DAV Class 8 Maths Chapter 14 Worksheet 6
Mensuration Worksheet 6
1. Find the volume of a right circular cylinder if
(i) Radius of its base = 7 cm, height = \( 1.5 \) dm
Solution
\[ \begin{align*} 1 \ dm & = 10 \ cm \\ 1.5 \ dm & = 15 \ cm \\ \\ \text{Height } (h) &= 15 \, \text{cm} \\ \text{Radius } (r) &= 7 \, \text{cm} \\ \\ \color{green} \text{Volume of cylinder} &= \color{green} \pi r^2 h \\[6pt] &= \frac{22}{\cancel{7}_{1}} \times \cancel{7}^{1} \times 7 \times 15 \\[6pt] &= 22 \times 105 \\ &= 2310 \, \text{cm}^3 \end{align*} \]
Answer Volume of the cylinder \( = \color{red} 2310 \, \text{cm}^3 \)
(ii) Area of base = 154 cm2, height = \( 1.5 \) cm
Solution
\[ \begin{align*} \text{Area of base } ( \pi r^2) &= 154 \ cm^2 \\ \text{Height } (h) &= 1.5 \ cm \\ \\ \color{green} \text{Volume of cylinder} &= \color{green} \pi r^2 h \\[6pt] &= 154 \times 1.5 \\ &= 231 \, \text{cm}^3 \end{align*} \]
Answer Volume of the cylinder \( = \color{red} 231 \, \text{cm}^3 \)
(iii) Circumference of base = \( 1.32 \) m, height = 25 cm
Solution
\[ \begin{align*} 1 \ m & = 100 \ cm \\ 1.32 \ m & = 132 \ cm \\ \\ \text{Circumference} &= 132 \ cm \\ 2 \pi r &= 132 \ cm \\[6pt] 2 \times \frac{22}{7} \times r &= 132 \\[6pt] r &= \frac{\cancel{132}^6 \times 7}{2 \times \cancel{22}_1} \\[6pt] r &= \frac{42}{2} \\[6pt] r &= 21 \ cm \\ \text{Height } (h) &= 25 \ cm \\ \\ \color{green} \text{Volume of cylinder} &= \color{green} \pi r^2 h \\[6pt] &= \frac{22}{\cancel{7}_{1}} \times \cancel{21}^{3} \times 21 \times 25 \\[6pt] &= 66 \times 21 \times 25 \\ &= 34650 \, \text{cm}^3 \end{align*} \]
Answer Volume of the cylinder \( = \color{red} 34650 \, \text{cm}^3 \)
2. How many cubic metres of earth must be dug out to sink a well 16 m in depth and 7 m in diameter?
Solution
\[ \begin{aligned} \text{Height } (h) &= 16 \ \text{m} \\ \text{Diameter} &= 7 \ \text{m} \\ r &= \frac{7}{2} \text{m} \\[6pt] \color{green}\text{Volume of the cylinder} &= \color{green}\pi r^2 h \\[4pt] &= \frac{\cancel{22}^{11}}{\cancel7_1} \times \frac{\cancel7^1}{\cancel2_1} \times \frac{7}{\cancel2_1} \times \cancel{16}^8 \\[4pt] &= 77 \times 8 \\ &= 616 \ \text{m}^3 \end{aligned} \]
Answer Earth to be dug out \(=\ \color{red}{616 \ m^3}\)
3. A water tank is cylindrical in shape and the diameter of its base is 28 m. If it is 7 metres deep, how many kilolitres of water can it hold?
Solution
\[ \begin{aligned} d &= 28 \ m \\[6pt] r &= \frac{28}{2} \implies 14 \ m \\[6pt] \text{Height }(h) &= 7 \ m \\[6pt] \color{green}\text{Volume of tank} &= \color{green}\pi r^2 h \\[6pt] &= \frac{22}{\cancel{7}_{1}} \times 14 \times 14 \times \cancel{7}^{\,1} \\[6pt] &= 22 \times 196 \\[4pt] &= 4312 \ m^3 \\[8pt] 1 \ m^3 & = 1 \ kl \\ \text{Capacity} & = 4312 \ kl \end{aligned} \]
Answer The tank can hold \( \color{red}{4312 \ kilolitres} \) of water
4. The radius of the base of a cylindrical oil can is 4 m. Find its height if it can contain 1408 kilolitres of oil.
Solution
\[ \begin{aligned} 1 \ kl &= 1 \ m^3 \\ 1408 \ kl & = 1408 \ m^3 \\ \\ r &= 4 \ m \\ Volume &= 1408 \ m^3 \\ \\ \pi r^2 h &= 1408 \\ \frac{22}{7} \times 4^2 \times h & = 1408 \\ h &= \frac{\cancel{1408}^{ \cancel{\ 88}^{ \ 4}} \times 7}{\cancel{22}_{1} \times \cancel{16}_{1}} \\[6pt] h & = 4 \times 7 \\ h & = 28 \ m \end{aligned} \]
Answer Height of the oil can \( = \color{red}{28 \ m}\)
