Samacheer Class 8 Maths - Measurements

(i) central angle 45° r = 16 cm
Length of the arc l = \(\frac{\theta^{\circ}}{360^{\circ}}\) × 2πr units
l = \(\frac{45^{\circ}}{360^{\circ}}\) × 2 × 3.14 × 16 cm
l = \(\frac{1}{8}\) × 2 × 3.14 × 16 cm
l = 12.56 cm
Area of the sector = \(\frac{\theta^{\circ}}{360^{\circ}}\) × πr 2 sq. units
A = \(\frac{45^{\circ}}{360^{\circ}}\) × 3.14 × 16 × 16
A = 100.48 cm 2
Perimeter of the sector P = l + 2r units
P = 12.56 + 2(16) cm
p = 44.56 cm
(ii) central angle 120°, d = 12.6 cm
∴ r = \(\frac{12.6}{2}\) cm
r = 6.3cm
Length of the arc l = \(\frac{\theta^{\circ}}{360^{\circ}}\) × 2πr units
l = \(\frac{120^{\circ}}{360^{\circ}}\) × 2 × 3.14 × 63 cm
l = 13.188cm
I = 13.19cm
Area of the sector A = \(\frac{\theta^{\circ}}{360^{\circ}}\) × πr 2 sq. units
A = \(\frac{120^{\circ}}{360^{\circ}}\) × 3 14 × 6.3 × 6.3 cm 2
A = 3.14 × 6.3 × 2.1 cm 2
A = 41.54 cm 2
Perimeter of the sector P = l + 2r cm
P = 13.19 + 2(6.3) cm
= 13.19 + 1.2.6 cm
P = 25.79 cm

Area of the sector A = \(\frac{l r}{2}\) sq. units
l = 48m
r = 10m
= \(\frac{48 \times 10}{2}\) m 2
= 24 × 10m 2
= 240 m 2
Area of the sector = 240 m 2
(ii) length of the arc = 50 cm, r = 13.5 cm
Length of the arc l = 12.5 cm
Radius r = 6 cm
Area of the sector A = \(\frac{l r}{2}\) sq. units
A = \(\frac{12.5 \times 6}{2}\)
A = 12.5 × 3cm 2
A = 37.5 cm 2
Area of the sector A = 37.5 cm 2

area = 462 cm 2 , r = 21 cm
Radius of the Sector = 21 cm
Area of the sector = 462 cm 2
\(\frac{l r}{2}\) = 462
\(\frac{l \times 21}{2}\) = 462
l = \(\frac{462 \times 2}{21}\)
l = 22 × 2
Length of the arc l = 44 cmθ° = 120°
∴ Central angle of the sector = 120°
(ii) length of the arc = 44 m, r = 35 m
Radius of the sector = 8.4cm
Area of the sector = 18.48 cm 2
\(\frac{l r}{2}\) = 18.48
\(\frac{1 \times 8.4}{2}\) = 18.48
l = \(\frac{18.48 \times 2}{8.4}\)
Hint:Length of the arc l = 4.4 cmHint:θ° = 30°
Central angle = 30°

Radius of the circle r = 120 m
Number of equal sectors = 8
∴ Central angle of each sector = \(\frac{360^{\circ}}{n}\)
θ° = \(\frac{360^{\circ}}{8}\)
θ° = 45°
Length of the arc l = \(\frac{\theta^{\circ}}{360^{\circ}}\) × 2πr units
= \(\frac{45^{\circ}}{360^{\circ}}\) × 2π × 120 m
Length of the arc = 30 × πm
Another method:
l = \(\frac{1}{n}\) × 2πr = \(\frac{1}{8}\) × 2 × π × 120 = 30 π m
Length of the arc = 30 π m

Radius of the sector r = 70 cm
Number of equal sectors = 5
∴ Central angle of each sector = \(\frac{360^{\circ}}{n}\)
θ° = 360°
θ° = 72°
Area of the sector = \(\frac{\theta^{\circ}}{360^{\circ}}\) × πr 2 sq.units
= \(\frac{72^{\circ}}{360^{\circ}}\) × π × 70 × 70cm 2
Hint:= 14 × 70 × πcm 2
= 980 πcm 2
Note: We can solve this problem using A = \(\frac{1}{n}\) πr 2 sq. units also.

