DAV Class 8 Maths Chapter 14 Practice Worksheet
Mensuration Practice Worksheet
1. A cuboidal tin open at the top has dimensions \( 20 \ cm \times 16 \ cm \times 14 \ cm \) respectively. Find the Total Surface Area of metal sheet required to make 10 such boxes in sq. m.
Solution
\[ \begin{aligned} l &=20 \ cm \\ b &=16 \ cm \\ h &=14 \ cm \\[6pt] \color{green} \text{TSA of cuboid open top} &= \color{green} \text{LSA}+\text{base} \\ &= \color{green}2h(l+b) + lb \\ &= 2 \times 14 \times (20+16) + 20 \times 16 \\ &= 28 \times 36 + 320 \\ &= 1008+320 \\ &=1328 \ cm^2\\[6pt] \text{TSA for }10\text{ boxes} &= 10 \times 1328 \\ &=13280 \ cm^2 \\[6pt] 1 \ cm^2 & = \frac{1}{10000} \ m^2 \\[6pt] 13280 \ cm^2 &= \frac{13280}{10000} \\[6pt] &= 1.328 \ m^2 \end{aligned} \]Answer Total Surface Area of metal sheet required to make 10 boxes \( = \color{red} {1.328 \ m^2} \)
2. Rohith is painting the walls and ceiling of a cuboidal hall with length \( 12 \ m \), breadth \( 10 \ m \) and height \( 6 \ m \) respectively. From each can of paint, \( 192 \ m^2 \) of area can be painted. How many cans of paint will he need to paint the room?
Solution
\[ \begin{aligned} l &= 12 \ m \\ b &= 10 \ m \\ h &= 6 \ m \\[6pt] \color{green}\text{Area of 5 walls} &= \color{green} LSA + \text{Ceiling area} \\ &= \color{green}2h(l+b) + lb \\ &= 2 \times 6 \times (12+10) + 12 \times 10\\ &= 12 \times 22 + 120 \\ &= 264 + 120 \\ &= 384 \ m^2 \\[4pt] \text{Area covered by 1 can} &= 192 \ m^2 \\[6pt] \text{No. of cans needed} &= \dfrac{384}{192} \implies 2 \ cans \end{aligned} \]Answer Rohith will need \( \color{red}{2 \ \text{cans}} \) of paint.
3. The dimensions of a metallic cuboid are \( 100 \ cm \times 80 \ cm \times 64 \ cm \) respectively. It is melted and recast into a cube. Find the surface area of the cube.
Solution
\[ \begin{aligned} l &= 100 \ cm \\ b &= 80 \ cm \\ h &= 64 \ cm \\[6pt] \color{green} \text{Volume of cuboid} &= l \times b \times h \\ & = 100 \times 80 \times 64 \\ \text{Volume of cuboid} &= 512000 \ cm^3 \\ \\ \color{green} \text{Volume of Cuboid} &= \color{green} \text{Volume of Cube} \\[6pt] \text{Volume of Cube} &= 512000 \ cm^3 \\[6pt] (side)^3 &= 512000 \\[6pt] side &= \sqrt[3]{512000} \\[6pt] side &= \sqrt[3]{512 \times 1000} \\[6pt] side &= \sqrt[3]{8^3 \times 10^3} \\[6pt] side &= 8 \times 10 \\[6pt] side &= 80 \ cm \\ \\ \color{green}\text{TSA of cube} &= \color{green} 6l^2 \\ &= 6 \times 80^2 \\ &= 6 \times 6400 \\ TSA &= 38400 \ cm^2 \end{aligned} \]Answer Surface area of the cube \( = \color{red}{38400 \ cm^2} \)
