Solution

\[ \begin{align*} \text{Number of pillars} &= 15 \\ \text{Diameter } (d) &= 48 \, \text{cm} = 0.48 \, \text{m} \\ \implies \text{Radius } (r) &= \frac{d}{2} = 0.24 \, \text{m} \\ \text{Height } (h) &= 7 \, \text{m} \\ \text{Rate} &= \text{₹}2 \, \text{per sq.m} \\ \\ \text{CSA of one pillar} &= 2\pi rh \\ &= 2 \times \frac{22}{7} \times 0.24 \times 7 \\ &= 44 \times 0.24 \\ &= 10.56 \, \text{m}^2 \\ \\ \text{CSA of 15 pillars} &= 15 \times 10.56 \\ &= 158.40 \, \text{m}^2 \\ \\ \text{Cost of painting} &= \text{Total Area} \times \text{Rate} \\ &= 158.40 \times 2 \\ &= \text{₹}316.80 \end{align*} \]

Answer The cost of painting is \(\color{red} \text{₹}316.80 \).

Solution

\[ \begin{align*} \text{Let side along road } (a) &= x \, \text{m} \\ \implies \text{Side along river } (b) &= 2x \, \text{m} \\ \text{Distance } (h) &= 100 \, \text{m} \\ \text{Area} &= 10,500 \, \text{m}^2 \\ \\ \text{Area of trapezium} &= \frac{1}{2} \times (a + b) \times h \\ 10500 &= \frac{1}{2} \times (x + 2x) \times 100 \\ 10500 &= \frac{1}{\cancel{2}_1} \times 3x \times \cancel{100}^{50} \\ 10500 &= 150x \\ x &= \frac{10500}{150} \\ x &= 70 \, \text{m} \end{align*} \]

Answer Length of the side along the road is \(\color{red} 70 \, \text{m} \).

Solution

\[ \begin{align*} l &= 12 \, \text{m}, \, b = 10 \, \text{m}, \, h = 6 \, \text{m} \\ \text{Area of one can} &= 192 \, \text{m}^2 \\ \\ \text{Area to be painted} &= \text{Area of 4 walls} + \text{Area of ceiling} \\ &= 2h(l + b) + lb \\ &= 2 \times 6(12 + 10) + (12 \times 10) \\ &= 12 \times 22 + 120 \\ &= 264 + 120 \\ &= 384 \, \text{m}^2 \\ \\ \text{Number of cans} &= \frac{\text{Total area}}{\text{Area per can}} \\ &= \frac{384}{192} \\ &= 2 \end{align*} \]

Answer He will need \(\color{red} 2 \) cans of paint.

Solution

\[ \begin{align*} \text{Height } (h) &= 140 \, \text{cm} = 1.4 \, \text{m} \\ \text{Diameter } (d) &= 84 \, \text{cm} \\ \text{Radius } (r) &= 42 \, \text{cm} = 0.42 \, \text{m} \\ \\ \text{Area leveled in 1 rev} &= \text{CSA of cylinder} \\ &= 2\pi rh \\ &= 2 \times \frac{22}{7} \times 0.42 \times 1.4 \\ &= 44 \times 0.06 \times 1.4 \\ &= 3.696 \, \text{m}^2 \\ \\ \text{Area leveled in 1000 revs} &= 1000 \times 3.696 \\ &= 3696 \, \text{m}^2 \\ \\ \text{Cost} &= \frac{3696 \times 50}{100} \\[6pt] &= 36.96 \times 50 \\ &= \text{₹}1848 \end{align*} \]

Answer Cost of levelling the road is \(\color{red} \text{₹}1848 \).

Solution

\[ \begin{align*} l &= 30 \, \text{cm} = 0.3 \, \text{m} \\ b &= 40 \, \text{cm} = 0.4 \, \text{m} \\ h &= 50 \, \text{cm} = 0.5 \, \text{m} \\ \\ \text{Surface Area of 1 tin} &= \text{Area of 4 walls} + \text{Area of base} \\ &= 2h(l + b) + lb \\ &= 2 \times 0.5(0.3 + 0.4) + (0.3 \times 0.4) \\ &= 1.0 \times 0.7 + 0.12 \\ &= 0.82 \, \text{m}^2 \\ \\ \text{Area for 20 tins} &= 20 \times 0.82 \\ &= 16.4 \, \text{m}^2 \\ \\ \text{Cost} &= 16.4 \times 20 \\ &= \text{₹}328 \end{align*} \]

Answer The cost of tin sheet required is \(\color{red} \text{₹}328 \).

