
∠RPQ = 90° [Angle of semi-circle]
Now in ∆PQR,
QR² = RP² + PQ²
QR = \(\sqrt{(7)^{2}+(24)^{2}}\)
QR = \(\sqrt{49+576}\)
QR = \(\sqrt{625}\)
∴ QR = 25 cm.
∴ Diameter of circle (QR) = 25 cm
So, Radius of circle (R) = \(\frac{25}{2}\) cm.
Area of the shaded region = Area of semi-circle - Area of ∆RPQ

Hence area of the shaded region = \(\frac{4523}{28}\) or 161.53 cm²

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

Solution:
According to the question,
Radius of the smaller circle (r) = 7 cm.
and radius of the larger circle (R) = 14 cm.
Central angle ∠AOC (θ) = 40°
Area of the shaded region = Area of larger sector OAC - Area of smaller sector OBD

Question 3.
Find the area of the shaded region in Fig., if ABCD is a square of side 14 cm and APD and BPC are semicircles.

Solution:
According to the question,
Side of the square = 14 cm.
Diameter of the semicircle (AD or BC) = 14 cm
Radius of the semicircle (R) = 7 cm
Area of square = (Side)²
= 14 × 14
= 196 cm²
∵ Area of one semicircle = \(\frac{1}{2} \pi R^{2}\)
= \(\frac{1}{2} \times \frac{22}{7}\) × 7 × 7
= 77 cm²
∴ Area of two semicircle = 2(77) = 154 cm²
∴ Area of the shaded region = Area of square ABCD - Area of two semicircles
= (196 - 154)
= 42 cm²
∴ Area of the shaded region = 42 cm²

Question 4.
Find the area of the shaded region in Fig., 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:
According to the question,
Radius of arc (R) = 6 cm
Side of equilateral triangle OAB = 12 cm
OA = OB = AB = 12 cm
Central angle of sector = 60°
∵ Each angle of an equilateral triangle = 60°
∴ Area of larger sector of circle = Area of circle - Area of smaller sector

∴ Area of shaded region = Area of equilateral triangle OAB + Area of larger sector of circle = (36√3 + \(\frac{660}{7}\)) cm²

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

Solution:
According to the question,
Side of square = 4 cm
Radius of each quadrant cut (r) = 1 cm
Diameter of circle (R) = 2 cm
∴ Radius of circle (R) = 1 cm
Area of square = (side)²
= (4)²
= 16 cm²

∵ Radius of the middle circle = 1 cm.
∴ Area of the required shaded region = Area of square - Area of 4 quadrants - Area of circle

Question 6.
In a circular table covering of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in Fig. Find the area of the design (shaded region).

Solution:
According to the question,
Radius of the table cover (R) = 32 cm
OA = OB = OC = 32 cm
∴ ∆ABC is an equilateral trianlg
∴ ∠B = 60°
OB and OC are the radii of the circular table cover
∴ OB = 32 cm
and ∠OBM = \(\frac{1}{2}\) ∠B = 30°
From centre O draw OM ⊥ BC
∴ BM = \(\frac{1}{2}\) BC
⇒ BC = 2BM
In right ∆OBM
cos 30° = \(\frac{\mathrm{BM}}{\mathrm{BO}}\)
⇒ \(\frac{\sqrt{3}}{2}=\frac{\mathrm{BM}}{32}\)
⇒ 2BM = 32√3
But side (BC) = 2BM
∴ Side (BC)= 32√3 cm
∴ Side of equilateral ∆ABC = 32√3 cm
∴ Area of equilateral ∆ABC

Hence area of the shaded region = Area of table cover - Area of equilateral triangle = (\(\frac{22528}{7}-768 \sqrt{3}\)) cm²

Question 7.
In Fig., ABCD is a square of a side of 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:
According to the question,
Side of square ABCD = 14 cm
and radius of circle (R) = 7 cm
Angle of sector (θ) = 90°
Area of square = (Side)²
= 14 × 14
= 196 cm²

