NCERT Solutions for Class 9 Maths Chapter 14 — Exercise Set 14.1#

Cube of side $a$: surface area $6a^2$, volume $a^3$.

Cuboid $l \times w \times h$: surface area $2(lw + wh + hl)$, volume $lwh$.

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

Solution
$$a^3 = 64$$

$$\Rightarrow a = 4 \text{ cm}$$$$\text{TSA} = 6a^2$$

$$\text{TSA} = 6 \times 16$$

$$\text{TSA} = 96 \text{ cm}^2$$

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

Solution

$2 \text{ m} = 200 \text{ cm}$, so along each edge of the box we can fit $\dfrac{200}{20} = 10$ small cubes.

$$\text{Number of cubes} = 10 \times 10 \times 10$$

$$\text{Number of cubes} = 1000$$

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. #

Solution
$$\text{Number of boxes} = \dfrac{40 \times 25 \times 10}{2 \times 1.25 \times 1}$$

$$\text{Number of boxes} = \dfrac{10000}{2.5}$$

$$\text{Number of boxes} = 4000$$

(The boxes fit exactly: 20 along 40 m, 20 along 25 m and 10 along 10 m.)

4. Two cubes each of volume 125 cm³ are joined end to end. Find the surface area of the resulting cuboid. #

Solution

Each cube has side $\sqrt[3]{125} = 5$ cm. Joined end to end, they form a cuboid $10 \times 5 \times 5$ cm:

$$\text{SA} = 2(10 \times 5 + 5 \times 5 + 5 \times 10)$$

$$\text{SA} = 2(50 + 25 + 50)$$

$$\text{SA} = 250 \text{ cm}^2$$

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? #

Solution
$$\text{Original cube} = 6 \times 4^2 = 96 \text{ cm}^2$$

There are $4^3 = 64$ small cubes, each with area $6 \times 1^2 = 6$ cm²:

$$\text{Small cubes} = 64 \times 6 = 384 \text{ cm}^2$$$$\text{Ratio} = 96 : 384 = 1 : 4$$

The volume is the same (64 cm³), but cutting into smaller pieces makes the surface area 4 times larger. (This is why small animals lose heat faster, and why crushed food dissolves faster.)

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? #

Solution

Let the edges be $l$, $w$, $h$, with $lw = 6$, $wh = 15$, $hl = 10$. Multiplying the three:

$$(lw)(wh)(hl) = 6 \times 15 \times 10$$

$$\Rightarrow (lwh)^2 = 900$$

$$\Rightarrow lwh = 30$$

∴ The volume is 30 cm³. (The edges are 2 cm, 3 cm and 5 cm.)

*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? #

Solution

The cube is cut into $5 \times 5 \times 5 = 125$ small cubes.

Painted faces Where they are Number
(i) Three the corners 8
(ii) Two along the 12 edges, not at corners: $12 \times (5 - 2)$ 36
(iii) One in the middle of each face: $6 \times (5 - 2)^2$ 54
(iv) None the inside cube $(5 - 2)^3$ 27

Check: $8 + 36 + 54 + 27 = 125$ ✓

*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. #

Solution

Let the edges be integers $a \leq b \leq c$:

$$2(ab + bc + ca) = 100$$

$$\Rightarrow ab + bc + ca = 50$$

(i) Yes, there is more than one. (ii) There are exactly two:

Edges $ab + bc + ca$
$1 \times 2 \times 16$ $2 + 32 + 16 = 50$ ✓
$2 \times 4 \times 7$ $8 + 28 + 14 = 50$ ✓

(iii) Why these are all. Since $a$ is the smallest, $ab + bc + ca \geq 3a^2$, so $3a^2 \leq 50$ and $a \leq 4$. We check each $a$, writing the condition as $a(b + c) + bc = 50$, that is, $(b + a)(c + a) = 50 + a^2$:

$a$ $(b + a)(c + a)$ Factor pairs with $a \leq b \leq c$ Cuboid
1 51 $3 \times 17$ $1, 2, 16$
2 54 $6 \times 9$ $2, 4, 7$
3 59 59 is prime — none with $b \geq 3$ —
4 66 $6 \times 11$ needs $b = 2 < 4$; none with $b \geq 4$ —

So the only cuboids are $1 \times 2 \times 16$ and $2 \times 4 \times 7$.