4. The radius and height of a cylinder are in the ratio 5 : 7 and its volume is 550 cm3. Find its radius and height.
Solution
\[ \begin{aligned} r : h &= 5 : 7 \\[6pt] \text{Let }r &= 5x \\[2pt] h &= 7x \\[6pt] \color{green}V &= \color{green} \pi r^2 h \\[6pt] Volume &= 550 \ cm^3 \\ \pi r^2 h &= 550 \\ \frac{22}{7} \times (5x)^2 \times (7x) &= 550 \\[6pt] \frac{22}{\cancel7_1} \times 25x^2 \times \cancel7^1x &= 550 \\[6pt] 22 \times 25x^2 \times x &= 550 \\[6pt] 550 x^3 &= 550 \\[6pt] x^3 &= \frac{550}{550} \\[6pt] x^3 &= 1 \\ x&= 1 \\ \\ radius &= 5x \\ radius &= 5 \ cm \\ \\ height &= 7x \\ height &= 7 \ cm \\ \\\end{aligned} \]
Answer Radius \(= \color{red}{5 \ cm } \), Height \(= \color{red}{7 \ cm} \)
5. The dimensions of the cuboid are in the ratio \( 1 : 2 : 3 \) respectively and its Total Surface Area is \( 88 \, \text{m}^2 \). Find the dimensions.
Solution
\[ \begin{align*} l : b : h &= 1 : 2 : 3 \\[6pt] \text{Let } l &= x \\ b &= 2x \\ h &= 3x \\ \\ \color{green}\text{T.S.A of cuboid} &= 88 \, \text{m}^2 \\ \color{green} 2(lb + bh + hl) &= 88 \\ 2\Big( x \times 2x + 2x \times 3x + 3x \times x \Big) & = 88 \\[6pt] 2(2x^2 + 6x^2 + 3x^2) & = 88 \\[6pt] 2(11x^2) & = 88 \\[6pt] 22x^2 & = 88 \\[6pt] x^2 &= \dfrac{{\cancel{88}}^{\color{green}4}}{{\cancel{22}}_{\color{green}1}} \\ x^2 &= 4 \\ x &= \sqrt{4} \\ x &= 2 \ \text{m} \\ \\ \color{magenta} \text{Dimensions} &\color{magenta} \text{ of the cuboid} \\[6pt] l &= 2 \ \text{m} \\ b &= 4 \ \text{m} \\ h &= 6 \ \text{m} \end{align*} \]
Answer The dimensions of the cuboid \( \color{red} l = 2 \ \text{m},b = 4 \ \text{m}, h= 6 \ \text{m} \)
6. The length and width of the hall is 20 m and 16 m respectively. The sum of the areas of the floor and the flat roof is equal to the sum of the areas of four walls. Find the height of the hall.
Solution
\[ \begin{align*} l &= 20 \ \text{m} \\ b &= 16 \ \text{m} \\ h &= ? \\[6pt] \color{green}\text{Area of floor} &= \color{green} lb \\ \color{green}\text{Area of flat roof} &= \color{green} lb \\ \text{Sum} &= 2lb \\[6pt] \color{green}\text{Area of four walls} &= \color{green} 2h(l+b) \\[6pt] \text{Area of four walls} &= \text{Area of (floor + flat roof)} \\[6pt] 2h(l+b) & = 2lb \\[6pt] h & = \frac{\cancel{2}lb}{\cancel{2}(l+b)} \\[6pt] h &= \dfrac{lb}{l+b} \\[6pt] h &= \dfrac{20 \times 16}{20 + 16} \\[6pt] h &= \dfrac{{\cancel{320}}^{\color{green}80}}{{\cancel{36}}_{\color{green}9}} \\[6pt] h &= \dfrac{80}{9} \ \text{m} \\[6pt] h &= 8.89 \ m \\[6pt] \end{align*} \]
Answer Height of the hall \( = \color{red} 8.89 \ m \)
7. If the area of the trapezium is \( 1586 \, \text{cm}^2 \) and two parallel sides are of lengths \( 10 \, \text{dm} \) and \( 22 \, \text{cm} \) respectively, then find the altitude.