5. The radius and height of a cylinder are in the ratio 5 : 7 and its volume is 550 cm3. Find its radius.
Solution
\[ \begin{aligned} r : h &= 5 : 7 \\[6pt] \text{Let }r &= 5x \\[2pt] h &= 7x \\[6pt] \color{green}V &= \color{green} \pi r^2 h \\[6pt] Volume &= 550 \ cm^3 \\ \pi r^2 h &= 550 \\ \frac{22}{7} \times (5x)^2 \times (7x) &= 550 \\[6pt] \frac{22}{\cancel7_1} \times 25x^2 \times \cancel7^1x &= 550 \\[6pt] 22 \times 25x^2 \times x &= 550 \\[6pt] 550 x^3 &= 550 \\[6pt] x^3 &= \frac{550}{550} \\[6pt] x^3 &= 1 \\ x&= 1 \\ r &= 5x \\ r &= 5 \ cm \end{aligned} \]
Answer The radius of the cylinder \( = \color{red}{5 \ cm} \)
6. The thickness of a hollow metallic cylinder is 2 cm. It is 35 cm long and its inner radius is 12 cm. Find the volume of the metal required to make the cylinder, assuming it is open at either end.
Solution
\[ \begin{aligned} \text{Thickness }(t) &= 2 \ cm \\ \text{Inner radius }(r) &= 12 \ cm \\ \text{Outer radius }(R) &= 12 + 2 \implies 14 \ cm \\ \text{Height }(h) &= 35 \ cm \\[6pt] \color{green} \text{Volume of metal} &= \color{green} \text{Volume of (outer cylinder - inner cylinder)} \\[6pt] &= \pi R^2 h - \pi r^2h \\[6pt] &= \pi h(R^2 - r^2) \\[6pt] &= \frac{22}{7} \times 35 \times (14^2 - 12^2)\\[6pt] &= \frac{22}{\cancel7_1} \times \cancel{35}^5 \times (196 - 144)\\[6pt] &= 110 \times 52 \\ &= 5720 \ cm^3 \end{aligned} \]
Answer Volume of metal required \(= \color{red}{5720 \ cm^3}\)
7. A rectangular sheet of paper is rolled along its length to make a cylinder. The sheet is 33 cm long and 32 cm wide. A circular sheet of paper is attached to the bottom of the cylinder formed. Find the capacity of the cylinder so formed.
Solution
\[ \begin{aligned} \text{Circumference} & = \text{Length of sheet} \\[2pt] 2\pi r &= 33 \\[6pt] 2 \times \frac{22}{7} \times r &= 33 \\[6pt] \frac{44}{7} \times r &= 33 \\[6pt] r &= \frac{\cancel{33}^{3} \times 7}{\cancel{44}_4} \\[6pt] r &= \frac{21}{4} \ cm \\[6pt] h &= 32 \ cm \\[6pt] \color{green}\text{Volume} &= \color{green}\pi r^2 h \\[6pt] &= \frac{22}{\cancel7_1} \times \frac{\cancel{21}^{3}}{\cancel4_1} \times \frac{21}{\cancel4_1} \times \cancel{32}^{ \ \cancel8^{ \ 2}} \\[6pt] &= 66 \times 42 \\[6pt] &= 2772 \ cm^3 \\ \\ 1000 \ cm^3 &= 1 \ L \\ & \implies 2.772 \ L \end{aligned} \]
Answer Capacity of the cylinder \(= \color{red}{2772 \ cm^3} \text{ or } 2.772 \ L \)
8. Two solids of right cylindrical shape are 49 cm and 35 cm high and their base diameters are 16 cm and 14 cm respectively. Both are melted and moulded into a single cylinder, 56 cm high. Find its base diameter.
Solution
\[ \begin{aligned} & \color{magenta} \text{Cylinder - } 1 \\ d_1 &= 16 \ \text{cm} \\ r_1 &= 8 \ \text{cm} \\ h_1 &= 49 \ \text{cm} \\ & \color{magenta} \text{Cylinder - } 2 \\ d_2 &= 14 \ \text{cm} \\ r_2 &= 7 \ \text{cm} \\ h_2 &= 35 \ \text{cm} \\ \\ \color{green} \text{New volume} &= \color{green} \text{Volume of (Cylinder 1 + Cylinder 2 )} \\[6pt] & = \color{green}\pi r_1^2 h_1 + \pi r_2^2 h_2 \\[6pt] &= \left(\frac{22}{7} \times 8^2 \times 49\right) + \left( \frac{22}{7} \times 7^2 \times 35 \right) \\[6pt] &= \left(\frac{22}{7} \times 64 \times 49\right) + \left( \frac{22}{7} \times 49 \times 35 \right) \\[6pt] &= \frac{22}{\cancel7_1} \times \cancel{49}^7 \times (64+35) \\[6pt] &= 22 \times 7 \times 99 \\ &=15246 \ cm^3 \\[6pt] \color{green} \text{New Volume} &= \color{green} 15246 \ cm^3 \\ \text{New Height} &= 56 \ cm \\ \text{Let }R &= \text{new radius} \\[6pt] \pi R^2 H &= 15246 \\[6pt] \frac{22}{\cancel7_1} \times R^2 \times \cancel{56}^8 &= 15246 \\ R^2 &= \frac{\cancel{15246}^{ \ 693}}{\cancel{22}_1 \times 8} \\[6pt] R^2 &= \frac{693}{8} \\[6pt] R^2 &= 86.625 \\[6pt] R &= \sqrt{86.625} \\[6pt] R &= 9.3 \ cm \\[6pt] D &= 2R \\ &= 2 \times 9.3 \\ D & = 18.6 \ cm \end{aligned} \]
Answer The base diameter of the new cylinder \(=\ \color{red}{18.6 \ cm } \)
9. The volume of a metallic cylindrical pipe is 748 cm3. Its length is 14 cm and its external radius is 9 cm. Find its thickness.