Area of the field = Area of trapezium FBCH + Area of ∆DHC + Area of ∆EGD + Area of ∆EGA + Area of ∆BFA
Area of the triangle = \(\frac { 1 }{ 2 }\) bhsq.units
Area of the trapezium = \(\frac { 1 }{ 2 }\) × h × (a + b) sq.units
Area of the trapezium FBCH = \(\frac { 1 }{ 2 }\) × (10 + 8) × (8 + 3)m 2 = 9 × 11 = 99 m 2 ….. (1)
Area of the ∆DHC = \(\frac { 1 }{ 2 }\) × 8 × 5 m 2 = 20 m 2 ….. (2)
Area of ∆EGD = \(\frac { 1 }{ 2 }\) × 8 × 15m 2 = 60 m 2 …….. (3)
Area of ∆EGA = \(\frac { 1 }{ 2 }\) × 8 × (8 + 6)m 2 = 4 × 14 m 2
= 56m 2
Area of ∆BFA = \(\frac { 1 }{ 2 }\) × 3 × 6m 2 = 9 m 2
∴ Area of the field = 99 + 20 + 60 + 56 + 9 m 2
= 244 m 2
Area of the field = 244 m 2

length, breadth, height
(ii) The meeting point of more than two edges in a polyhedron is called as ______ .
Vertex(iii) A cube has _________ faces.
six
(iv) 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 ______ .
cube

Radius of the circular track = 150m
Distance covers in 9 minutes = Perimeter of the circle = 2 × π × r units
Distance covered in 9 min = 2 × 3.14 × 150m
Distance covered in 1 min = \(\frac{2 \times 3.14 \times 150}{9} \mathrm{m}\)Distance he covers in 3min = 314m

Width of the door that Guna fixed = 3 feet.
When the door is open the radius of the sector = 3 feetAngle covered = 120°
∴ Area required to open the door= 3π feet 2
(b) Width of the double doors that Nathan fixed = 1 \(\frac { 1 }{ 2 }\) feet.
Angle described to open = 120°
Area required to open = 2 × Area of the sectorz
= \(\frac { 1 }{ 2 }\) (3π) feet 2
∴ The double door requires the minimum area.

Area of the field where none of the cow can graze = Area of the rectangle – [Area of 4 quadrant circles] – Area of a circleArea of the rectangle = l × b units 2
= 15 × 8m 2 = 120m 2
Area of 4 quadrant circles = 4 × \(\frac { 1 }{ 4 }\) πr 2 units
Radius ofthe circle = 3m
Area of 4 quadrant circles = 4 × \(\frac { 1 }{ 4 }\) × 3.14 × 3 × 3 = 28.26 m 2
Area of the circle at the middle = πr 2 units
= 3.14 × 3 × 3m 2 = 28.26m 2
∴ Area where none of the cows can graze
= [120 – 28.26 – 28.26]m 2 = 120 – 56.52 m 2 = 63.48 m 2

So, \(\frac{180^{\circ}}{360^{\circ}}\) = \(\frac{1}{2}\)
\(\frac{120^{\circ}}{360^{\circ}}\) = \(\frac{1}{3}\)
\(\frac{90^{\circ}}{360^{\circ}}\) = \(\frac{1}{4}\)
Think (Text Book Page No. 57)
If the radius of a circle is doubled, what will happen to the area of the new circle so formed?
If r = 2r 1 ⇒ Area of the circle = πr 2 = π(2r 1 ) 2 = π4r 1 2
Area = 4 × old area
Think (Text Book Page No. 61)
All the sides of a rhombus are equal. It is a regular polygon?
For a regular polygon all sides and all the angles must be equal. But in a rhombus all the sides are equal. But all the angles are not equal
∴ It is not a regular polygon.Try This (Text Book Page No. 64)
In the above example split the given mat into two trapeziums and verify your answer.
Area of the mat = Area of I trapezium + Area of II trapezium
= [\(\frac { 1 }{ 2 }\) × h 1 × (a 1 + b 1 )] + [\(\frac { 1 }{ 2 }\) × h 2 × (a 2 + b 2 )] sq. units
= [\(\frac { 1 }{ 2 }\) × 2 × (7 + 5)] + \(\frac { 1 }{ 2 }\) × 2 × (9 + 7) sq.feet
= 12 + 16 = 28 sq.feet
∴ Cost per sq.feet = ₹ 20
Cost for 28sq. feet = ₹ 20 × 28 = ₹ 560
∴ Total cost for the entire mat = ₹ 560
Both the answers are the same.Try This (Text Book Page No. 68)
Tabulate the number of faces (F), vertices (V) and edges (E) for the following polyhedrons. Also find F + V – EWhat do you observe from the above table? We observe that, F + V – E = 2 in all the cases. This is true for any polyhedron and this relation F + V – E = 2 is known as Euler’s formula.From the table F + V – E = 2 for all the solid shapes.