Solution
\[ \begin{align*} 10 \,\text{dm} &= 100 \,\text{cm} \\[6pt] a &= 100 \,\text{cm} \\ b &= 22 \,\text{cm} \\ h &= ? \\[6pt] Area &= 1586 \ cm^2 \\ \color{green}\text{Area of the trapezium} &= \color{green} \frac{1}{2}\times(a+b)\times h \\[6pt] 1586 &= \frac{1}{2}\times {{(100+22)}}\times h \\[6pt] 1586 &= \frac{1}{\cancel2_{\color{green} 1}}\times {\cancel{122}}^{\color{green} \ 61}\times h \\[6pt] 1586 &= 61 \times h \\[6pt] \frac{{\cancel{1586}}^{\color{green}26}}{{\cancel{61}}_{\color{green}1}} &= h \\[6pt] h &= 26 \,\text{cm} \end{align*} \]
Answer Altitude \( = \color{red}{26 \, \text{cm}} \)
8. Find the height of a cuboid which has a base area of \( 180 \ \text{cm}^2 \) and Volume is \( 900 \ \text{cm}^3 \).
Solution
\[ \begin{align*} \text{Base area } (l \times b) &= 180 \ \text{cm}^2 \\ \color{green} \text{Volume of cuboid} &= 900 \ \text{cm}^3 \\[6pt] \color{green} l \times b \times h &= 900 \\[6pt] 180 \times h & = 900\\[6pt] h & = \frac{\cancel{900}^{ \ 5}}{\cancel{180}_1} \\[6pt] h &= 5 \,\text{cm} \end{align*} \]
Answer Height of the cuboid \( = \color{red}{5 \ \text{cm}} \)
9. A cuboidal box of dimensions \( 1 \, \text{m} \times 2 \, \text{m} \times 1.5 \, \text{m} \) respectively is to be painted except its bottom. Find the area of the box that has to be painted.
Solution
\[ \begin{align*} l &= 1 \ \text{m} \\ b &= 2 \ \text{m} \\ h &= 1.5 \ \text{m} \\[6pt] \color{green}\text{Area to be painted} &= \color{green}\text{LSA} + \text{Top area} \\[4pt] &= \color{green}2h(l+b) + lb \\[6pt] &= 2 \times 1.5 \times (1+2) + 1 \times 2 \\[4pt] &= 3 \times 3 + 2 \\[4pt] &= 9 + 2 \\[4pt] &= 11 \ \text{m}^2 \end{align*} \]
Answer Area to be painted \( = \color{red}{11 \ \text{m}^2} \)
10. The parallel sides of the trapezium are 40 cm and 20 cm respectively. Find the area of the trapezium, if its non-parallel sides are equal having the lengths of 26 cm.
Solution
\[ \begin{align*} \text{Draw } & DF \parallel CB \\ \implies DFBC & \text{ is a parallelogram} \\ \text{In } &\triangle ADF \\ DA = DF & = 26 \, cm \\ AB &= 40 \, \text{cm} \\ AF &= 40 - 20 \implies 20 \, \text{cm} \\ \\ DE \perp AF \implies & \text{E is the midpoint of AF} \\ \\ EF = \frac{AF}{2} \\[6pt] = \frac{20}{2} \\[6pt] \implies EF = 10 \, cm \\ \\ \text{In } \triangle DEF & \text{ (right angled triangle)} \\ (DE)^2 + (EF)^2 &= (DF)^2 \\ (DE)^2 + 10^2 &= 26^2 \\ (DE)^2 + 100 &= 676 \\ (DE)^2 &= 676 - 100 \\ (DE)^2 &= 576 \\ DE &= \sqrt{576} \\ DE &= 24 \, cm \\ \\\text{In trapezium } & DABC \\ a &= 40 \, \text{cm} \\ b &= 20 \, \text{cm} \\ h &= 24 \, \text{cm} \\ \\ \color{green} \text{Area of the trapezium} &= \color{green} \frac{1}{2} \times (a + b) \times h \\ \\ &= \frac{1}{2} \times (40 + 20) \times 24 \\ \\ &= \frac{1}{{\cancel{2}}_{\color{green}1}} \times {{\cancel{60}}^{\color{green}30}} \times 24 \\ \\ &= 30 \times 24 \\ \text{Area} &= 720 \, \text{cm}^2 \end{align*} \]
Answer Area of the trapezium \( = \color{red}{720 \ \text{cm}^2} \).
11. A cuboidal box is of dimensions \( 60 \, \text{cm} \times 54 \, \text{cm} \times 30 \, \text{cm} \) respectively. How many small cubes of side \( 6 \, \text{cm} \) can be placed in the cuboidal box?