∴ Required shaded area = Area of square - Area of four quadrants
= 196 - 154
= 42 cm²

Question 8.
Fig. 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:
(i)

Here AB = DC = 106 m
AF = BE = CG = HD = 10 m
Internal diameter of circle = 60 m
Internal radius of semicircle (r) = 30 m
External radius of semicircle (R) = r + 10
= 30 + 10
= 40 m
Distance covered around the track along its inner edge = AB + Perimeter of semicircle BRC + CD + Perimeter of semicircle DPA
= 2AB + 2[Area of semicircle BRC]

(ii) Area of the track = Area of rectangle ABEF + Area of region BEMGCRB + Area of rectangle CGHD + Area of region
= 2 Area of rectangle ABCD + 2 Area of region
= 2(AB × AF) + 2
[∵ Area of the semicircle whose radius is 60 cm - Area of the semicircle whose radius is 30 cm]
= 2[106 × 10] + 2 \(\left[\frac{\pi \mathrm{R}^{2}}{2}-\frac{\pi r^{2}}{2}\right]\)

∴ Area of the Track = 4320 m²

Question 9.
In Fig., 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:
According to the question,
Diameter of larger circle = 14 cm
∴ Radius of larger circle = 7 cm
Diameter of smaller circle = 7 cm
∴ Radius of smaller circle = \(\frac{7}{2}\) cm
AB and CD are two perpendicular diameters of a circle.
∴ AD ⊥ CD

= \(\frac{49}{2}\) cm²
∴ Area of segment BPC = \(\left(\frac{77}{2}-\frac{49}{2}\right)\) cm² = 14 cm²
Similarly area of segment AQC = 14 cm²
Area of smaller circle with diameter
DO = πr²
= \(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\)
= \(\frac{77}{2}\) cm²
Hence the total area of the shaded region

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 a radius equal to half the length of the side of the triangle (see Fig.). Find the area of the shaded region. (Use π = 3.14 and √3 = 1.73205)

Solution:
Area of equilateral triangle ABC = 17320.5 cm²

Side = 2 × 100 = 200 cm
∴ AB = BC = AC = 200 cm [∵ Triangle is equilateral]
Radius of circle (R) = \(\frac{\mathrm{AB}}{2}=\frac{200}{2}\) = 100 cm
Angle of sefetor (θ) = 60°
There are three sectors in the figure.
So area of the three sectors = \(3 \times \frac{\pi R^{2\theta}}{360^{\circ}}\)
= \(\frac{3 \times 3.14 \times 100 \times 100 \times 60^{\circ}}{360^{\circ}}\)
= 3.14 × 50 × 100
= 15700 cm²
∴ Required shaded area = Area of triangle - Area of three sectors
= 17320.5 - 15700
= 1620.5 cm²
∴ Shaded area = 1620.5 cm²

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

Solution:
Radius of circle (R) = 7 cm
Diameter of circle = 2 × R = 2 × 7 = 14 cm
Since there are three circles along the side of the square
∴ Side of the square = 3[14] = 42 cm
Total area of the handkerchief = Area of square
= (Side)²
= (42)²
= 1764 cm²
Area of a circular desings = 9πR²
= 9 × \(\frac{22}{7}\) × (7)²
= 9 × \(\frac{22}{7}\) × 7 × 7
= 9 × 154
= 1386 cm²
∴ Required area of the remaining portion = Area of square - Area of circular designs
= 1764 - 1386
= 378 cm²

Question 12.
In Fig. OACB is a quadrant of a circle with centre O and a radius of 3.5 cm. If OD = 2 cm, find the area of the
(i) quadrant OACB
(ii) shaded region.

