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

The pyramid's square base \(\displaystyle ABCD\) has its corners on the base circle of the cylinder (top view). Its apex is on the top face, so its height is \(\displaystyle h\).The diagonals are diameters, \(\displaystyle OA = OB = r\), and they meet at right angles at \(\displaystyle O\).\[AB^2 = OA^2 + OB^2 = r^2 + r^2 = 2r^2 \] \[\text{base area} = AB^2 = 2r^2 \] \[V_{\text{pyramid}} = \tfrac{1}{3} \times 2r^2 \times h = \tfrac{2}{3} r^2 h \] \[V_{\text{cylinder}} = \pi r^2 h \] \[\frac{V_{\text{pyramid}}}{V_{\text{cylinder}}} = \frac{\tfrac{2}{3} r^2 h}{\pi r^2 h} = \frac{2}{3\pi} \approx 0.21 \]Answer: \(\displaystyle V_{\text{pyramid}} : V_{\text{cylinder}} = 2 : 3\pi\).

(i) New volume minus old volume: \[(l+1)\,wh - lwh = wh \] The volume increases by \(\displaystyle wh\) cubic units: option (c).(ii) \[\pi (r+1)^2 h - \pi r^2 h = \pi h\,(2r + 1) = 2\pi r h + \pi h \] The volume increases by \(\displaystyle 2\pi r h + \pi h\) cubic units: option (e).(iii) The radius \(\displaystyle r\) becomes \(\displaystyle r - 1\) (with \(\displaystyle r \ge 1\)): \[\tfrac{4}{3}\pi (r-1)^3 - \tfrac{4}{3}\pi r^3 = \tfrac{4}{3}\pi\,(-3r^2 + 3r - 1) \] \[= -\left(4\pi r^2 - 4\pi r + \tfrac{4\pi}{3}\right) \] The volume decreases by \(\displaystyle 4\pi r^2 - 4\pi r + \tfrac{4\pi}{3}\) cubic units.Answer: (i) (c) \(\displaystyle wh\); (ii) (e) \(\displaystyle 2\pi r h + \pi h\); (iii) a decrease of \(\displaystyle 4\pi r^2 - 4\pi r + \tfrac{4\pi}{3}\) cubic units.

Let the edge of the cube be \(\displaystyle a\).\[V = a^3 \ \Rightarrow\ a = V^{1/3} \] \[S = 6a^2 = 6\left(V^{1/3}\right)^2 = 6V^{2/3} \]Check with \(\displaystyle a = 2\): \(\displaystyle V = 8\) and \(\displaystyle 6\cdot 8^{2/3} = 6 \cdot 4 = 24 = 6 \cdot 2^2\).Answer: \(\displaystyle S = 6V^{2/3}\).

Let the edge of the cube be \(\displaystyle a\).\[S = 6a^2 \ \Rightarrow\ a = \sqrt{\frac{S}{6}} \] \[V = a^3 = \left(\frac{S}{6}\right)^{3/2} = \frac{S\sqrt{S}}{6\sqrt{6}} \]Check with \(\displaystyle S = 24\): \(\displaystyle \left(\tfrac{24}{6}\right)^{3/2} = 4^{3/2} = 8 = 2^3\).Answer: \(\displaystyle V = \left(\dfrac{S}{6}\right)^{3/2}\).

Guess: a few dozen marbles.The marbles sink, so the water rises from \(\displaystyle EF\) to \(\displaystyle GH\), a rise of \(\displaystyle 20 - 16 = 4\) cm. The volume of this extra layer equals the volume of the marbles.\[V_{\text{rise}} = \pi (4)^2 (4) = 64\pi\ \text{cm}^3 \] \[V_{\text{marble}} = \tfrac{4}{3}\pi (1)^3 = \tfrac{4\pi}{3}\ \text{cm}^3 \] \[n = \frac{64\pi}{4\pi/3} = 48 \]The water at $\displaystyle 20$ cm stays below the rim at $\displaystyle 25$ cm.Answer: $\displaystyle 48$ marbles.