Solution
\[ \begin{aligned} \text{Volume of box} & = 60 \times 54 \times 30 \\ \text{Volume of cube} & = 6 \times 6 \times 6 \\[6pt] \text{No. of cubes} &= \frac{\text{Volume of box}}{\text{Volume of small cube}} \\ \\ &= \frac{\cancel{60}^{10} \times \cancel{54}^{9} \times \cancel{30}^{5}}{\cancel6_1 \times \cancel6_1 \times \cancel6_1} \\ \\ &= 10 \times 9 \times 5 \\[6pt] &= 450 \end{aligned} \]
Answer Number of small cubes \( = \color{red}{450} \)
12. A suitcase with dimensions \( 80 \, \text{cm} \times 48 \, \text{cm} \times 24 \, \text{cm} \) respectively, is to be covered with a tarpaulin cloth. How many meters of tarpaulin of width \( 96 \, \text{cm} \) is required to cover 100 such suitcase?
Solution
\[ \begin{align*} l &= 80 \ \text{cm} \\ b &= 48 \ \text{cm} \\ h & = 24 \ \text{cm} \\[6pt] \color{green}\text{T.S.A} &= \color{green} 2(lb + bh + hl) \\[2pt] &= 2(80 \times 48 + 48 \times 24 + 24 \times 80) \\[2pt] &= 2(3840 + 1152 + 1920) \\[2pt] &= 2 \times 6912 \\[2pt] &= 13824 \ \text{cm}^2 \\[8pt] \text{Area for 100 suitcase} &= 13824 \times 100 \\ \implies & 1382400 \ cm^2 \\[6pt] \text{Width of cloth} &= 96 \ \text{cm} \\[6pt] \text{Cloth required} &= \frac{1382400}{96} \\[4pt] &= \frac{{\cancel{1382400}}^{\color{green}14400}}{{\cancel{96}}_{\color{green}1}} \\[2pt] &= 14400 \ \text{cm} \\[6pt] \text{Length (m)} &= \dfrac{14400}{100} \\[6pt] \implies & 144 \ \text{m} \end{align*} \]
Answer Tarpaulin required \( = \color{red}{144 \ \text{m}} \)
13. Find the number of cuboidal boxes measuring \( 2 \, \text{cm} \times 3 \, \text{cm} \times 10 \, \text{cm} \) respectively, can be stored in a carton of length \( 40 \, \text{cm} \), breadth \( 36 \, \text{cm} \) and height \( 24 \, \text{cm} \).
Solution
\[ \begin{aligned} \text{Volume of carton} & = 40 \times 36 \times 24 \\ \text{Volume of cuboidal box} & = 2 \times 3 \times 10 \\[6pt] \text{No. of cuboidal boxes} &= \frac{\text{Volume of carton}}{\text{Volume of cuboidal box}} \\ \\ &= \frac{\cancel{40}^{ \ 4} \times \cancel{36}^{12} \times \cancel{24}^{12}}{\cancel2_1 \times \cancel3_1 \times \cancel{10}_1} \\ \\ &= 4 \times 12 \times 12 \\[6pt] &= 576 \end{aligned} \]
Answer Number of cuboidal boxes \( = \color{red}{576} \)
14. How many soap cakes can be placed in a box of size \( 56 \, \text{cm} \times 0.4 \, \text{m} \times 0.25 \, \text{m} \), if the size of a soap cake is \( 7 \, \text{cm} \times 5 \, \text{cm} \times 2.5 \, \text{cm} \) respectively?
Solution
\[ \begin{aligned} & \textbf{Convert to } (cm)\\& \color{magenta} \text{Box} \\ l & = 56 \ cm \\ b & = 0.4 \ m \implies 40 \ cm\\ h & = 0.25 \ m \implies 25 \ cm \\ \\& \color{magenta} \text{Soap cake} \\ l & = 7 \ cm \\ b & = 5 \ cm \\ h & = 2.5 \ cm \\ \\\color{green} \text{No. of soap cakes} &= \color{green} \frac{\text{Volume of Box}}{\text{Volume of Soap cake}} \\ \\&= \frac{\cancel{56}^{\ 8} \times \cancel{40}^8 \times \cancel{25}^{10}}{\cancel{7}_1 \times \cancel{5}_1 \times \cancel{2.5}_1} \\ \\ &= 8 \times 8 \times 10 \\ & = 640 \end{aligned} \]
Answer Number of soap cakes \( = \color{red}{640} \)
15. The length, breadth and height of a room are \( 5 \, \text{m}, 4.5 \, \text{m} \) and \( 3 \, \text{m} \) respectively. Find the volume of the air present inside the room.
