NCERT Solutions for Class 10 Maths Areas Related to Circles Exercise 12.3
In this page we have NCERT Solutions for Class 10 Maths Areas Related to Circles for
Exercise 12.3 . Hope you like them and do not forget to like , social share
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Unless stated otherwise, use π =22/7 Question 1
Find the area of the shaded region in below figure, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle. Solution
PQ = 24 cm and PR = 7 cm
∠P = 90° (Angle in the semi-circle)
By Pythagoras theorem,
QR2 = PR2 + PQ2
QR2 = 72 + 242
QR = 25 cm
Now Diameter of the circle=QR
Therefore, Radius of the circle = 25/2 cm
Area of the semicircle = (π R2)/2
= (22/7 × 25/2 × 25/2)/2 cm2
= 13750/56 cm2 = 245.54 cm2
Area of the ΔPQR = 1/2 × PR × PQ
= 1/2 × 7 × 24 cm2
= 84 cm2 Area of the shaded region = 245.54 - 84 = 161.54 cm2
Question 2
Find the area of the shaded region in below figure, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠AOC = 40°. Solution
Radius inner circle = 7 cm
Radius of outer circle = 14 cm
Angle made by sector = 40°
Area of the sector OAC = (40°/360°) × π r2
= 1/9 × 22/7 × 142 = 68.44 cm2
Area of the sector OBD = (40°/360°) × π r2
= 1/9 × 22/7 × 72 = 17.11 cm2
Area of the shaded region ABDC = Area of the sector OAC - Area of the sector circle OBD
= 68.44 - 17.11 = 51.33 cm2
Question 3
Find the area of the shaded region in below figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles. Solution
There are two semicircles in the figure.
Side of the square = 14 cm
Therefore, Diameter of the semicircle = 14 cm
Radius of the semicircle = 7 cm
Area of the square = 14 × 14 = 196 cm2
Area of the semicircle = (π R2)/2
= (22/7 × 7 × 7)/2 cm2 = 77 cm2
Area of two semicircles = 2 × 77 = 154 cm2
Area of the shaded region = 196 - 154 = 42 cm2
Question 4
Find the area of the shaded region in below 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
OAB is an equilateral triangle with each angle equal to 60°.
Area of the sector is common in both.
Radius of the circle = 6 cm.
Side of the triangle = 12 cm.
Area of the equilateral triangle = √3/4 × (OA)2 = √3/4 × 122 = 36√3 cm2
Area of the circle = π R2 = 22/7 × 62 = 792/7 cm2
Area of the sector making angle 60° = (60°/360°) × π r2
= 1/6 × 22/7 × 62= 132/7 cm2 Area of the shaded region = Area of the equilateral triangle + Area of the circle - Area of the sector
= 36√3 + 792/7 - 132/7
= (36√3 + 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 a circle of diameter 2 cm is cut as shown in below figure. Find the area of the remaining portion of the square. Solutions
Side of the square = 4 cm
Radius of the circle = 1 cm
Four quadrant of a circle are cut from corner and one circle of radius are cut from middle.
Area of square = (side)2 = 42 = 16 cm2
Area of the quadrant = (π R2)/4 cm2 = (22/7 × 12)/4 = 11/14 cm2
Therefore, Total area of the 4 quadrants = 4 × (11/14) cm2 = 22/7 cm2
Area of the circle = π R2 cm2 = (22/7 × 12) = 22/7 cm2 Area of the shaded region = Area of square - (Area of the 4 quadrants + Area of the circle)
= 16 cm2 - (22/7 + 22/7) cm2
= 68/7 cm2
Question 6
In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in below figure. Find the area of the design. Solution
Radius of the circle = 32 cm
Draw a median AD of the triangle passing through the centre of the circle.
Since, AD is the median of the triangle
Therefore, AO = Radius of the circle = 2/3 AD
2/3 AD = 32 cm
AD = 48 cm
In ΔADB,
By Pythagoras theorem,
AB2 = AD2 + BD2
AB2 = 482 + (AB/2)2 ( As BD is the mid-point )
3/4 (AB2) = 2304
AB = 32√3 cm
Area of ΔABC = √3/4 × (32√3)2 = 768√3 cm2
Area of circle = π R2 = 22/7 × 32 × 32 = 22528/7 cm2
Area of the design = Area of circle - Area of ΔBC
= (22528/7 - 768√3) cm2 Question 7
In below figure, ABCD is a square of side 14 cm. With centers 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
Side of square = 14 cm
Four quadrants are included in the four sides of the square.
