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

Guess: about $\displaystyle 40$ scoops.Assume a scoop is a ball of radius $\displaystyle 2.5$ cm (diameter $\displaystyle 5$ cm) and the whole block can be scooped out.\[V_{\text{container}} = 10 \times 15 \times 20 = 3000 \ \text{cm}^3 \] \[V_{\text{scoop}} = \tfrac{4}{3}\pi (2.5)^3 \approx 65.4 \ \text{cm}^3 \] \[\text{number of scoops} \approx \frac{3000}{65.4} \approx 45.8 \]Only whole scoops count, so 45. A different scoop size gives a different count; any reasonable assumption is valid.Answer: about $\displaystyle 45$ scoops (scoop of radius $\displaystyle 2.5$ cm).

A cube of side \(\displaystyle a\) has \(\displaystyle a^3\) unit cubes and a cube of side \(\displaystyle a+1\) has \(\displaystyle (a+1)^3\).\[(a+1)^3 - a^3 = (a^3 + 3a^2 + 3a + 1) - a^3 \] \[= 3a^2 + 3a + 1 \]The added layer is $\displaystyle 3$ slabs \(\displaystyle a \times a \times 1\), $\displaystyle 3$ rods \(\displaystyle a \times 1 \times 1\) and $\displaystyle 1$ corner cube:\[3(a^2) + 3(a) + 1 = 3a^2 + 3a + 1 \]Check for \(\displaystyle a = 2\):\[3^3 - 2^3 = 19 = 3(4) + 3(2) + 1 \]Answer: \(\displaystyle 3a^2 + 3a + 1\) unit cubes.

Assume a classroom \(\displaystyle 8\ \text{m} \times 6\ \text{m} \times 3\ \text{m}\); other classrooms give different numbers.(i) Guess: yes. Take $\displaystyle 2$ L of water a day for $\displaystyle 70$ years, and \(\displaystyle 1\ \text{m}^3 = 1000\ \text{L}\).\[V_{\text{room}} = 8 \times 6 \times 3 = 144\ \text{m}^3 = 144\,000\ \text{L} \] \[\text{water drunk} \approx 2 \times 365 \times 70 = 51\,100\ \text{L} \] \[51\,100 < 144\,000 \]Even $\displaystyle 3$ L a day for $\displaystyle 80$ years gives only $\displaystyle 87$ $\displaystyle 600$ L.(ii) Walls $\displaystyle 3$ m high and $\displaystyle 20$ cm thick, doors and windows about $\displaystyle 10$ m\(\displaystyle ^2\), brick \(\displaystyle 20 \times 10 \times 10\) cm.\[\text{wall area} = 2(8 + 6) \times 3 - 10 = 74\ \text{m}^2 \] \[\text{wall volume} = 74 \times 0.2 = 14.8\ \text{m}^3 \] \[\text{number of bricks} = \frac{14.8}{0.2 \times 0.1 \times 0.1} = 7400 \]Answer: (i) No; about $\displaystyle 51$ $\displaystyle 000$ L against $\displaystyle 144$ $\displaystyle 000$ L. (ii) About $\displaystyle 7400$ bricks.

A quick guess may favour B, since $\displaystyle 100$ pieces sounds like more. Compare the volumes (same chocolate).\[V_A = 50^3 = 125\,000\ \text{mm}^3 \] \[V_B = 100 \times 10^3 = 100\,000\ \text{mm}^3 \] \[V_A - V_B = 25\,000\ \text{mm}^3 > 0 \] \[\frac{V_A}{V_B} = 1.25 \]Answer: Option A gives more, $\displaystyle 25$% more than B.

Guess: about the area of a large city.Assume $\displaystyle 8$ billion people, each standing in a \(\displaystyle 0.5\ \text{m} \times 0.5\ \text{m}\) space (a tight crowd), and \(\displaystyle 1\ \text{km}^2 = 10^6\ \text{m}^2\).\[\text{area per person} = 0.5 \times 0.5 = 0.25\ \text{m}^2 \] \[A = 8 \times 10^{9} \times 0.25 = 2 \times 10^{9}\ \text{m}^2 = 2000\ \text{km}^2 \] \[\text{side of a square of this area} = \sqrt{2000} \approx 45\ \text{km} \]Allowing \(\displaystyle 0.5\ \text{m}^2\) each doubles this to $\displaystyle 4000$ km\(\displaystyle ^2\); other assumptions are valid.Answer: about $\displaystyle 2000$ km\(\displaystyle ^2\), a square roughly $\displaystyle 45$ km on a side.

