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**Surface Area of Cube, Cuboid and Cylinder**

**Cube: **A cube has 6 faces, all of which are congruent squares.

Total surface area of a cube having length of one side as ‘l’ = 6l^{2} |

**Remember**

- For a cube, length, breadth and height are all equal.
- If the length of the edge of a cube is doubled, it surface area increases four times

**Cuboid: **A cuboid too have 6 faces but they may or may not be equal. All the six faces are rectangular and the opposite faces are identical, Thus, in a cuboid, we get 3 pairs of identical faces.

For a cuboid having length, breadth and height as l, b and h respectively, we have:

Total surface area = 2(lb + bh + hl) |

Note: Area of 4 walls of a room = [perimeter of the room] *height of the room

**Cylinder:** A cylinder has three surfaces: (i) curved surface, (ii) base surface and (iii) Top surface.

**Note: **

- The top and bottom (base) surfaces are circular.
- The top surface and bottom surface are identical.

Total surface area of a cylinder = 2πr[h + r] |

**Question: **Why is it incorrect to call the solid shown here a cylinder?

**Solution: **We know that a cylinder has two identical (congruent) circular faces, parallel to each other. Therefore, it is incorrect to call the given solid as a cylinder, because its opposite faces are not congruent.

**Question: **Find the total surface area of the following cuboids

**Solution:** (i) ∵ l = 6 cm, b = 4 cm and h = 2 cm

∴ Total surface area of the cuboids = 2(lb + bh + hl)

= 2(6 * 4 + 4 * 2 + 2 * 6cm^{2}

= 2(24 + 8 + 12) cm^{2}

= 2* 44 cm^{2} = 88 cm^{2}

**(ii)** ∵ l = 4 cm, b = 4 cm and h = 10 cm

∴ Total surface area of the cuboid = 2(lb + bh + hl)

= 2(4*4 + 4*10 + 10*4) cm^{2}

= 2(16 + 40 + 40) cm^{2}

= 2*96 cm2 = 192 cm^{2}

**Question 1**. Can we say that the total surface area of cuboid = lateral surface area + 2 * area of base?

**Solution: **Yes, or a cuboid,

Total surface area = Lateral surface area + 2 * Area of the base [∵ Opposite faces are congruent]

**Question 2. **If we interchange the lengths of the base and the height of a cuboid [Fig. (i)] to get another cuboid [Fig. (ii)], will its lateral surface area change?

**Solution: **Lateral surface area of figure (i) = 2(l + b) * h

Lateral surface area of figure (ii) = 2(h + b) * l

Since, the two results are different.

∴ By changing the position of a cuboid, its lateral surface will change.

**Question: **Find the surface area of cube A and lateral surface area of cube B.

**Solution:** For cube A:

Side (l) = 10 cm

∴ Total surface area of the cube = 6l^{2}

= 6 * (10)^{2} cm^{2} = 600 cm^{2}

For cube B:

Side (l) = 8 cm

∴ Total surface area of the cube = 6l^{2}

= 6 * (8)^{2} cm^{2}

= 6 * 64 cm^{2} = 256 cm^{2}

**Question 1. **Two cubes each with side b are joined to form a cuboid figure. What is the surface area of this cuboid? Is it 12b2? Is the surface area of cuboid formed by joining three such cubes, 18b^{2}? Why?

**Solution:** When 2 cubes are joined end to end:

Length (L) = b + b = 2b

Breadth (B) = b

Height (H) = b (∵ Height and breadth do not chang.)

∴ Total surface area = 2[LB + BH + HL]

= 2[(2b * b) + (b * b) + (b * 2b)]

= 2[2b^{2} + b^{2} + 2b^{2}]

= 2[5b^{2}] = 10b^{2}

∴ The total surface area is not 12b^{2}.

When 3 cubes are joined end to end:

Length (L) = 3b

Breadth (B) = b

Height (H) = b (Here too breadth and height do not change.)

∴ Total surface area of the cuboid = 2[LB + BH + HL]

= 2[(3b * b) + (b * b) + (b * 3b)]

= 2[3b^{2} + b^{2} + 3b^{2}]

= 2[7b^{2}] = 14b^{2}

∴ The surface area is not 18b^{2}.

**Question 2. **How will you arrange 12 cubes of equal length to form a cuboid of smallest surface area?

**Solution:** **Case I**

Length (L) = 12b

Breadth (B) = b

Height (H) = b

∴ Total surface area of the cuboid = 2[LB + BH + HL]

= 2(12b * b) + (b * b) + (b * 12b)]

= 2[12b^{2} + b^{2} + 12b^{2}]

= 2[25b^{2}] = 50b^{2}

**Case II**

Length (L) = 6b

Breadth (B) = 2b

Height (H) = b

∴ Surface area of the cuboid = 2[LB + BH + HL]

= 2[(6b * 2b) + (2b * b) + (b * 6b)]

= 2[12b^{2} + 2b^{2} + 6b^{2}]

= 2[20b^{2}] = 40b^{2}

**Case III**

Length (L) = 3b

Breadth (B) = 2b

Height (H) = 2b

∴ Surface area of the cuboid = 2(LB + BH + HL)

= 2[(3b * 2b) + (2b * 2b) + (2b * 3b)

= 2[6b^{2 }+ 4b^{2} + 6b^{2}]

= 2[16b^{2}] = 32b^{2}

Obviously, the case III arrangement of 12 cubes, has the minimum surface area.

**Question 3.** After the surface area of cube is painted, the cube is cut into 64 smaller cubes of same dimensions. How many have no face painted? 1 face painted? 2 faces painted? 3 faces painted?

**Solution: **In the given figure, we have:

(a) Number of cubes having no face painted = 16

(b) Number of cubes having 1 face painted = 16

(c) Number of cubes having 2 faces painted = 24

(d) Number of cubes having 3 faces painted = 8

**Question: **Find total surface area of the following cylinders.

**Solution:**

**(i) **We have

Radius (r) = 14 cm

Height (h) = 8 cm

∴ Total surface area = 2πr(r + h)

**(ii) **We have r = 2/2 1 m and h = 2 m

∴ Total surface area = 2πr(r + h)

**Remember**

If the cylinder is open then the total surface area is equal to the curved surface area of the cylinder

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