Guess: well under $\displaystyle 1$ mL.Treat the ink in the refill as a thin cylinder. Measured with a ruler on a transparent refill: ink column length \(\displaystyle h \approx 10\) cm, inner diameter \(\displaystyle \approx 2\) mm, so \(\displaystyle r \approx 0.1\) cm.\[V = \pi r^2 h \] \[V \approx 3.14 \times (0.1)^2 \times 10 = 0.314\ \text{cm}^3 \] \[0.314\ \text{cm}^3 \approx 0.3\ \text{mL} \quad (1000\ \text{cm}^3 = 1\ \text{litre}) \]The ball at the tip is ignored. Other pens and measurements give somewhat different values; any estimate from measured \(\displaystyle r\) and \(\displaystyle h\) is valid.Answer: about \(\displaystyle 0.3\ \text{cm}^3\) ($\displaystyle 0.3$ mL) from \(\displaystyle V = \pi r^2 h\), as one worked example.

(i) Guess: about 20. Model the ball as a sphere and a chapati as a thin disc, diameter $\displaystyle 15$ cm and thickness $\displaystyle 2$ mm.\[V_{\text{ball}} = \tfrac{4}{3}\pi (6)^3 = 288\pi \ \text{cm}^3 \] \[V_{\text{chapati}} = \pi (7.5)^2 (0.2) = 11.25\pi \ \text{cm}^3 \] \[\frac{288\pi}{11.25\pi} = 25.6 \]Round down: a 26th chapati would fall short.(ii) Guess: about $\displaystyle 400$ cm\(\displaystyle ^3\). Suppose the cut face of a coconut shows outer radius \(\displaystyle R = 6\) cm and flesh $\displaystyle 1$ cm thick, so \(\displaystyle r = 5\) cm. Each half's flesh is a hemispherical shell.\[V_{\text{half}} = \tfrac{2}{3}\pi R^3 - \tfrac{2}{3}\pi r^3 = \tfrac{2}{3}\pi (216 - 125) = \tfrac{182\pi}{3} \approx 190.6 \ \text{cm}^3 \] \[V_{\text{coconut}} = 2 \times 190.6 \approx 381 \ \text{cm}^3 \]Other measurements give other valid answers. For a muskmelon, \(\displaystyle r\) is the seed cavity's radius.Answer: (i) about $\displaystyle 25$ chapatis; (ii) about $\displaystyle 380$ cm\(\displaystyle ^3\) of flesh for the coconut measured above.

(i) Guess: no. Take $\displaystyle 100$ sheets ($\displaystyle 200$ pages) of $\displaystyle 20$ cm \(\displaystyle \times\) $\displaystyle 27$ cm and a classroom $\displaystyle 8$ m \(\displaystyle \times\) $\displaystyle 6$ m.\[A_{\text{sheets}} = 100 \times 20 \times 27 = 54\,000 \ \text{cm}^2 = 5.4 \ \text{m}^2 \] \[A_{\text{floor}} = 8 \times 6 = 48 \ \text{m}^2 \] \[\frac{5.4}{48} \approx \frac{1}{9} \]Counting each of the $\displaystyle 200$ pages separately:\[200 \times 540 = 108\,000 \ \text{cm}^2 = 10.8 \ \text{m}^2 < 48 \ \text{m}^2 \]Other page sizes give other values, but nowhere near 48.(ii) Guess: about $\displaystyle 5$ lakh. Book: $\displaystyle 20$ \(\displaystyle \times\) $\displaystyle 27$ \(\displaystyle \times\) $\displaystyle 1$ cm; with $\displaystyle 1$ ft \(\displaystyle \approx\) $\displaystyle 30$ cm the room is $\displaystyle 450$ \(\displaystyle \times\) $\displaystyle 600$ \(\displaystyle \times\) $\displaystyle 900$ cm. Volume bound:\[\frac{450 \times 600 \times 900}{20 \times 27 \times 1} = \frac{2.43 \times 10^{8}}{540} = 450\,000 \]Stack the $\displaystyle 1$, $\displaystyle 20$, $\displaystyle 27$ cm edges along $\displaystyle 450$, $\displaystyle 600$, $\displaystyle 900$ cm:\[\frac{450}{1} = 450, \qquad \frac{600}{20} = 30, \qquad \left\lfloor \frac{900}{27} \right\rfloor = 33 \] \[450 \times 30 \times 33 = 445\,500 \]Other book sizes give other values.Answer: (i) No: $\displaystyle 5.4$ m\(\displaystyle ^2\) of paper ($\displaystyle 10.8$ m\(\displaystyle ^2\) even counting each page separately) against a $\displaystyle 48$ m\(\displaystyle ^2\) floor. (ii) About $\displaystyle 4.5$ lakh books ($\displaystyle 445,500$ stacked; never more than $\displaystyle 450,000$).