Diameter $\displaystyle 7$ cm gives radius \(\displaystyle r = 3.5\) cm. Take \(\displaystyle \pi = \tfrac{22}{7}\).\[V_{\text{glass}} = \pi r^2 h = \tfrac{22}{7} \times (3.5)^2 \times 12 = 462\ \text{cm}^3 \] \[V_{\text{total}} = 1600 \times 462 = 739\,200\ \text{cm}^3 \] \[\frac{739\,200}{1000} = 739.2\ \text{litres} \]Answer: $\displaystyle 739.2$ litres.

\[\text{sphere: } 4\pi r^2 = 4\pi (5)^2 = 100\pi\ \text{cm}^2 \] \[\text{cone: } \pi r l = \pi (4)\, l = 4\pi l \] \[100\pi = 5 \times 4\pi l \] \[l = 5\ \text{cm} \]In the right triangle \(\displaystyle AOB\), with \(\displaystyle AO = h\), \(\displaystyle OB = r\), \(\displaystyle AB = l\):\[h = \sqrt{l^2 - r^2} = \sqrt{25 - 16} = 3\ \text{cm} \] \[V = \tfrac{1}{3}\pi r^2 h = \tfrac{1}{3}\pi (16)(3) = 16\pi\ \text{cm}^3 \approx 50.3\ \text{cm}^3 \]Answer: \(\displaystyle h = 3\) cm and \(\displaystyle V = 16\pi \approx 50.3\ \text{cm}^3\).

Both are spheres, so \(\displaystyle V = \tfrac{4}{3}\pi r^3\) and the \(\displaystyle \tfrac43\pi\) cancels in a ratio:\[\frac{V_1}{V_2} = \frac{\tfrac{4}{3}\pi r_1^3}{\tfrac{4}{3}\pi r_2^3} = \left(\frac{r_1}{r_2}\right)^3 \](i) \[\frac{V_{\text{Jupiter}}}{V_{\text{Earth}}} = \left(\frac{69\,900}{6370}\right)^3 \approx (10.97)^3 \approx 1321 \](ii) \[\frac{V_{\text{Sun}}}{V_{\text{Earth}}} = \left(\frac{695\,700}{6370}\right)^3 \approx (109.2)^3 \approx 1.30 \times 10^{6} \]Answer: (i) about $\displaystyle 1300$ ($\displaystyle 1321$); (ii) about \(\displaystyle 1.3 \times 10^{6}\).

The smallest enclosing cylinder touches the sphere at its base, top and curved surface. So its base radius is \(\displaystyle r\) and its height is the diameter, \(\displaystyle PR = 2r\) (see figure).\[V_{\text{sphere}} = \tfrac{4}{3}\pi r^3 \] \[V_{\text{cylinder}} = \pi r^2 \cdot 2r = 2\pi r^3 \] \[\frac{V_{\text{sphere}}}{V_{\text{cylinder}}} = \frac{\tfrac{4}{3}\pi r^3}{2\pi r^3} = \frac{2}{3} \]Answer: \(\displaystyle V_{\text{sphere}} = \tfrac{2}{3}\,V_{\text{cylinder}}\).

Guess: the bigger the sphere, the smaller the gap.Let \(\displaystyle R\) be the radius and \(\displaystyle h\) the height of the second string; each string's length is the circumference of its circle (\(\displaystyle OA = R\), \(\displaystyle AB = h\) in the figure).\[2\pi R = C \quad \text{(tight string)} \] \[2\pi (R + h) = C + 1 \quad \text{(longer string)} \] \[2\pi h = 1 \ \Rightarrow\ h = \frac{1}{2\pi} \approx 0.159\ \text{m} \]\(\displaystyle R\) has cancelled, so the height is the same for every sphere. For the Earth, \(\displaystyle R \approx 6.37\times10^{6}\) m and \(\displaystyle C \approx 4.00\times10^{7}\) m.(i) Moon. \(\displaystyle R \approx 1.74\times10^{6}\) m: \[C \approx 1.09\times10^{7}\ \text{m}, \qquad h = \frac{(C+1) - C}{2\pi} \approx 0.159\ \text{m} \](ii) Jupiter. \(\displaystyle R \approx 6.99\times10^{7}\) m: \[C \approx 4.39\times10^{8}\ \text{m}, \qquad h = \frac{(C+1) - C}{2\pi} \approx 0.159\ \text{m} \](iii) Volleyball. \(\displaystyle R \approx 0.105\) m: \[C \approx 0.66\ \text{m}, \qquad h = \frac{(C+1) - C}{2\pi} \approx 0.159\ \text{m} \]Surprised? Yes at first: one extra metre on a string $\displaystyle 40,000$ km long lifts it $\displaystyle 16$ cm, exactly as it does on a volleyball. The extra length alone fixes \(\displaystyle h\).Answer: \(\displaystyle h = \dfrac{1}{2\pi}\ \text{m} \approx 16\) cm for the Earth, the Moon, Jupiter and the volleyball alike.