Therefore, Radius of the circles = 14/2 cm = 7 cm
Area of the square ABCD = 142 = 196 cm2
Area of the quadrant = (π R2)/4 = (22/7 × 72)/4 cm2
= 77/2 cm2
Total area of the quadrant = 4 × 77/2 cm2 = 154 cm2
Area of the shaded region = Area of the square ABCD - Area of the quadrant
= 196 - 154
= 42 cm2 Question 8
Below 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
Width of track = 10 m
Distance between two parallel lines = 60 m
Length of parallel tracks = 106 m
DE = CF = 60 m
Radius of inner semicircle, r = OD = O'C
= 60/2 m = 30 m
Radius of outer semicircle, R = OA = O'B
= 30 + 10 m = 40 m
Also, AB = CD = EF = GH = 106 m
Distance around the track along its inner edge = CD + EF + 2 × (Circumference of inner semicircle)
= 106 + 106 + (2 × πr) m = 212 + (2 × 22/7 × 30) m
= 2804/7 m
Area of the track = Area of ABCD + Area EFGH + 2 × (area of outer semicircle) - 2 × (area of inner semicircle)
= (AB × CD) + (EF × GH) + 2 × (πr2/2) - 2 × (πR2/2) m2
= (106 × 10) + (106 × 10) + 2 × π/2 (r2 -R2) m2
= 4320 m2
Question 9
In below 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
Radius of larger circle, R = 7 cm
Radius of smaller circle, r = 7/2 cm
Height of ΔBCA = OC = 7 cm
Base of ΔBCA = AB = 14 cm
Area of ΔBCA = 1/2 × AB × OC = 1/2 × 7 × 14 = 49 cm2
Area of larger circle = πR2 = 22/7 × 72 = 154 cm2
Area of larger semicircle = 154/2 cm2 = 77 cm2
Area of smaller circle = πr2 = 22/7 × 7/2 × 7/2 = 38.5 cm2
Area of the shaded region
= Area of larger circle - Area of triangle - Area of larger semicircle + Area of smaller circle
Area of the shaded region = (154 - 49 - 77 + 39.5)
= 66.5 cm2
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. Find the area of the shaded region. (Use π = 3.14 and √3 = 1.73205) Solution
ABC is an equilateral triangle.
Therefore, ∠A = ∠B = ∠C = 60°
There are three sectors each making 60°.
Let x be the side of the equilateral triangle
Area of ΔABC = 17320.5 cm2
√3/4 × (x)2 = 17320.5
x = 200 cm
Therefore, Radius of the circles = 200/2 cm = 100 cm
Area of the sector = (60°/360°) × π r2 cm2
= 1/6 × 3.14 × (100)2 cm2
= 15700/3 cm2
Area of the shaded region = Area of equilateral triangle ABC – 3 × Area of sector
= 17320.5 - 15700 cm2 = 1620.5 cm2
Question 11
On a square handkerchief, nine circular designs each of radius 7 cm are made . Find the area of the remaining portion of the handkerchief. Solution
Number of circular design = 9
Radius of the circular design = 7 cm
There are three circles in one side of square handkerchief.
Therefore, Side of the square = 3 × diameter of circle = 3 × 14 = 42 cm
Area of the square = 42 × 42 cm2 = 1764 cm2
Area of the circle = π r2 = 22/7 × 7 × 7 = 154 cm2
Area of the remaining portion of the handkerchief
= Area of the square - Total area of the design
= Area of the square - 9× Area of the circle
= 1764 - 1386 = 378 cm2
Question 12
In below 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 the quadrant = 3.5 cm
(i) Area of quadrant OACB = (πR2)/4 cm2
= (22/7 × 3.52)/4 cm2
= 9.625 cm2
(ii) Area of triangle BOD = 1/2 × 3.5 × 2 cm2
= 3.5 cm2
Area of shaded region = Area of quadrant - Area of triangle BOD
= (9.625 – 3.5) cm2
= 6.125 cm2 Question 13
In below 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
Side of square = OA = AB = 20 cm
Radius of the quadrant = OB
Now OB is the diagonal of the square
OB = 20√2 cm
Area of the quadrant = (πR2)/4 = 3.14/4 × (20√2)2 = 628 cm2
Area of the square = 20 × 20 = 400 cm2
Area of the shaded region = Area of the quadrant - Area of the square
= 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 .If ∠AOB = 30°, find the area of the shaded region. Solution
Radius of the larger circle, R = 21 cm
Radius of the smaller circle, r = 7 cm
Angle made by sectors of both concentric circles = 30°
Area of the larger sector = (30°/360°) × π R2 cm2
= 1/12 × 22/7 × 212 cm2
= 231/2 cm2
Area of the smaller circle = (30°/360°) × π r2 cm2
= 1/12 × 22/7 × 72 cm2
= 77/6 cm2
Area of the shaded region = 231/2 - 77/6 cm2
= 616/6 cm2 = 308/3 cm2
Question 15
In below 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
Radius of the quadrant ABC of circle = 14 cm
AB = AC = 14 cm
BC is diameter of semicircle.
ABC is right angled triangle.
By Pythagoras theorem in ΔABC,
BC2 = AB2 + AC2
BC2 = 142 + 142
BC = 14√2 cm
Radius of semicircle = 14√2/2 cm = 7√2 cm
Area of ΔABC = 1/2 × 14 × 14 = 98 cm2
Area of quadrant = 1/4 × 22/7 × 14 × 14 = 154 cm2
Area of the semicircle = 1/2 × 22/7 × 7√2 × 7√2 = 154 cm2
Area of the shaded region = Area of the semicircle –(Area of quadrant - Area of ΔABC )
= 154 + 98 - 154 cm2 = 98 cm2
Question 16
Calculate the area of the designed region in below figure common between the two quadrants of circles of radius 8 cm each. Solution
AB = BC = CD = AD = 8 cm
Area of ΔABC = Area of ΔADC = 1/2 × 8 × 8 = 32 cm2
Area of quadrant AECB = Area of quadrant AFCD = 1/4 × 22/7 × 82
= 352/7 cm2
Area of shaded region =
(Area of quadrant AECB - Area of ΔABC) + (Area of quadrant AFCD - Area of ΔADC)
= (352/7 - 32) + (352/7 -32) cm2
= 2 × (352/7 -32) cm2
= 256/7 cm2
