Samacheer Kalvi Class 8 Maths Chapter 2 Measurements Solutions
Welcome to NCTB Solutions. Here with this post we are going to help 8th class students by providing Solutions for Samacheer Kalvi Class 8 Maths chapter 2 Measurements. Here students can easily find all the solutions for Measurements Exercise 2.1, 2.2, 2.3 and 2.4. Also here our Expert Maths Teacher’s solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 2 solutions. Here all the solutions are based on Tamil Nadu State Board latest syllabus.
Measurements Exercise 2.1 Solutions :
(1) Fill in the blanks :
(i) The ratio between the circumference and diameter of any circle is ___
Solution :
Circumference : Diameter
= 2πr : 2r
= π : 1
= 22/7 : 1
= 22/7 × 7 : 1 × 7
= 22 : 7
∴ The ratio between the circumference and diameter of any circle is 22 : 7.
(ii) A line segment which joins any two points on a circle is a ___
Solution :
A line segment which joins any two points on a circle is a Chord.
(iii) The longest chord of a circle is ___
Solution :
The longest chord of a circle is “Diameter”.
(iv) The radius of a circle of diameter 24 cm is ___
Solution :
Diameter = 24 cm
∴ Radius = 24/2 = 12 cm
∴ The radius of a circle of diameter 24 cm is 12 cm.
(v) A part of circumference of a circle is called as ___
Solution :
A part of circumference of a circle is called as Circular arc.
(2) Match the following :
| (i) Area of a circle | (a) 1/4 πr² |
| (ii) Circumference of a circle | (b) (π+2)r |
| (iii) Area of the sector of a circle | (c) πr² |
| (iv) Circumference of a semicircle | (d) 2 π r |
| (v) Area of a quadrant of a circle | (e) 0°/360° × πr² |
Solution :
(i) → (e) 0°/360° × πr²
(ii) → (d) 2 π r
(iii) → (c) πr²
(iv) → (b) (π+2)r
(v) → (a) 1/4 πr²
Question no – (4)
For the sectors with given measures, find the length of the arc, area and perimeter.
(i) central angle 45°, r = 16 cm
(ii) central angle 120°, d =12.6 cm
Solution :
(i) central angle 45°, r = 16 cm
Length of the arc = 0°/360° × 2πr
= 45°/360 × 2 × 3.14 × 16
= 3.14 × 4 cm
= 12.56 cm
Area of the arc = 0°/360° × πr² sq.cm
= 45°/360° × 3.14 × 16 × 16 sq.cm
= 3.14 × 321 sq.cm
= 100.48 sq.cm
Perimeter of the arc = l + 2r
= 12.56 + (2×16) cm
= 12.56 + 32
= 44.56 cm
(ii) central angle 120°, d = 12.6 cm
∴ r = 12.6/2 = 6.3 cm
Length = 120/360 × 2 × 3.14 × 6.3
= 13.188 cm
Area = 12.0/360 × 3.14 × 6.3× 6.3
= 41.5422 sq.cm
Perimeter = l +2r
= 13.88 + (2×6.3)
= 13.188 + 12.6
= 25.788 cm
Question no – (5)
From the measures given below, find the area of the sectors.
(i) length of the arc = 48 m, r = 10 m
(ii) length of the arc = 50 cm, r = 13.5 cm
Solution :
(i) length of the arc = 48 m, r = 10 m
Area of the sector = length of the arc × π²
= 48 × 10²
= 4800 sq.cm
(ii) length of the arc = 50 cm, r = 13.5 cm
= Area of the sector = length of the arc × π²
= 50 × (13.5)²
= 9112.5 sq.cm
Question no – (6)
Find the central angle of each of the sectors whose measures are given below
(i) Area = 462 cm², r = 21 cm
(ii) Length of the arc = 44 m, r = 35 m
Solution :
(i) Given area 462 cm², r = 21 cm
Area = θ°/360° πr²
= 462 = θ/360 × 22/7 × 21 × 21
= θ = 120°
(ii) Length of the arc = 44 m, r = 35 m
∴ Length of the arc = θ/360 × 2πr
= 44 = θ/360 × 2 × 2/7 × 35
= θ = 72°
Question no – (7)
A circle of radius 120 m is divided into 8 equal sectors. Find the length of the arc of each of the sectors.
Solution :
r = 120 m, n = 8
Length of the arc = 1/n × 2πr
= 1/8 × 2 × 3.14/100 × 120
= 94.2 m
Hence, the length of the arc of each of the sectors are 94.2 m.
Question no – (8)
A circle of radius 70 cm is divided into 5 equal sectors. Find the area of each of the sectors.
Solution :
r = 70 cm, n = 5
Area of each of sector = 1/n πr²
= 1/5 × 3.14/100 × 70 × 70
= 341/10 × 14 × 7
= 3077.2 sq.cm
So, the area of each of the sectors are 3077.2 sq.cm
Question no – (9)
Dhamu fixes a square tile of 30 cm on the floor. The tile has a sector design on it as shown in the figure. Find the area of the sector.
