Step-by-step solutions for Class 9 Mathematics Ganita Manjari Part 1I Chapter 14 Exercise Set 14.1 (Questions 1–8, Page 126). Covers calculating total surface area from cube volume, 3D packing of cubes into a cubical box, godown capacity with cuboidal boxes, surface area of cubes joined end-to-end, surface-area-to-volume ratio scaling, finding cuboid volume from three meeting face areas, painted cube subdivision analysis, and integer cuboid dimensions.
The volume V of a cube of edge length a is the measure of three-dimensional space enclosed by it, calculated as side cubed.
A cube has 6 congruent square faces, each of area a². Its total surface area (TSA) is the sum of the areas of all six faces.
For a cuboid of length l, width (breadth) w, and height h, volume is the product of its three dimensions, and total surface area is the sum of the areas of its three pairs of opposite rectangular faces.
When packing smaller items into a larger volume, all dimensions must be expressed in the same unit before calculation (e.g. 1 m = 100 cm). When dimensions divide evenly along length, width, and height, total count is (N_l × N_w × N_h) = Volume_container / Volume_item.
When a solid is partitioned into smaller similar pieces, the total volume remains constant, but the total exposed surface area increases in inverse proportion to the edge length ratio.
If the three faces of a cuboid meeting at a vertex have areas A₁ = lw, A₂ = wh, and A₃ = hl, their product is (lwh)² = V². Hence, the volume is the square root of the product of the three face areas: V = √(A₁A₂A₃).
When an n × n × n cube painted on all outer faces is divided into unit cubes: exactly 3 faces painted = 8 (the corners), exactly 2 faces painted = 12(n − 2) (along edges), exactly 1 face painted = 6(n − 2)² (on faces), and no faces painted = (n − 2)³ (interior core).
The volume of a cube is 64 cm³. What is its total surface area?
Let the side of the cube be a cm.
The volume of a cube is
Since 4 × 4 × 4 = 64,
The total surface area of a cube is
How many small cubes with side 20 cm can be packed tight in a cubical box with side 2 m?
Side of the large cubical box:
Side of each small cube = 20 cm.
Number of small cubes that fit along each edge:
Since the box is cubical, 10 cubes fit along its length, 10 along its width, and 10 along its height.
Number of small cubes = 10 × 10 × 10
We can also check using the volume formula, V = a³:
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.
Dimensions of the godown:
40 m × 25 m × 10 m
Dimensions of each box:
2 m × 1.25 m × 1 m
Number of boxes along the length:
Number of boxes along the width:
Number of boxes along the height:
Number of boxes = 20 × 20 × 10
This can also be checked using the volume formula for a cuboid:
Volume of the godown:
40 × 25 × 10 = 10,000 m³
Volume of one box:
2 × 1.25 × 1 = 2.5 m³
Number of boxes = 10,0002.5 = 4000
Two cubes each of volume 125 cm³ are joined end to end. Find the surface area of the resulting cuboid.
Let the side of each cube be a cm.
The volume of a cube is
Since 5 × 5 × 5 = 125,
When the two cubes are joined end to end, the dimensions of the resulting cuboid are
l = 5 + 5 = 10 cm
The total surface area of a cuboid is
TSA = 2(10 × 5 + 5 × 5 + 5 × 10)
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.)
Side of the original cube = 4 cm.
The total surface area of a cube is
Surface area of the original cube:
Each small cube has side 1 cm. The number of small cubes formed is
Number of small cubes = 4³ / 1³
Surface area of one small cube:
Total surface area of all 64 small cubes, considered separately:
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?
Let the length, width and height of the cuboid be l, w, h, respectively.
The areas of the three faces are
lw = 6, wh = 15, hl = 10
Multiplying the three equations,
Since the volume of a cuboid is
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?
A 5 cm cube painted on all its faces and sliced into 125 small 1 cm³ cubes.
The side of the large cube is 5 cm, and each small cube has volume 1 cm³. Hence, each small cube has side 1 cm.
Number of small cubes along each edge:
Total number of small cubes:
Since the large cube is painted on all its outer faces before being sliced, the number of painted faces on each small cube depends on its position.
The corner cubes have three painted faces. A cube has 8 corners.
8 small cubes
The cubes along the edges, excluding the corner cubes, have exactly two painted faces.
A cube has 12 edges, and each edge contains 5 − 2 = 3 cubes between its corners.
36 small cubes
The cubes inside each face, excluding the edge cubes, have exactly one painted face.
Number of such cubes on each face:
Since a cube has 6 faces,
54 small cubes
The cubes completely inside the large cube have no painted faces.
Number of interior cubes along each dimension:
27 small cubes
Check:
8 + 36 + 54 + 27 = 125
This equals the total number of small cubes, so all the cubes have been accounted for.
Find 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.
Let the integer edge lengths of the cuboid be a, b, c cm, arranged in increasing order:
1 ≤ a ≤ b ≤ c
The total surface area of a cuboid is
Yes. For example, the dimensions 1 cm × 2 cm × 16 cm and 2 cm × 4 cm × 7 cm both satisfy the condition.
For 1 × 2 × 16,
TSA = 2(1 × 2 + 2 × 16 + 16 × 1)
For 2 × 4 × 7,
TSA = 2(2 × 4 + 4 × 7 + 7 × 2)
Starting with
adding a² to both sides gives
a² + ab + ac + bc = 50 + a²
Factorising,
Also, since a ≤ b ≤ c,
ab + bc + ca ≥ 3a²
3a² ≤ 50
The factor pair 3 × 17 = 51 gives
b + 1 = 3, c + 1 = 17
The dimensions are 1 × 2 × 16 cm.
The factor pair 6 × 9 = 54 gives
b + 2 = 6, c + 2 = 9
The dimensions are 2 × 4 × 7 cm.
Since 59 is a prime number, its only positive factor pair is 1 × 59. But b ≥ 3, so b + 3 ≥ 6. No valid factor pair exists.
Since b ≥ 4, we need b + 4 ≥ 8. But none of the factor pairs of 66, namely 1 × 66, 2 × 33, 3 × 22, and 6 × 11, has a smaller factor of at least 8. Therefore, no solution exists in this case.
We have considered every possible value of the smallest edge, a = 1, 2, 3, 4. Only the first two cases give valid integer dimensions.
1 cm × 2 cm × 16 cm
2 cm × 4 cm × 7 cm
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