Solution

\[ \begin{align*} \text{Volume } (V) &= 748 \, \text{cm}^3 \\ \text{Height } (h) &= 14 \, \text{cm} \\ \text{External Radius } (R) &= 9 \, \text{cm} \\ \text{Let Internal Radius} &= r \\ \\ \text{Volume of pipe} &= \pi(R^2 - r^2)h \\ 748 &= \frac{22}{7} \times (9^2 - r^2) \times 14 \\ 748 &= 22 \times (81 - r^2) \times 2 \\ 748 &= 44 \times (81 - r^2) \\ 81 - r^2 &= \frac{748}{44} \\ 81 - r^2 &= 17 \\ r^2 &= 81 - 17 = 64 \\ r &= 8 \, \text{cm} \\ \\ \text{Thickness} &= R - r \\ &= 9 - 8 \\ &= 1 \, \text{cm} \end{align*} \]

Answer Thickness of the pipe is \(\color{red} 1 \, \text{cm} \).

Solution

\[ \begin{align*} \text{Height } (h) &= 70 \, \text{cm} \\ \text{Outer Radius } (R) &= 14 \, \text{cm} \\ \text{Thickness} &= 2 \, \text{cm} \\ \implies \text{Inner Radius } (r) &= R - \text{thickness} \\ &= 14 - 2 \\ &= 12 \, \text{cm} \\ \\ \text{Volume of metal} &= \pi(R^2 - r^2)h \\ &= \frac{22}{7} \times (14^2 - 12^2) \times 70 \\ &= 22 \times (196 - 144) \times 10 \\ &= 22 \times 52 \times 10 \\ &= 11,440 \, \text{cm}^3 \\ \\ \text{Weight} &= \text{Volume} \times \text{Density} \\ &= 11440 \times 8 \\ &= 91,520 \, \text{gm} \\ &= 91.52 \, \text{kg} \end{align*} \]

Answer Volume is \(\color{red} 11,440 \, \text{cm}^3 \) and weight is \(\color{red} 91.52 \, \text{kg} \).

Solution

\[ \begin{align*} \text{Area} &= 168 \, \text{m}^2 \\ a &= 36 \, \text{m} \\ b &= 20 \, \text{m} \\[6pt] \color{green} \text{Area of the trapezium} &= \frac{1}{2} \times (a + b) \times h \\ 168 &= \frac{1}{2} \times (36 + 20) \times h \\ 168 &= \frac{1}{\cancel{2}_1} \times \cancel{56}^{28} \times h \\ 168 &= 28h \\ h &= \frac{168}{28} \\[6pt] h &= 6 \, \text{m} \\ \\ \text{Draw } & DF \parallel CB \\ \implies DFBC & \text{ is a parallelogram} \\ \text{In } &\triangle ADF \\ DA = DF & = x \\ AB &= 36 \, \text{m} \\ AF &= 36 - 20 = 16 \, \text{m} \\ \\ \text{DE } \perp \text{ AF} \implies & \text{E is the midpoint of AF} \\ EF &= \frac{AF}{2} = \frac{16}{2} = 8 \, \text{m} \\ \\ \text{In } \triangle DEF & \text{ (right angled triangle)} \\ (DE)^2 + (EF)^2 &= (DF)^2 \\ 6^2 + 8^2 &= x^2 \\ 36 + 64 &= x^2 \\ x^2 &= 100 \\ x &= \sqrt{100} \\ x &= 10 \, \text{m} \end{align*} \]

Answer Length of each non-parallel side is \(\color{red} 10 \, \text{m} \).

