Solid means having three dimensions (length, breadth and thickness), as a geometrical body or figure. Or it may also be defined as relating to bodies or figures of three dimensions.
Prisms are the solid object that has two opposite faces congruent and parallel.
The above figures are different solid figures which have congruent opposite face.
Each congruent face of a prism is called its cross-section. There are infinite numbers of imaginary surfaces that cut the prism perpendicular to its height or length with congruent surfaces.
Total surface area of prism
Let's take a cartoon box. While unfolding the cartoon box, there we can find many faces of the boxes. Measure all the surfaces and find the area of each surface. If the length is “l” breadth is “b” and height is ‘h’, what will be the total surface area?
In a box, altogether there are 6 surfaces. Among them three pairs are congruent. Thus, total surface area (A) = 2(l \(\times\) b) + 2(b \(\times\)h) + 2(l \(\times\) h) square units
= 2(lb + bh + lh) square units.
In the case of irregular shape, area of surface are calculated separately and added to find out the total surface area.
Lateral surface area of prism
The sum of areas of four lateral surfaces of the prism is called the lateral surface area of the prism.
Lateral surface area (S) = 2(l \(\times\) h) + 2(b \(\times\) h)
= 2lh + 2bh
= h \(\times\) 2(l+b)
= h \(\times\) p
\(\therefore\) lateral surface area (S) = height \(\times\) perimeter of base
Total surface area of the prism (TSA) = Lateral surface area + 2 \(\times\) area of cross section
i.e. TSA =LSA + 2A
Volume of prism
let's take a cuboidal prism.
For a cuboid | For a cube |
Volume (V) = l \(\times\) b \(\times\) h V = A \(\times\) H Where, A = l \(\times\) b |
Volume (V) = l \(\times\) l \(\times\) l (V) = l^{2} \(\times\) l (V) = A \(\times\) l |
Thus, Volume = Area of cross-section × height
S.N |
Solid Figures |
Area of base or cross section |
Lateral Surface Area |
Total surface Area |
Volume |
1. | Cuboid | A = l \(\times\) b | 2h(l + B) | 2(lb + bh + lh) | V = l\(\times\) b\(\times\) h |
2. | Cube | A = l^{2} | 4l^{2 }or 4A | 6l^{2}or 6A | V = l^{3 } |
3. | Prism | A = base area or A = area of cross section | h \(\times\) p | p\(\times\) h + 2A | V = A\(\times\) h |
V = A \(\times\) H
Where, A = l \(\times\) b
(V) = l^{2} \(\times\) l
(V) = A \(\times\) l
.
Solution:
Here, length (l) = 8cm
Total surface area (TSA) = 6l^{2}
= 6× (8cm)^{2}
= 6× 64cm^{2}
= 384cm^{2}
Again,
Volume (V) = l^{3}
= (8cm)^{3}
= 512cm^{3}
Hence, TSA = 384cm^{2} and volume = 512cm^{3}
Solution:
Here,
Lateral Surface Area (LSA) = perimeter× height
= (1cm +cm + 4cm + 1cm + 8cm + 6cm + 8cm) × 6cm
= 34cm× 6cm
= 204cm^{2}
Also,
Volume (V) = Area of cross section× height
= (8cm× 6cm - 4cm× 3cm)× 6cm
= (48cm^{2} - 12cm^{2})× 6cm
= 36cm^{2}× 6cm
= 216cm^{3}
Solution:
Here,
Length (l) = 10cm
Breadth (b) = 5cm
Heigth (h) = 8cm
Now,
LSA = 2h(l + b)
= 2× 8cm (10cm + 5cm)
= 16cm× 15cm
= 240cm^{2}
Again,
TSA = 2(lb + bh + lh)
= 2 (10cm× 5cm + 5cm× 8cm +10cm× 8cm)
= 2 (50cm^{2} + 40cm^{2} + 80cm^{2})
= 2× 170cm^{2}
= 340cm^{2}
Also,
Volume (V) = l× b× h
= 10cm× 5cm× 8cm
= 400cm^{3}
Solution:
Here,
Length of box (l) = 30cm
Breadth of box (b) = 20cm
Height of box (h) = 10cm
Length of soap (l_{1}) = 3cm
Breadth of soap (b_{1}) = 2cm
Height of soap (h_{1}) = 2cm
Now,
Volume of box (V) = l× b× h
= 30cm× 20cm× 10cm
= 6000cm^{3}
Volume of soap (V_{1}) =l_{1}× b_{1}× h_{1}
= 3cm × 2cm ×2cm
= 12cm^{3}
∴ Number of soap that can be fitted on the box (N) = \(\frac{V}{V_1}\)
= \(\frac{600cm^3}{12cm^3}\)
= 50
Hence, 50 soap can be fitted on the box.
Solution:
Here,
Length (l) = 16m
Breadth (b) = 15m
Heigth (h) = 13m
Area of 4 walls (A) = 2h(l + b)
= 2× 13m (16m + 15m)
= 26m× 31m
= 806m^{2}
Cost of painting wall (C) = Rs 120m^{2}
Total Cost (T) = ?
∴ Total Cost (T) = C× A
= Rs 120 ×806m^{2}
= Rs 96,720
LSA | TSA |
The lateral surface area of a three-dimensional object is the surface area of the object minus the area of its bases. | The total surface area of a three- dimensional object is the surface area of the object adding all of its areas. |
For example, dice. We found that the surface area of a six-sided dice.Since the dice has two bases, we subtractthe area of the two bases. | For example,dice. We found that the surface area of a six-sided dice. We add all the areas of a dice. |
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