Solution :
r = 30 cm
Area of the sector = 1/4πr²
= 1/4 × 3.14 30 × 30
= 47.571 sq.cm
Hence, the area of the sector is 47.571 sq.cm.
Question no – (10)
A circle is formed with 8 equal granite stones as shown in the figure each of radius 56 cm and whose central angle is 45°. find the area of each of the granite stones.
Solution :
n = 8 and r = 56 cm, θ = 45°
Area of the granite stone = 1/n πr²
= 1/8 × 22/7 × 56 × 56
= 1232 sq.cm
Therefore, the area of each of the granite stones are 1232 sq.cm.
Measurements Exercise 2.2 Solutions :
Question no – (1)
Find the perimeter and area of the figures given below.
Solution :
Figure – (i)
Area of the figure = area of the rectangle – area of the half circle
= (l × b) – 1/2πr²
= (10 × 7) 1/2 π (3.5)²
= 70 – 1/2 × 22/7 × 35 × 35
= 70 – 19.25
= 57.75 sq.m
Figure – (ii)
Area of the figure = area of the rectangle + 2 × quadrants of the circle
= (l × b) + (2 × 1/4πr²)
= (6 × 3.5) + (1/2 × 22/7 × 3.5 × 3.5)
= 21.0 + 19.25
= 40.25 sq.m
Question no – (2)
Find the area of the shaded part in the following figures.
Solution :
Figure – (i)
a = 10 cm
r = 5 cm
Area of the shaded part = area of the square – 4 × area of the quadrant of the circle
= a² – (4 × 1/4πr²) sq,.m
= (10)² – (3.14 × 55)
= 100 – 78.5
= 21.5 sq.m
Figure – (ii)
r = 7 cm
a = 7 cm
Area of the shaded part = area of the half circle – area of the triangle
= 1/2πr² – √3/4 a²
= 1/2 × 3.14 × 7 × 7 – √3/4 × 7 × 7
= 49/2 (3.14 – √3/4)
= 49/4 (6.28 – √3).
Question no – (3)
Find the area of the combined figure given which is got by the joining of two parallelograms.
Solution :
Area of the parallelograms = area of the rectangle – area of I triangle + area of II triangle
= (l + b) – √3/4 a² + √3/4 a²
= (6 × 8) – √3/4 a² + √3/4 a²
= 48 sq.cm
Hence, the area of the combined figure is 48 sq. cm.
Question no – (4)
Find the area of the combined figure given, formed by joining a semicircle of diameter 6 cm with a triangle of base 6 cm and height 9 cm.
Solution :
Area of the combined figure = area of the semicircle + area of the triangle
= 1/2 πr² + 1/2 × h × b
= 1/2 × 3.14 × 3 × 3 + 1/2 × 9 × 6
= 1/2 (28.26 + 54)
= 41.13 sq.cm
Hence, the area of the combined figure is 41.13 sq.cm.
Question no – (5)
The door mat which is hexagonal in shape has the following measures as given in the figure. Find its area.
Solution :
a = 70 cm
h = 90 – 70/2 = 20/2 = 10 cm
Area of the hexagonal = area of the square + 2× area of same triangle
= a² + 2 × 1/2 × h × b
= a² + h × b
= 70 × 70 + 10 × 70
= 70 × 80
= 560 sq.cm
Therefore, its area is 560 sq. cm.
Question no – (6)
A rocket drawing has the measures as given in the figure. Find its area.
Solution :
Length of the green rectangle l = 120 – (20+20)
= 120 – 40
= 80 cm
∴ l = 80 cm, b = 30 cm, a = 50 cm, h = 20 cm
Area of the rocket = area of the trapezium + area of the rectangle + area of the triangle
= 1/2 × h × (a+b) + (l×b) + 1/2 × h × b
= 1/2 × 20 × 80 + (80×30) + 1/2 × 20 × 30
= 800 + 2400 + 300
= 3500 sq.cm
Hence, its area is 3500 sq.cm.
Measurements Exercise 2.3 Solutions :
(1) Fill in the blanks :
(i) The three dimensions of a cuboid are __, __ and __
Solution :
The three dimensions of a cuboid are Length, Breadth and Height.
(ii) The meeting point of more than two edges in a polyhedron is called as __
Solution :
The meeting point of more than two edges in a polyhedron is called as Vertex.
(iii) A cube has __ faces.
Solution :
A cube has 6 faces.
(iv) The cross section of a solid cylinder is __
Solution :
The cross section of a solid cylinder is Circle.
(v) If a net of a 3-D shape has six plane squares, then it is called __
Solution :
If a net of a 3-D shape has six plane squares, then it is called Cube.
