These NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Questions and Answers are prepared by our highly skilled subject experts.
NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Exercise 12.3
Question 1.
Find the area of the shaded region in the given figure, if PQ = 24cm, PR = 7cm and O is the centre of the circle.
Solution:
Question 2.
Find the area of the shaded region in the given figure, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠AOC = 400.
Solution:
∠AOC = 40° (given)
Radius of the sector AOC = 14 cm
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:
ABCD is a square
Given: side of the square = 14 cm
∴ Area of the square = (side)² = (14)² = 196 cm²
Radius of the semicircle APD = \(\frac { 1 }{ 2 }\)(side of square) = \(\frac { 1 }{ 2 }\) x 14 = 7 cm
Area of the semicircle APD = \(\frac { 1 }{ 2 }\) πr² = \(\frac { 1 }{ 2 }\) × \(\frac { 22 }{ 7 }\) × 7 × 7 = 11 × 7 = 77cm²
Similarly, area of the semicircle BPC = 77 cm²
Total area of both the semicircles = 77 + 77 = 154 cm²
Area of the shaded region = Area of square – area of both semicircles
= 196 – 154 = 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.
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 figure. Find the area of the remaining portion of the square.
Solution:
In the figure, area of the shaded region = area of the square ABCD with side 4 cm – area of the circle of radius 1 cm, centrally placed – area of the four quarter circles of radii 1 cm each, placed | at each corner of the square ABCD.
∴ Area of the remaining portion of the square = \(\frac { 68 }{ 7 }\) cm²
Question 6.
In a circular table cover of the radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in the figure. Find the area of the design (shaded region).
Solution:
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:
Edge of the square ABCD = 14 cm
Area of square ABCD = (14)² = 196 cm²
Here radius of each circle is 7 cm.
Area of shaded region = Area of square – Area of 4 sectors
196 – 154 = 42cm².
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:
(i) ABCD and EFGH are two rectangles and corner has two semicircles. The distance around the track along its inner edge
= 2 x length of rectangle + 2 x circumference of semicirlce
(ii) Area of the track = 2 [Area of rectangles] + 2 [Area of outer semicircle] – [Area of inner semicircles]
Question 9.
In the 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.
Solution:
Since AB ⊥ CD,
therefore, ∠COB = ∠COA
∠DOA = ∠DOB = 90°
In the figure, area of the shaded region, area of the small circle of diameter (OD = OA = 7cm) + (area of the segment BMC with central angle BOC = 90c and radius (OB = OC = 7cm) + area of the segment ANC with central angle AOC = 90° and radius (OA = OC) = 7cm
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:
∆ABC is an equilateral triangle. About the angular points A, B and C as centres three circles half the length of the side of the triangle are described.
Area of equilateral ∆ABC(given) = 17320.5 cm²
Since ABC is an equilateral triangle, therefore ∠A = ∠B = ∠C = 60°
= 17320.5 cm²
There are 3 equal sectors in the figure of central angles 60° and radii 100 cm [∵ \(\frac { 200 }{ 2 }\) = 10 cm]
Then, area of the shaded region = area of ∆ABC – 3 (area of one sector of central angle 60° and radius 100 cm)
= 17320.5 cm² – [3 x \(\frac { 60° }{ 360° }\) x π x (100)²]
= 17320.5 cm² – (\(\frac { 3 }{ 6 }\) x 3.14 x 1000) cm²
= 17320.5 cm² – (5000 x 3.14) cm²
= (17320.5 – 15700) cm² = 1620.5 cm²
∴ area of the shaded region is 1620.5 cm²
Question 11.
On a square handkerchief, nine circular designs each of the radius 7 cm are made (see figure). Find the area of the remaining portion of the handkerchief.
Solution:
ABCD is a square
Edge of the square ABCD = 3 x diameter of circle
= 3 x 4 = 42 cm
Area of square ABCD = (42)²
= 1764 cm²
Radius of one circle = 7 cm
Area of one circle = πr²
= \(\frac { 22 }{ 7 }\) x (7)²
= 154 cm²
Area of nine circles = 9 x Area of one circle
= 9 x 154 = 1386 cm²
Area of remaining portion = Area of square – Area of circle
= 1764 – 1386 = 378 cm²
Question 12.
In the 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:
(i) Let r be the radius of the circle and r = 35 cm \(\frac { 35 }{ 10 }\) = \(\frac { 7 }{ 2 }\) cm
Radius of the circle = Radius of quadrant of circle
= r = \(\frac { 7 }{ 2 }\) cm
Area of the quadrant of the circle = \(\frac { 1 }{ 4 }\)πr²
Since it is quadrant of circle, therfore the central angle of it θ = ∠AOB = 90°
(ii) Area of the shaded portion = Area of the sector for area of the quadrant of circle) – area of ∆AOD
Question 13.
In the figure, 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:
OABC is a square and OA is 20 cm. Join OB. Now we have a triangle QAB.
By Pythagoras theorem
(OB)² = (OA)² + (AB)²
(OB)² = (20)² + (20)²
= 400 + 400
Area of square
OABC = (20)² = 400 cm².
Area of shaded region = Area of sector OPBQ – Area of square OABC
= 628 – 400
= 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.
Solution:
Question 15.
In the 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.
Solution:
Let us find BC By Pythagoras theorem BC² = AB² + AC²
BC² = 14² + 14²
BC = 14\(\sqrt{2}\) cm
Required Area = Area of semicircle BRC – [Area of quadrant – area of ∆ABC)
Required Area = Area BCQB – (Area BACQB – Area of ∆ABC)
Question 16.
Calculate the area of the designed region in the figure common between the two quadrants of the circles of the radius 8 cm each.
Solution:
Area of square = (8)² = 64 cm²
Area of shaded region = Area of both sectors – Area of square
= \(\frac { 704 }{ 7 }\) – 64
= \(\frac { 704-448 }{ 7 }\) = \(\frac { 256 }{ 7 }\) cm²