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 cm

θ° = 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 cm

Hint:

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.

Number of equal sectors ‘n’ = 8
Radius of the sector ‘r’ = 56 cm
Area of each sector = \(\frac{1}{n}\) πr 2 sq. units
= \(\frac{1}{8} \times \frac{22}{7}\) × 56 × 56cm 2 =1232 cm 2
Area of each sector = 1232 cm 2 (approximately)
Posted in Class 8 on September 10, 2024 September 11, 2024
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Area of the doormat = Area of 2 trapezium
Height of the trapezium h =\(\frac { 70 }{ 2 }\) cm: a = 90 cm; b = 70 cm
∴ Area of the trapezium = \(\frac { 1 }{ 2 }\) h (a + b) sq. units
Area of the door mat = 2 × \(\frac { 1 }{ 2 }\) × 35 (90 + 70)cm 2
= 35 × 160 cm 2 = 5600 cm 2
∴ Area of the door mat = 5600 cm 2

Area = Area of a rectangle + Area of a triangle + Area of a trapezium
For rectangle length l = 120 – 20 – 20 cm = 80 cm
Breadth b = 30 cm
For the triangle base = 30 cm
Height = 20 cm
For the trapezium height h = 20 cm
Parallel sided a = 50 cm
b = 30cm
∴ Area of the figure (l × b) + (\(\frac { 1 }{ 2 }\) × base × height) + \(\frac { 1 }{ 2 }\) × h × (a + b)sq. units
= (80 × 30) + (\(\frac { 1 }{ 2 }\) × 30 × 20) + \(\frac { 1 }{ 2 }\) × 20 × (50 + 30) cm 2
= 2400 + 300 + 800 cm 2 = 3500 cm 2
Area of the figure = 3500 cm 2

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
Posted in Class 8 on September 10, 2024 September 11, 2024
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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 feet

Angle 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 sector

z
= \(\frac { 1 }{ 2 }\) (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 circle

Area 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 – E

What 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.
Posted in Class 8 on September 11, 2024 September 12, 2024
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