Guess: about $\displaystyle 1$ km. Take \(\displaystyle 8 \times 10^{9}\) people, each of mass $\displaystyle 60$ kg and about the density of water (\(\displaystyle 1000\ \text{kg/m}^3\)), packed without gaps.\[v = \frac{60}{1000} = 0.06 \ \text{m}^3 \] \[V = 8 \times 10^{9} \times 0.06 = 4.8 \times 10^{8} \ \text{m}^3 \] \[s = \sqrt[3]{4.8 \times 10^{8}} \approx 783 \ \text{m} \]With standing room of \(\displaystyle 0.5\ \text{m}^3\) each:\[s = \sqrt[3]{8 \times 10^{9} \times 0.5} \approx 1587 \ \text{m} \]Answer: about $\displaystyle 0.8$ km; about $\displaystyle 1.6$ km with standing room.

(i) Guess: a few km. The water is a shell between spheres of radius \(\displaystyle R\) and \(\displaystyle R + t\), with \(\displaystyle V = 1.38 \times 10^{9}\ \text{km}^3\).\[V = \tfrac{4}{3}\pi (R + t)^3 - \tfrac{4}{3}\pi R^3 \] \[(R + t)^3 = R^3 + \frac{3V}{4\pi} \] \[t = \sqrt[3]{R^3 + \frac{3V}{4\pi}} - R \](ii) With \(\displaystyle R = 6371\) km:\[R^3 \approx 2.5860 \times 10^{11}, \qquad \frac{3V}{4\pi} \approx 3.2945 \times 10^{8} \] \[t = \sqrt[3]{2.58926 \times 10^{11}} - 6371 \approx 6373.70 - 6371 \approx 2.7 \ \text{km} \]Check: \(\displaystyle t\) is tiny beside \(\displaystyle R\), so \(\displaystyle t \approx \dfrac{V}{4\pi R^2} = \dfrac{1.38 \times 10^{9}}{5.101 \times 10^{8}} \approx 2.7\ \text{km}\).Answer: (i) \(\displaystyle t = \sqrt[3]{R^3 + \tfrac{3V}{4\pi}} - R\); (ii) about $\displaystyle 2.7$ km.

Guess: a thick block reshaped into a flat slab, since a slab has much surface for little volume. Take block A and slab B (in cm):\[V_A = 8 \times 14 \times 14 = 1568, \qquad S_A = 2(112 + 196 + 112) = 840 \] \[V_B = 1 \times 28 \times 28 = 784, \qquad S_B = 2(28 + 784 + 28) = 1680 \] \[V_B = \tfrac{1}{2} V_A, \qquad S_B = 2 S_A \]Other pairs work, e.g. \(\displaystyle 8 \times 12 \times 16\) and \(\displaystyle 1 \times 16 \times 48\); multiplying every edge of both by the same \(\displaystyle k\) keeps the ratios.Answer: \(\displaystyle 8 \times 14 \times 14\) cm, reshaped to \(\displaystyle 1 \times 28 \times 28\) cm.

Treat each room as a cuboid, take the ceiling height as $\displaystyle 3$ m and ignore wall thickness. Measure each room's floor and add the volumes. Sample plan (metres):\[\begin{array}{|l|c|c|c|c|} \hline \text{Room} & l & w & h & V = lwh \ (\text{m}^3) \\ \hline \text{Living room} & 5 & 4 & 3 & 60 \\ \text{Bedroom 1} & 4 & 4 & 3 & 48 \\ \text{Bedroom 2} & 3.5 & 3 & 3 & 31.5 \\ \text{Kitchen} & 3 & 3 & 3 & 27 \\ \text{Baths and passage} & 2.5 & 3 & 3 & 22.5 \\ \hline & & & \text{Total} & 189 \\ \hline \end{array} \]Check with the whole house as one box:\[9 \times 7 \times 3 = 189 \ \text{m}^3 = 1\,89\,000 \ \text{litres} \]Another house gives other numbers; the method is the same.Answer: for the sample house, about \(\displaystyle 189\ \text{m}^3\) ($\displaystyle 1,89,000$ litres).