DAV Class 8 Maths Chapter 14 HOTS
Mensuration HOTS
1. In a cylindrical pipe (as shown in the figure), the areas of the surfaces are given. Find the radius and height of the cylinder.
Solution
\[ \begin{aligned} \color{magenta} \text{Area of circle} &= \color{magenta}154 \ sq. units \\[6pt] \pi r^2 &= 154 \\[6pt] \frac{22}{7} \times r^2 &= 154 \\[6pt] r^2&= \cancel{154}^{7} \times \frac{7}{\cancel{22}_1}\\[6pt] r^2 &= 49 \\[6pt] \color{green} r &= \color{green} 7 \text{ units} \\[8pt] \color{magenta} \text{CSA of cylinder} &= \color{magenta} 880 \ sq. units \\[6pt] 2\pi r h &= 880 \\[6pt] 2 \times \frac{22}{\cancel{7}_1}\times \cancel{7}^1 \times h &= 880 \\[6pt] 44 \times h &= 880\\[6pt] h &= \frac{\cancel{880}^{20}}{\cancel{44}_1}\\[6pt] \color{green} h &= \color{green} 20 \text{ units} \end{aligned} \]
Answer Radius \(=\ \color{red}{7 \ units}\) , Height \(=\ \color{red}{20} \ units \)
2. The length of a room is 50% more than its breadth. The cost of carpeting the room at the rate of ₹ 38.50 m² is ₹ 924 and the cost of painting the walls at the rate of ₹ 5.50 per m² is ₹ 1,320. Find the dimensions of the room.
Solution
\[ \begin{aligned} \text{Let breadth } \ b &= x \\ \text{length } \ell &= x + 50\% \text{ of } x \\[6pt] &= x + \frac{\cancel{50}^1}{\cancel{100}_2} \times x \\[6pt] &= x + \frac{x}{2} \\[6pt] &=\frac{2x+x}{2} \\[6pt] \ell &=\frac{3x}{2} \\ \\ \color{magenta} \text{Base area} &= \frac{\text{Total cost of carpeting}}{\text{Cost of carpeting}} \\[6pt] &= \frac{924 \times {\color{green}10}}{38.50 \times {\color{green}10}} \\[6pt] &= \frac{9240}{385} \\[6pt] \color{green} \text{Base area } &= 24\ \text{m}^2 \\[6pt] \color{green} \ell \times b &= 24 \\[6pt] \frac{3}{2}x \times x &= 24 \\[6pt] x^2 &= \cancel{24}^{8} \times \frac{2}{\cancel3_1} \\[6pt] x^2 &= 16 \\[6pt] x &= 4 \\[6pt] \color{green}b &= \color{green}4\ \text{m} \\[6pt] \ell &=\frac{3x}{2} \\[6pt] & = \frac{3}{\cancel2_1} \times \cancel4^2 \\[6pt] \color{green} \ell &= \color{green} 6\ \text{m} \\[10pt] \text{Area of 4 walls} & = \frac{\text{Total cost of painting}}{\text{Cost of painting}} \\[6pt] & = \frac{1320 \times {\color{green}10} }{5.50 \times {\color{green}10}} \\[6pt] & = \frac{13200}{55} \\[6pt] \color{green} \text{Area of 4 walls} & = 240 \ m^2 \\ \\ 2h(\ell+b) & = 240 \\ 2h(6+4) &= 240 \\[6pt] h &=\frac{240}{20} \\[6pt] \color{green}h &=\color{green}12\ \text{m} \end{aligned} \]
Answer Length \(=\ \color{red}{6\ \text{m}}\), Breadth \(=\ \color{red}{4\ \text{m}}\), Height \(=\ \color{red}{12\ \text{m}}\).