NCERT Solutions · Class 9 Mathematics Math of Space: Surface Area and Volume

Let the side be \(\displaystyle a\) cm.\[a^3 = 64 \Rightarrow a = \sqrt[3]{64} = 4 \]\[\text{TSA} = 6a^2 = 6 \times 4^2 = 96 \]Answer: \(\displaystyle 96\ \text{cm}^2\)

\[2\ \text{m} = 200\ \text{cm} \]\[\text{cubes along one edge} = \frac{200}{20} = 10 \]\[\text{cubes in the box} = 10 \times 10 \times 10 = 1000 \]Check by volumes:\[\frac{200^3}{20^3} = \frac{8\,000\,000}{8000} = 1000 \]Answer: \(\displaystyle 1000\) cubes

The boxes fit exactly along each edge of the godown:\[\frac{40}{2} = 20, \qquad \frac{25}{1.25} = 20, \qquad \frac{10}{1} = 10 \]\[\text{number of boxes} = 20 \times 20 \times 10 = 4000 \]Check by volumes:\[\frac{40 \times 25 \times 10}{2 \times 1.25 \times 1} = \frac{10\,000}{2.5} = 4000 \]Answer: \(\displaystyle 4000\) boxes

Each cube has side \(\displaystyle a\) cm.\[a^3 = 125 \Rightarrow a = 5 \]Joined end to end, the cubes form the cuboid in the figure:\[l = 10,\quad w = 5,\quad h = 5 \]\[\text{TSA} = 2(lw + wh + hl) = 2(10 \cdot 5 + 5 \cdot 5 + 5 \cdot 10) = 2 \times 125 = 250 \]Check: $\displaystyle 12$ faces of area \(\displaystyle 25\), two of them hidden where the cubes touch:\[(12 - 2) \times 25 = 250 \]Answer: \(\displaystyle 250\ \text{cm}^2\)

\[\text{TSA of the original cube} = 6 \times 4^2 = 96 \]\[\text{number of small cubes} = \frac{4^3}{1^3} = 64 \]\[\text{TSA of one small cube} = 6 \times 1^2 = 6 \]\[\text{TSA of all small cubes} = 64 \times 6 = 384 \]\[\text{ratio} = 96 : 384 = 1 : 4 \]Answer: \(\displaystyle 1 : 4\)

Let the edges of the cuboid in the figure be \(\displaystyle l = AB,\ w = BF,\ h = BC\). The three faces at \(\displaystyle C\) have areas \(\displaystyle lw,\ wh,\ lh\) (which is which does not affect the volume).\[lw = 6, \qquad wh = 15, \qquad lh = 10 \]\[(lw)(wh)(lh) = (lwh)^2 = 6 \times 15 \times 10 = 900 \]\[V = lwh = \sqrt{900} = 30 \]Check: \(\displaystyle l = \dfrac{30}{15} = 2,\ w = \dfrac{30}{10} = 3,\ h = \dfrac{30}{6} = 5\), and\[lw = 6, \qquad wh = 15, \qquad lh = 10, \qquad lwh = 30 \]Answer: \(\displaystyle 30\ \text{cm}^3\)

(i) Only the corner cubes have three faces on the outside; a cube has $\displaystyle 8$ corners.\[8 \](ii) Each of the $\displaystyle 12$ edges has \(\displaystyle 5 - 2 = 3\) cubes that are not corners.\[12 \times (5 - 2) = 36 \](iii) Each of the $\displaystyle 6$ faces has a \(\displaystyle (5-2) \times (5-2)\) block of cubes clear of its edges.\[6 \times (5 - 2)^2 = 54 \](iv) The unpainted cubes form the inner \(\displaystyle 3 \times 3 \times 3\) cuboid.\[(5 - 2)^3 = 27 \]Check:\[8 + 36 + 54 + 27 = 125 = 5^3 \]Answer: (i) $\displaystyle 8$, (ii) $\displaystyle 36$, (iii) $\displaystyle 54$, (iv) $\displaystyle 27$

With integer edges \(\displaystyle l \le w \le h\):\[2(lw + wh + hl) = 100 \Rightarrow lw + wh + hl = 50 \](i) Yes. Two examples:\[1 \times 2 \times 16: \quad 2(2 + 32 + 16) = 100 \]\[2 \times 4 \times 7: \quad 2(8 + 28 + 14) = 100 \](ii) Yes, these are the only two; (iii) shows why.(iii) Adding \(\displaystyle l^2\) to both sides factorises the equation:\[(l + w)(l + h) = 50 + l^2 \]Since \(\displaystyle l \le w \le h\), both factors are at least \(\displaystyle 2l\):\[2l \le l + w \le l + h \]Also \(\displaystyle 50 = lw + wh + hl \ge 3l^2\), so \(\displaystyle l \le 4\). Check each \(\displaystyle l\):\[\begin{array}{|c|c|l|c|} \hline l & 50 + l^2 & \text{pairs } p \times q \text{ with } 2l \le p \le q & (w, h) \\ \hline 1 & 51 & 3 \times 17 & (2, 16) \\ \hline 2 & 54 & 6 \times 9 & (4, 7) \\ \hline 3 & 59 & \text{none (59 is prime)} & - \\ \hline 4 & 66 & \text{none (66 = } 6 \times 11 \text{ but } 6 < 8) & - \\ \hline \end{array} \]Answer: (i) yes; (ii), (iii) exactly two cuboids: \(\displaystyle 1 \times 2 \times 16\) and \(\displaystyle 2 \times 4 \times 7\) (cm)