Solution

\[ \begin{align*} \text{Volume of cuboid} &= 4.4 \times 2.6 \times 1 \\ &= 11.44 \, \text{m}^3 \\ & \implies 11,440,000 \, \text{cm}^3 \\ \\ \text{Internal radius } (r) &= 30 \, \text{cm} \\ \text{Thickness} &= 5 \, \text{cm} \\ \implies \text{External radius } (R) &= 30 + 5 = 35 \, \text{cm} \\ \\ \text{Volume of hollow pipe} &= \pi \times h \times (R^2 - r^2) \\[6pt] 11440000 &= \frac{22}{7} \times (35^2 - 30^2) \times h \\[6pt] 11440000 &= \frac{22}{7} \times (1225 - 900) \times h \\[6pt] 11440000 &= \frac{22}{7} \times 325 \times h \\[6pt] h &= \frac{11440000 \times 7}{22 \times 325} \\[6pt] h &= \frac{520000 \times 7}{325} \\[6pt] h &= 1600 \times 7 \\ &= 11,200 \, \text{cm} \\[6pt] h &= 112 \, \text{m} \end{align*} \]

Answer Length of the pipe is \(\color{red} 112 \, \text{m} \).

Solution

\[ \begin{align*} \text{Draw } & DF \parallel CB \\ \implies DFBC & \text{ is a parallelogram} \\ \text{In } &\triangle ADF \\ DA = DF & = 13 \, \text{cm} \\ AB &= 20 \, \text{cm} \\ AF &= 20 - 10 \implies 10 \, \text{cm} \\ \\ DE \perp AF \implies & \text{E is the midpoint of AF} \\ \\ EF = \frac{AF}{2} &= \frac{10}{2} = 5 \, \text{cm} \\ \\ \text{In } \triangle DEF & \text{ (right angled triangle)} \\ (DE)^2 + (EF)^2 &= (DF)^2 \\ (DE)^2 + 5^2 &= 13^2 \\ (DE)^2 + 25 &= 169 \\ (DE)^2 &= 169 - 25 \\ (DE)^2 &= 144 \\ DE &= \sqrt{144} \\ DE &= 12 \, \text{cm} \\ \\ \text{In trapezium } & ABCD \\ a &= 20 \, \text{cm} \\ b &= 10 \, \text{cm} \\ h &= 12 \, \text{cm} \\ \\ \color{green} \text{Area of the trapezium} &= \frac{1}{2} \times (a + b) \times h \\ &= \frac{1}{2} \times (20 + 10) \times 12 \\ &= \frac{1}{\cancel{2}_1} \times 30 \times \cancel{12}^6 \\ &= 180 \, \text{cm}^2 \end{align*} \]

Answer Area of the trapezium is \(\color{red} 180 \, \text{cm}^2 \).

Solution

\[ \begin{align*} \text{Perimeter of base } &= 14 \, \text{cm} \\ 2(l+b) &= 14 \, \text{cm} \\ \text{LSA} &= 231 \, \text{cm}^2 \\ \\ \text{LSA of cuboid} &= 2h(l + b) \\ 231&= h \times 2(l+b) \\ 231 &= h \times 14 \\[6pt] h &= \frac{231}{14} \\[6pt] h &= 16.5 \, \text{cm} \end{align*} \]

Answer Height of the cuboid is \(\color{red} 16.5 \, \text{cm} \).

Solution

\[ \begin{align*} \text{Volume } (V) &= 150\pi \, \text{cm}^3 \\ \text{Height } (h) &= 6 \, \text{cm} \\ \\ \text{Volume} &= \pi r^2 h \\ 150\pi &= \pi r^2 \times 6 \\ r^2 &= \frac{150}{6} \\ r^2 &= 25 \\ r &= 5 \, \text{cm} \\ \\ \text{CSA} &= 2\pi rh \\ &= 2 \times \pi \times 5 \times 6 \\ &= 60\pi \, \text{cm}^2 \\ \\ \text{TSA} &= 2\pi r(r + h) \\ &= 2 \times \pi \times 5 \times (5 + 6) \\ &= 10\pi \times 11 \\ &= 110\pi \, \text{cm}^2 \end{align*} \]

Answer CSA is \(\color{red} 60\pi \, \text{cm}^2 \) and TSA is \(\color{red} 110\pi \, \text{cm}^2 \).