NCERT Solutions Class 9 Mathematics Ganita Manjari 2 Chapter 14 Math of Space Surface Area and Volume Exercise 14.1

Exercise Set 14.1

Question 1. The volume of a cube is 64 cm³. What is its total surface area?
Answer:

Let each edge of the cube be a cm.
a³ = 64, so a = ∛64 = 4 (because 4 × 4 × 4 = 64). The edge is 4 cm.
Total surface area = 6a² = 6 × 4² = 6 × 16 = 96 cm²
The total surface area of the cube is 96 cm².

Question 2. How many small cubes with side 20 cm can be packed tight in a cubical box with side 2 m?
Answer:

Step 1: Use one unit. 2 m = 2 × 100 cm = 200 cm.
Step 2: Along one edge of the box, the number of small cubes = 200 ÷ 20 = 10.
Step 3: The box is a cube, so 10 cubes fit along the length, 10 along the width and 10 along the height.
Number of small cubes = 10 × 10 × 10 = 1000
Check with volumes: \( \frac{200^3}{20^3} \) = \( \frac{8000000}{8000} \) = 1000 ✓
1000 small cubes can be packed tightly in the box.

Teacher's Note:
1. Convert both lengths to the same unit before dividing (2 m is 200 cm, not 2 cm).
2. Counting how many fit along each edge is safer than dividing volumes, because it also shows whether the cubes fit exactly.

Question 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 the godown = 40 × 25 × 10 = 10000 m³
Volume of one box = 2 × 1.25 × 1 = 2.5 m³
Number of boxes = \( \frac{10000}{2.5} \) = 4000
Check that the boxes fit exactly: along the length 40 ÷ 2 = 20, along the width 25 ÷ 1.25 = 20, along the height 10 ÷ 1 = 10; and 20 × 20 × 10 = 4000, with no gap left.
The godown can hold 4000 boxes.

Question 4. Two cubes each of volume 125 cm³ are joined end to end. Find the surface area of the resulting cuboid.
Answer:
Let the side of each cube be a cm. Then a³ = 125, so a = ∛125 = 5 cm.

Joined end to end, the two cubes form a cuboid with
l = 5 + 5 = 10 cm, w = 5 cm, h = 5 cm.
Surface area = 2(lw + wh + hl) = 2(10 × 5 + 5 × 5 + 5 × 10) = 2(50 + 25 + 50) = 2 × 125 = 250 cm²
Check: the two cubes have 12 square faces of 25 cm² each; the 2 faces pressed together are hidden, leaving 10 × 25 = 250 cm².
The surface area of the resulting cuboid is 250 cm².

Teacher's Note:
1. A common slip is to add the two cube surface areas (2 × 150 = 300 cm²); the two glued faces are no longer on the outside.
2. The face-counting check is a quick way to catch this error.

Question 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:

Surface area of the original cube = 6 × 4² = 6 × 16 = 96 cm²
Number of 1 cm cubes = \( \frac{4^3}{1^3} \) = 64
Surface area of one small cube = 6 × 1² = 6 cm², so all 64 together have 64 × 6 = 384 cm².
Ratio = 96 : 384 = 1 : 4
The ratio is 1 : 4. The volume stays 64 cm³, but the total surface becomes four times as large. That is why living things use many tiny units (cells, the small air sacs of the lungs) to get a huge surface for exchange.

Question 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:

Let the edges be l, w and h. The three faces meeting at a corner have areas lw, wh and hl, so
lw = 6, wh = 15, hl = 10
Multiply the three equations: (lw)(wh)(hl) = 6 × 15 × 10
l²w²h² = 900, i.e. (lwh)² = 900
lwh = √900 = 30
(The edges themselves are l = 30 ÷ 15 = 2 cm, w = 30 ÷ 10 = 3 cm, h = 30 ÷ 6 = 5 cm; check: 2 × 3 = 6, 3 × 5 = 15, 5 × 2 = 10.)
The volume of the cuboid is 30 cm³.

Teacher's Note:
1. The key step is to multiply the three face areas: the product is (volume)².
2. Students do not need to find l, w and h separately, but doing so gives a good check.

Question 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:
Slicing gives 5 × 5 × 5 = 125 small cubes, 5 along each edge.

(i) Three painted faces: only the corner cubes. A cube has 8 corners, so 8 cubes.
(ii) Two painted faces: cubes along an edge but not at its ends. Each edge has 5 − 2 = 3 of them and there are 12 edges, so 12 × 3 = 36 cubes.
(iii) One painted face: cubes in the middle of a face, away from its edges; they form a 3 × 3 square of 9 on each of the 6 faces, so 6 × 9 = 54 cubes.
(iv) No painted face: the hidden inner cube of side 5 − 2 = 3, so 3 × 3 × 3 = 27 cubes.
Check: 8 + 36 + 54 + 27 = 125 ✓

Teacher's Note:
1. For a cube cut into n × n × n pieces the counts are 8, 12(n − 2), 6(n − 2)² and (n − 2)³.
2. Always add the four counts to check that they give n³.

Question 8. 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.
Answer:
Let the edges be the whole numbers l ≤ w ≤ h (in cm). Then
2(lw + wh + hl) = 100, i.e. lw + wh + hl = 50.
Limit the smallest edge: each of lw, wh, hl is at least l², so 3l² ≤ 50, giving l² ≤ 16.6 and l ≤ 4. So only l = 1, 2, 3, 4 need checking.
Factorising trick: adding l² to both sides, lw + wh + hl + l² = (w + l)(h + l), so the condition becomes (w + l)(h + l) = 50 + l².
l = 1: (w + 1)(h + 1) = 51 = 3 × 17 (the pair 1 × 51 would make w = 0). So w = 2, h = 16: cuboid 1 cm × 2 cm × 16 cm.
l = 2: (w + 2)(h + 2) = 54 with 4 ≤ w + 2 ≤ h + 2. The factor pairs of 54 are 1 × 54, 2 × 27, 3 × 18, 6 × 9; only 6 × 9 is allowed, so w = 4, h = 7: cuboid 2 cm × 4 cm × 7 cm.
l = 3: (w + 3)(h + 3) = 59, which is prime, so there is no pair with both factors at least 6. No cuboid.
l = 4: (w + 4)(h + 4) = 66 with 8 ≤ w + 4 ≤ h + 4. Then (w + 4)² ≤ 66 forces w + 4 = 8, but 8 does not divide 66. No cuboid.

Check: 1 × 2 + 2 × 16 + 16 × 1 = 2 + 32 + 16 = 50 and 2 × 4 + 4 × 7 + 7 × 2 = 8 + 28 + 14 = 50, so both have TSA = 100 cm².
(i) Yes, there is more than one such cuboid.
(ii) There are exactly two: 1 cm × 2 cm × 16 cm and 2 cm × 4 cm × 7 cm.
(iii) The smallest edge can only be 1, 2, 3 or 4 (since 3l² ≤ 50), and for each value every factor pair of 50 + l² was tested. Only the two cuboids above work, so the list is complete. (Rearranging the same three edges gives the same cuboid.)

Teacher's Note:
1. Ordering the edges as l ≤ w ≤ h avoids counting the same cuboid twice.
2. The step (w + l)(h + l) = 50 + l² turns a guessing game into a short factor search.