Solution:
According to the question,
Radius of quadrant (R) = 3.5 cm
Angle of sector (θ) = 90°
OD = 2 cm
(i) Area of quadrant OACB = \(\frac{\pi \mathrm{R}^{2} \theta}{360^{\circ}}\)
= \(\frac{22}{7} \times \frac{3.5 \times 3.5 \times 90^{\circ}}{360^{\circ}}\)
= \(\frac{77}{8}\) cm²
(ii) Area of ∆ODB = \(\frac{1}{2}\) × Base × Height
= \(\frac{1}{2}\) × 3.5 × 2 
= 3.5 cm²
∴ Shaded area = Area of quadrant OACB - Area of ∆ODB
= \(\frac{77}{8}\) - 3.5
= \(\frac{77-28}{8}\)
= \(\frac{49}{8}\) cm²
Hence area of the shaded region = \(\frac{49}{8}\) cm² or 3.125 cm²

Question 13.
In Fig., 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:
According to the question,
Side of square ABCO = 20 cm
∠AOC = 90°
AB = OA
∴ From ∆OAB,

= √800
OB = 20√2 cm.
Area of square OABC = (Side)² = (20)²
Area of square = 400 cm²
Radius of quadrant (R) = 20√2 cm
Angle of sector (θ) = 90°
∴ Angle of sector = \(\frac{\pi \mathrm{R}^{2} \theta}{360^{\circ}}\)
= \(\frac{3.14 \times 20 \sqrt{2} \times 20 \sqrt{2} \times 90^{\circ}}{360^{\circ}}\)
= 2 × 314 cm²
= 628 cm²
∴ Required area of shaded region = Area of sector - Area of square
= (628 - 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 Fig.). If ∠AOB = 30°, find the area of the shaded region.

Solution:
Radius of sector OBA (R) = 21 cm
Radius of sector ODC (r) = 7 cm
Angle of sector (θ) = 30°

= 12.83 cm²
Now area of the shaded region = Area of larger sector OAB - Area of smaller sector OCD
= 115.5 - 12.83
= 102.66 cm²
Area of the shaded region = 102.66 cm²

Question 15.
In Fig. 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:
Radius of sector ACPB (r) = 14 cm
Angle of sector (θ) = 90°
AB = AC = 7 cm
Area of triangle = \(\frac{1}{2}\) × AB × AC
= \(\frac{1}{2}\) × 14 × 14
= 98 cm²

∴ Area of BOCPB = Area of sector ABPC - Area of ∆ABC
= 154 cm² - 98 cm²
= 56 cm²
In ∆BAC,
AB² + AC² = BC²
(14)² + (14)² = BC²

Required Area = Area of semicircle - [Area of sector - Area of ∆BAC]
= 154 - [154 - 98]
= 154 - 56
= 98 cm²
∴ Area of shaded region = 98 cm²

Question 16.
Calculate the area of the designed region in Fig. common between the two quadrants of a circle with a radius of 8 cm each.

Solution:
Side of square = 8 cm
Area of square = (8)² = 64 cm²
Line BD divides square ABCD into equal parts

∴ Area of ∆ABD = Area of ∆BDC
Angle of sector (θ) = 90°
Area of sector ABPD = \(\frac{\pi \mathrm{R}^{2} \theta}{360^{\circ}}\)
= \(\frac{22}{7} \times \frac{8 \times 8 \times 90^{\circ}}{360^{\circ}}\)
Area of sector ABPD = \(\frac{352}{7}\) cm²
So area of sector ABPD = \(\frac{352}{7}\) cm²
Now area of ∆ABD = \(\frac{1}{2}\) × AB × AD
= \(\frac{1}{2}\) × 8 × 8
= 32 cm²
So area of ∆ABD = 32 cm²
∴ Area of segment DMBPD = Area of sector ABPD - Area of ∆ABD
= \(\frac{352}{7}-\frac{32}{1}\)
= \(\frac{128}{7}\) cm²
∴ Area of shaded region = 2 × Area of segment DMBPD
= 2 × \(\frac{128}{7}\)
= \(\frac{256}{7}\) cm²
Hence area of the shaded region = \(\frac{256}{7}\) cm²