Solution
\[ \begin{align*} l &= 5 \ \text{m} \\ b &= 4.5 \ \text{m} \\ h &= 3 \ \text{m} \\[6pt] \color{green}\text{Volume of room (air)} &= \color{green} l \times b \times h \\[2pt] &= 5 \times 4.5 \times 3 \\[2pt] &= 22.5 \times 3 \\[2pt] &= 67.5 \ \text{m}^3 \end{align*} \]
Answer Volume of air inside the room \( = \color{red}{67.5 \ \text{m}^3} \)
16. How many planks each of which is 3 m long, 15 cm broad and 5 cm thick can be prepared from a wooden block 6 m long, 75 cm broad and 45 cm thick?
Solution
\[ \begin{aligned} & \textbf{Convert to } (cm)\\ & \color{magenta} \text{Wooden block} \\ l &= 6 \ m \implies 600 \ cm \\ b &= 75 \ cm \\ h &= 45 \ cm \\ \\ & \color{magenta} \text{Plank} \\ l &= 3 \ m \implies 300 \ cm \\ b &= 15 \ cm \\ h &= 5 \ cm \\ \\ \color{green} \text{No. of planks} &= \color{green} \frac{\text{Volume of Wooden block}}{\text{Volume of Plank}} \\ \\ &=\frac{\cancel{600}^{2}\times\cancel{75}^{5}\times\cancel{45}^{9}} {\cancel{300}_{1}\times\cancel{15}_{1}\times\cancel{5}_{1}} \\[6pt] &= 2 \times 5 \times 9 \\ &= 90 \end{aligned} \]
Answer The number of planks that can be prepared \(=\ \color{red}{90}\)
17. The walls and ceiling of a room are to be plastered. The length, breadth and height of the room are \( 4.5 \, \text{m},\ 3 \, \text{m} \) and \( 350 \, \text{cm} \) respectively. Find the cost of plastering at the rate of ₹ \( 8 \) per square metre.
Solution
\[ \begin{aligned} l &=4.5 \ \text{m} \\ b &=3 \ \text{m} \\ h &= 350 \ \text{cm} \implies 3.5 \ \text{m} \\[4pt] \color{green}\text{Area to plaster} &= \color{green} \text{Area of 4 walls} + \text{Ceiling} \\ &= 2h(l+b) + lb \\ &= 2 \times 3.5 \times (4.5+3)+( 4.5 \times 3) \\ &= 7 \times 7.5 + 13.5 \\ &= 52.5 + 13.5 \\ &= 66 \ \text{m}^2 \\[6pt] \text{Cost} &= 66 \times 8 = \text{₹ } 528 \end{aligned} \]
Answer Cost of plastering \(=\color{red}{\text{₹ }528}\)
18. A cuboid has Total Surface Area of \(50 \, \text{m}^2\) and Lateral Surface Area of \(30 \, \text{m}^2\). Find the area of its base.
Solution
\[ \begin{aligned} \text{TSA} &= 50 \ \text{m}^2 \\ \text{LSA} &= 30 \ \text{m}^2 \\ \text{Base area} &= lb \\[6pt] \color{green}\text{TSA} &= \color{green}\text{LSA} + 2lb \\[4pt] 50 &= 30 + 2lb \\[2pt] 50 - 30 &= 2lb \\[2pt] 20 &= 2 lb \\[2pt] 2lb &= 20 \\ lb &= \frac{20}{2} \\ lb &= 10 \ \text{m}^2 \end{aligned} \]
Answer Area of base \(=\color{red}{10 \, \text{m}^2}\)
19. The thickness of a hollow metallic cylinder is 2 cm. It is 70 cm long with outer radius 14 cm. Find the volume of the metal used in making the cylinder, assuming that it is open at both ends. Also, find its weight if the metal weighs 8 gm/cm\(^3\).
