Exercise Set 14.1
1. The volume of a cube is 64 cm³. What is its total surface area?
Answer:
Let the side be a cm.
a³ = 64, so a = 4 cm.
TSA = 6a² = 6 × 4² = 96 cm².
Hence, the total surface area is 96 cm².
2. How many small cubes with side 20 cm can be packed tight in a cubical box with side 2 m?
Answer:
Side of the box = 2 m = 200 cm.
Volume of box = 200 × 200 × 200 = 80,00,000
Side of small cube = 20 cm.
Volume of small cube = 20 × 20 × 20 = 8000
So, number of cubes in the box = 80,00,000 ÷ 8000 = 1000.
Hence, 1000 small cubes with side 20 cm can be packed tight in the cubical box.
3. The dimensions of a godown are 40 m × 25 m × 10 m. If it is filled with cuboidal boxes, each of dimensions 2 m × 1.25 m × 1 m, then find the number of boxes.
Answer:
Volume of godown = 40 × 25 × 10 = 10000 m³.
Volume of one box = 2 × 1.25 × 1 = 2.5 m³.
Number of boxes = 10000 ÷ 2.5 = 4000.
4. Two cubes each of volume 125 cm³ are joined end to end. Find the surface area of the resulting cuboid.
Answer:
Side of each cube = ∛125 = 5 cm.
The joined cuboid has dimensions 10 cm × 5 cm × 5 cm.
TSA = 2(lw + lh + wh)
= 2(10 × 5 + 10 × 5 + 5 × 5)
= 250 cm².
Hence, the surface area of the resulting cuboid is 250 cm².
5. A cube of side 4 cm is cut into cubes of side 1 cm. What is the ratio of the surface areas of the original cube and all the cut-out cubes?
(Note that there is no change in volume but a big change in the surface area. This property has major consequences in the biological world.)
Answer:
Original surface area = 6 × 4² = 96 cm².
Number of small cubes = 4³ ÷ 1³ = 64.
Surface area of one small cube = 6 × 1² = 6 cm².
Total surface area of all separate small cubes = 64 × 6 = 384 cm².
Required ratio = 96 : 384 = 1 : 4.
6. The surface areas of the three faces of a cuboid that meet at one of the corners of the cuboid are 6 cm², 15 cm², and 10 cm² respectively. What is the volume of the cuboid?
Answer:
The three face areas are lw, wh and lh.
(lw)(wh)(lh) = (lwh)² = 6 × 15 × 10 = 900.
Volume = lwh = √900 = 30 cm³.
Hence, the volume of the cuboid is 30 cm³.
7. A cube of side 5 cm is painted on all its faces. If it is sliced into 1 cm³ cubes, how many of these 1 cm³ cubes have
(i) exactly three faces painted?
(ii) exactly two faces painted?
(iii) exactly one face painted?
(iv) no face painted?
Answer:
There are 5 small cubes along each edge.
(i) Exactly three faces painted:
These are the corner cubes. A cube has 8 corners.
So, exactly three faces painted = 8 cubes.
(ii) Exactly two faces painted:
These lie along the edges, excluding the corners.
Each edge has 5 − 2 = 3 such cubes. There are 12 edges.
Number = 12 × 3 = 36 cubes.
(iii) Exactly one face painted:
These lie in the middle of each face, away from the edges.
Each face has (5 − 2)² = 9 such cubes. There are 6 faces.
Number = 6 × 9 = 54 cubes.
(iv) No face painted:
These form the inner cube of side 5 − 2 = 3 cm.
Number = 3³ = 27 cubes.
8. Fine a cuboid with edges whose lengths are integers (in cm), given that it has a total surface area of exactly 100 cm².
(i) Is there more than one such cuboid?
(ii) Can you find them all?
(iii) Show that you have found them all.
Answer:
Let the positive integer edge lengths be a ≤ b ≤ c.
2(ab + bc + ca) = 100, so ab + bc + ca = 50.
(i) Yes, there is more than one such cuboid.
(ii) The two different sets of dimensions are:
First cuboid: 1 cm, 2 cm, 16 cm.
Surface Area: 2(1 × 2 + 2 × 16 + 1 × 16) = 2(50) = 100 cm².
Second cuboid: 2 cm, 4 cm, 7 cm.
Surface Area: 2(2 × 4 + 4 × 7 + 2 × 7) = 2(50) = 100 cm².
(iii) Proof that these are all:
Since a ≤ b ≤ c, each of ab, ac and bc is at least a².
Thus, 3a² ≤ 50, so a can only be 1, 2, 3 or 4.
Rearrange the equation as (b + a)(c + a) = 50 + a².
If a = 1: (b + 1)(c + 1) = 51.
The factor pairs are (1, 51) and (3, 17). Only (3, 17) gives positive b ≥ 1.
So b = 2 and c = 16.
If a = 2: (b + 2)(c + 2) = 54.
The factor pairs are (1, 54), (2, 27), (3, 18) and (6, 9).
Since b + 2 must be at least 4, only (6, 9) is allowed.
So b = 4 and c = 7.
If a = 3: (b + 3)(c + 3) = 59.
59 is prime, so no pair has both factors at least 6.
If a = 4: (b + 4)(c + 4) = 66.
The factor pairs are (1, 66), (2, 33), (3, 22) and (6, 11).
None has both factors at least 8.
Therefore, exactly two cuboids are possible, apart from rearranging their length, width and height.