NCERT Class Solutions
  • Home
  • 9th Solutions
    • Maths Solutions
    • Science Solutions
    • Social Science Solutions
  • 10th Solutions
    • Science Solutions
    • Maths Solutions
    • Social Science Solutions
    • English Solutions
    • Hindi Solutions
    • Sanskrit Solutions
  • NCERT Books
    • Class 10 Books PDF
    • Class 9 Books PDF
  • About Us
    • Write for Us
    • Contact Us
    • Privacy Policy
    • Disclaimer
  • MP Board
    • MP Board Solutions
    • Previous Year Papers
No Result
View All Result
  • Home
  • 9th Solutions
    • Maths Solutions
    • Science Solutions
    • Social Science Solutions
  • 10th Solutions
    • Science Solutions
    • Maths Solutions
    • Social Science Solutions
    • English Solutions
    • Hindi Solutions
    • Sanskrit Solutions
  • NCERT Books
    • Class 10 Books PDF
    • Class 9 Books PDF
  • About Us
    • Write for Us
    • Contact Us
    • Privacy Policy
    • Disclaimer
  • MP Board
    • MP Board Solutions
    • Previous Year Papers
No Result
View All Result
NCERT Class Solutions
No Result
View All Result
ADVERTISEMENT
Home Class 10th Solutions 10th Maths

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

by Sudhir
December 4, 2021
in 10th Maths, Class 10th Solutions
Reading Time: 11 mins read
0
NCERT Class 10th Maths Solutions
16
VIEWS
Share on FacebookShare on Twitter

In this post, we will share NCERT Class 10th Maths Book Solutions Chapter 12 Areas Related to Circles Ex 12.3. These solutions are based on new NCERT Syllabus.

NCERT Class 10th 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 the given figure, if PQ = 24 cm, PR = 7cm and O is the centre of the circle.
NCERT Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 1
Solution:
Since O is the centre of the circle,
∴ QOR is a diameter.
⇒ ∠RPQ = 90° [Angle in a semicircle]
Now, in right ∆RPQ, RQ2 = PQ2+ PR2 [Pythagoras theorem]
⇒ RQ2 = 242 + 72 = 576 + 49 = 625
⇒ RQ = \(\sqrt{625}\) = 25 cm
∴ Radius of a circle (r) = \(\frac{25}{2}\) cm
∴ Area of ∆RPQ
= \(\frac{1}{2}\) × PQ × RP = \(\frac{1}{2}\) × 24 × 7cm2
= 12 × 7 cm2 = 84 cm2
Now, area of semicircle
NCERT Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 2
∴ Area of the shaded portion
= 245.54 cm2 – 84 cm2 = 161.54 cm2

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°.
NCERT 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°
NCERT Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 4

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

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.
NCERT Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 5
Solution:
Side of the square = 14 cm
∴ Area of the square ABCD = 14 × 14 cm2
= 196 cm2
NCERT Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 6
Similarly, area of semi-circle BPC = 77 cm2
∴ Total area of two semi-circle = (77 + 77) cm2
= 154 cm2
Area of the shaded part
= Area of the square ABCD – Area of semi-circles
= (196 – 154) cm2 = 42 cm2

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.
NCERT Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 7
Solution:
Area of the equilateral triangle OAB
NCERT Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 8

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 figure. Find the area of the remaining portion of the square.
NCERT Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 9
Solution:
ABCD of the square ABCD = (4)2 cm2 = 16 cm2.
Area of circel inside the square = πr2
= π × (1)2 = π cm2.
NCERT Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 10

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

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

Question 7.
In the figure, 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:
Each side of the square ABCD = 14 cm
NCERT Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 13

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:
NCERT 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.
NCERT 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.

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

Question 9.
In the given figure, 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.
NCERT Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 16
Solution:
O is the centre of the circle and OA = 7 cm
⇒ AB = 2OA = 2 × 7 cm = 14 cm
∵ AB and CD are perpendicular to each other.
⇒ OC ⊥ AB and OC = OA = 7 cm
Area of ∆ABC = \(\frac{1}{2}\) × AB × OC
= \(\frac{1}{2}\) × 14 cm × 7cm = 49 cm2
Again, OD = OA = 7 cm
∴ Radius of the small circle = \(\frac{1}{2}\) (OD)
NCERT Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 17

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).
NCERT Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 18
Solution:
Area of ∆ABC = 17320.5 cm2
NCERT Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 19
⇒ (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
NCERT 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