Solution
Let \( r \) be internal radius and \( R \) be external radius. \[ \begin{aligned} h &= 14 \ cm \\ R & = 9 \ cm \\ \text{Volume of a metal pipe} &= 748 \ cm^3 \\[6pt] \color{green}\text{Volume of metal pipe} &= \color{green}\pi h\left(R^2 - r^2\right) \\[6pt] \frac{22}{\cancel{7}_1}\times \cancel{14}^{\,2} \times \left(9^2 - r^2\right) & = 748 \\[6pt] 44 \times \left(81 - r^2\right) & = 748 \\ 81 - r^2 &= \frac{\cancel{748}^{17}}{\cancel{44}_1} \\[6pt] 81 - r^2 &= 17 \\ 81 - 17 &= r^2 \\ 64 &= r^2 \\ r^2 &= 64 \\ r &= \sqrt{64} \\ r &= 8\ \text{cm} \\[8pt] \text{Thickness} &= R - r \\ &= 9 - 8 \\ \text{Thickness} &= 1\ \text{cm} \end{aligned} \]
Answer The thickness of the pipe \(=\ \color{red}{1 \ cm }\)
10. The volume of a cylinder is \(150\pi\) cm3 and its height is 6 cm. Find the areas of its total surface and lateral (curved) surface.
Solution
\[ \begin{aligned} \text{Volume }(V) &= 150\pi \ \text{cm}^3 \\ h &= 6 \ \text{cm} \\[6pt] \color{green}V &= \color{green}\pi r^2 h \\[6pt] \pi r^2h &= 150\pi \\[6pt] \pi \times r^2 \times 6 &= 150 \times \pi \\[6pt] r^2 &= \frac{150 \times \cancel{\pi}}{6 \times \cancel{\pi}} \\[6pt] r^2 &= 25 \\[6pt] r &= \sqrt{25} \\[6pt] r &= 5 \ \text{cm} \\[10pt] \color{green}\text{C.S.A} &= \color{green}2\pi r h \\[6pt] &= 2 \times 3.14 \times 5 \times 6 \\ &= 3.14 \times 10 \times 6 \\ &= 31.4 \times 6 \\ &= 188.4 \ \text{cm}^2 \\[10pt] \color{green}\text{T.S.A} &= \color{green}2\pi r(r + h) \\[6pt] &= 2 \times 3.14 \times 5 \times (5 + 6) \\ &= 10 \times 3.14 \times 11 \\ &= 31.4 \times 11 \\ &= 345.4 \ \text{cm}^2 \end{aligned} \]
Answer Lateral surface area \(=\ \color{red} {188.4 \ \text{cm}^2}\), Total surface area \(=\ \color{red}{ 345.4 \ \text{cm}^2}\)
11. The curved surface area of a cylindrical pillar is 264 m2 and its volume is 924 m3. Find the diameter and the height of the pillar.
Solution
\[ \begin{aligned} \text{Volume} &= 924 \ \text{m}^3 \\ \text{C.S.A} &= 264 \ \text{m}^2 \\[6pt] \color{green}\text{Volume} &= \color{green} \pi r^2 h \\ \color{green}\text{C.S.A} &= \color{green} 2\pi r h \\[6pt] \frac{\text{Volume}}{\text{C.S.A}} &= \frac{\pi r^2 h}{2\pi r h} \\[6pt] \frac{924}{264} &= \frac{r}{2} \\[6pt] \frac{\cancel{924}^{77}}{\cancel{264}_{22}} &= \frac{r}{2} \\[6pt] \frac{\cancel{77}^{7}}{\cancel{22}_{2}} &= \frac{r}{2} \\[6pt] \frac{7}{2} &= \frac{r}{2} \\[6pt] \frac{7 \times \cancel2^1}{\cancel2_1} &= r \\[6pt] r &= 7 \ \text{m} \\[10pt] 2\pi r h &= 264 \\[6pt] 2 \times \frac{22}{\cancel7_1} \times \cancel7^1 \times h &= 264 \\[6pt] 44 \times h &= 264 \\[6pt] h &= \frac{264}{44} \\[6pt] h &= 6 \ \text{m} \\[10pt] d &= 2r \\[6pt] d &= 2 \times 7 \\[6pt] d &= 14 \ \text{m} \end{aligned} \]
Answer Diameter \(= \color{red}{14 \ m }\), Height \(= \color{red}{6 \ m}\)