Question no – (5)
Verify Euler’s formula for the table given below.
| S.No. | Faces | Vertices | Edges |
| (i) | 4 | 4 | 6 |
| (ii) | 10 | 6 | 12 |
| (iii) | 12 | 20 | 30 |
| (iv) | 20 | 13 | 30 |
| (v) | 32 | 60 | 90 |
Solution :
Euler’s theorem : F + V – E = 2
When F = faces, V = vertices, E = Edge
(i) F= 4, V = 4, E = 6
∴ Euler’s formula
= 4 + 4 – 6
= 2 (satisfied)
(ii) F = 10, V = 6, E = 12
∴ F + V – E
= 10 + 6 – 12
= 4 (not satisfied)
(iii) F = 12, V = 20, E = 30
∴ F + V – E
= 12 + 20 – 30
= 2 (satisfied)
(iv) F = 20, V = 13, E = 30
∴ F + V – E
= 20 + 13 – 30
= 3 (not satisfied)
(v) F = 32, V = 60, E = 90
∴ F + V – E
= 32 + 60 – 90
= 2 (satisfied)
Measurements Exercise 2.4 Solutions :
Question no – (1)
Two gates are fitted at the entrance of a library. to open the gates easily, a wheel is fixed at 6 feet distance from the wall to which the gate is fixed. If one of the gates is opened to 90°, find the distance moved by the wheel.
Solution :
θ = 90°, r = 6 ft
Distance moved by the wheel,
= 1/4 πr
= 1/4 × 3.147 × 6
= 4.71 ft
Thus, the distance moved by the wheel is 4.71 ft.
Question no – (2)
With his usual speed, if a person covers a circular track of radius 150 m in 9 minutes, find the distance that he covers in 3 minutes
Solution :
Given, r = 150
In 9 min the person covers = 150 m
∴ In 1 min the person covers = 150/9 m
In 3 min the person covers
= 150 × 3/9
= 50 m.
Hence, the distance that he covers in 3 minutes is 50 m.
Question no – (3)
Find the area of the house drawing given in the figure.
Solution :
Area of the given house = area of the square + are of the triangle + area of rectangle + area of the parallelogram
= a² + (1/2 b × h) + (l + b) + (b × h)
= 6² + (1/2 × 6 × 4) + (8 × 6) + (8 × 4)
= 36 + 12 + 48 + 32
= 128 sq.cm
Therefore, the area of the house is 128 sq.cm.
Challenging Problem Solution :
Question no – (5)
Guna has fixed a single door of width 3 feet in his room where as Nathan has fixed a double door, each of width 1 1/2 feet in his room. from the closed position, if each of the single and double doors can open up to 120°, whose door takes a minimum area.
Solution :
Guna’s door,
r = 3 feet
Nathan’s door,
r = 1 1/2
= 3/2 ft.
Area of the single door.
= θ°/360 × πr
= 120/360 × 22/7 × 3
= 33/14 sq.ft.
= 2.357 (approximately)
Area of the double door = θ°/360 × πr
= (120 + 120)/360° × 22/7 × 3/2
= 240/360 × 22/7 × 3/2
= 22/7
= 3.14 sq. ft
Hence, Guna’s door takes minimum area.
Question no – (6)
In a rectangular field which measures 15 m × 8m, cows are tied with a rope of length 3m at four corners of the field and also at the centre. find the area of the field where no cows can graze.
Solution :
Area = l × b
= (15×8) m
∴ l = 15, b = 8
∴ area of the field where no cows can graze = area of the rectangle – area of the circle
= (15×8) – πr²
= 120 – (3.14) × 3²
= 91.74 sq.m
Hence, the area of the field where no cows can graze is 91.74 sq.m
Question no – (7)
Three identical coins each of diameter 6 cm are placed as shown. find the area of the shaded region between the coins.
Solution :
Given, d = 6 cm
∴ r = 3 cm, π = 3.147
Area of the shaded region = area of the triangle – area of the quadratic areas
= √3/4 a² – 1/3 πr² sq.cm
= √3/4 6² – 3.14/3 × 3²
= 1.732/4 × 36 – 3.14 × 3
= 6.168 sq.cm
Therefore, the area of the shaded region between the coins is 6.168 sq. cm.
Question no – (8)
Using Euler’s formula, find the unknowns.
| S.No. | Faces | Vertices | Edges |
| (i) | ? | 6 | 14 |
| (ii) | 8 | ? | 10 |
| (iii) | 20 | 10 | ? |
Solution :
Euler’s formula = F + V – E = 2
(I) F + V – E = 2
= F + 6 – 14 = 2
= F = 2 + 8
= 10
(II) F + V – E = 2
= 8 + V – 10 = 2
= V = 2 + 2
= 4
(III) F + V – E = 2
= 20 +10 – E = 2
= 30 – E = 2
= E = 30 – 2
= 28
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