Question 11.
On a square handkerchief, nine circular designs each of radius 7 cm are made (see figure). Find the area of the remaining portion of the handkerchief.
NCERT Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 21
Solution:
∵ The circles touch each other.
∴ The side of the square ABCD
= 3 × diameter of a circle = 3 × (2 × radius of a circle) = 3 × (2 × 7) cm = 42 cm
⇒ Area of the square ABCD = 42 × 42 cm2
= 1764 cm2
Now, area of one circle
= πr2 = \(\frac{22}{7}\) × 7 × 7 cm2 = 154cm2
∴ Total area of 9 circles = 9 × 154 cm2 = 1386 cm2
∴ Area of the remaining portion of the handkerchief = (1764 – 1386) cm2 = 378 cm2

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

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.
NCERT Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 22
Solution:
(i) Here, radius (r) = 3.5 cm
NCERT Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 23

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)
NCERT 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

Question 14.
AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see figure). If ∠AOB = 30°, find the area of the shaded region.
NCERT Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 25
Solution:
∵ Radius of bigger circle R = 21 cm and sector angle θ = 30°
∴ Area of the sector OAB
NCERT Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 26

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

Question 15.
In the given figure, 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.
NCERT Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 27
Solution:
Radius of the quadrant = 14 cm
Since, BC is a diameter in a semi-circle.
∴ ∠BAC = θ = 90°
Area of the semicircle ABPC
= [\(\frac{90^{\circ}}{360^{\circ}} \times \frac{22}{7}\) × 14 × 14] cm2
= 22 × 7 cm2 = 154 cm2
Area of right ∆ABC
= \(\frac{1}{2}\) × 14 × 14 cm2 = 98 cm2
⇒ Area of segment BPC = 154 cm2 – 98 cm2
= 56 cm2
NCERT Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 28
Now, area of the shaded region = [Area of semicircle BQC] – [Area of segment BRC]
= 154 cm2 – 56 cm2 = 98 cm2

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.
NCERT 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
NCERT Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 30

Previous Post

NCERT Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.2

Next Post

NCERT Class 10th Maths Solutions Chapter 12 वृतों से संबंधित क्षेत्रफल Examples and MCQs

Related

NCERT Class 10th Sanskrit Solutions
10th Sanskrit

Abhyasvan Bhav Sanskrit Class 10 Chapter 3 अनुच्छेदलेखमन्

NCERT Class 10th Sanskrit Solutions
10th Sanskrit

Abhyasvan Bhav Sanskrit Class 10 Chapter 4 चित्रवर्णनम्

Leave a Reply Cancel reply

Your email address will not be published. Required fields are marked *

Categories

  • Books
    • Class 10 Books PDF
  • Class 10th Solutions
    • 10th English
    • 10th Hindi
    • 10th Maths
    • 10th Sanskrit
    • 10th Science
    • 10th Social Science
  • Class 9th Solutions
    • 9th Maths
    • 9th Science
    • 9th Social Science
  • MP Board
  • Uncategorized

Recent

NCERT Class 10th Sanskrit Solutions

Abhyasvan Bhav Sanskrit Class 10 Chapter 4 चित्रवर्णनम्

NCERT Class 10th Sanskrit Solutions

Abhyasvan Bhav Sanskrit Class 10 Chapter 3 अनुच्छेदलेखमन्

NCERT Class 10th Sanskrit Solutions

Abhyasvan Bhav Sanskrit Class 10 Chapter 2 पत्रलेखनम्

NCERT Class Solutions

We provide NCERT Solutions

NCERT Class Solutions App Play Store

Follow Us

Browse By Category

  • Books
    • Class 10 Books PDF
  • Class 10th Solutions
    • 10th English
    • 10th Hindi
    • 10th Maths
    • 10th Sanskrit
    • 10th Science
    • 10th Social Science
  • Class 9th Solutions
    • 9th Maths
    • 9th Science
    • 9th Social Science
  • MP Board
  • Uncategorized
  • Write for Us
  • Privacy Policy
  • Contact Us

© 2022 NCERT Class Solutions .

No Result
View All Result
  • Home
  • 9th Solutions
    • Maths Solutions
    • Science Solutions
    • Social Science Solutions
  • 10th Solutions
    • Science Solutions
    • Maths Solutions
    • Social Science Solutions
    • English Solutions
    • Hindi Solutions
    • Sanskrit Solutions
  • NCERT Books
    • Class 10 Books PDF
    • Class 9 Books PDF
  • About Us
    • Write for Us
    • Contact Us
    • Privacy Policy
    • Disclaimer
  • MP Board
    • MP Board Solutions
    • Previous Year Papers

© 2022 NCERT Class Solutions .

This website uses cookies. By continuing to use this website you are giving consent to cookies being used. Visit our Privacy and Cookie Policy.