Solution
\[ \begin{aligned} \text{Outer radius } (R) &= 14 \ \text{cm} \\ \text{Thickness } (t) &= 2 \ \text{cm} \\ \text{Inner radius } (r) &= R - t \implies 12 \ \text{cm} \\ \text{Height } (h) &= 70 \ \text{cm} \\[6pt] \color{green}\text{Volume of metal} &= \color{green}\pi h (R^2 - r^2) \\[6pt] &= \frac{22}{\cancel7_1} \times \cancel{70}^{10} \times (14^2 - 12^2) \\[6pt] &= 22 \times 10 \times (196 - 144) \\[6pt] &= 220 \times 52 \\[6pt] &= 11440 \ \text{cm}^3 \\[8pt] \text{Weight} &= 11440 \times 8 \ \text{gm} \\ &= 91520 \ \text{gm} \\ \implies & 91.52 \ \text{kg} \end{aligned} \]
Answer Volume of metal \(=\ \color{red}{11440 \ \text{cm}^3}\), Weight \(= \color{red} 91.52 \ kg \)
20. A closed metallic cylinder has its base diameter 56 cm and it is 2.72 m high. Find the cost of the metal used to make it if it costs ₹ 80 per sq. m.
Solution
\[ \begin{aligned} \text{Diameter} &= 56 \,\text{cm} \\[6pt] r &= \frac{56}{2} \\[6pt] & = 28 \,\text{cm} \Rightarrow 0.28 \,\text{m} \\[6pt] h &= 2.72 \,\text{m} \\[6pt] \color{green} \text{T.S.A} &= \color{green} 2\pi r(h + r) \\[6pt] &= 2 \times \frac{22}{\cancel7_1} \times \cancel{0.28}^{.04} \times (2.72 + 0.28) \\[6pt] &= 44 \times 0.04 \times 3 \\[6pt] &= 1.76 \times 3 \\[6pt] &= 5.28 \,\text{m}^2 \\[6pt] \text{Total Cost} &= 5.28 \times 80 \\[6pt] &= \text{₹} 422.40 \end{aligned} \]
Answer The cost of the metal used to make the cylinder is \( \color{red} \text{₹} \, 422.40 \)
21. An iron pipe is 20 cm long and its external diameter is 25 cm. If the thickness of the pipe is 1 cm, find the Total Surface Area of the pipe.
Solution
\[ \begin{aligned} h &= 20 \ \text{cm} \\[6pt] R &=\frac{25}{2} \implies 12.5 \ \text{cm}\\[6pt] \text{Thickness} & = 1 \ cm \\[6pt] r&= 12.5 - 1 \implies 11.5 \ \text{cm}\\[6pt] \color{green}\text{T.S.A (pipe)} &= \color{green}\underbrace{2\pi Rh}_{\text{outer CSA}} + \underbrace{2\pi rh}_{\text{inner CSA}} + \underbrace{2\pi(R^2-r^2)}_{\text{two rings}}\\[6pt] &= 2\pi h(R+r) + 2\pi(R^2-r^2)\\[6pt] &= 2 \pi \left[h(R+r) + (R^2-r^2)\right]\\[6pt] &= 2 \times \frac{22}{7} \left[ 20 \times (12.5+11.5) + \big(12.5^2-11.5^2\big) \right]\\[6pt] &= \frac{44}{7} \left[ 20 \times 24 + 24\right]\\[6pt] &= \frac{44}{7} \left[ 480 + 24\right]\\[6pt] &= \frac{44}{\cancel7_1} \times \cancel{504}^{72} \\[6pt] &= 44 \times 72 \\[6pt] & = 3168 \ \text{cm}^2 \end{aligned} \]Answer T.S.A of the pipe \(= \color{red}3168 \ \text{cm}^2\)
22. Find the cost of plastering the inner surface area of the well at ₹ 9.50 per m2, if it is 21 m deep and diameter of its top is 6 m.
Solution
\[ \begin{aligned} d &= 6 \ \text{m} \\ r &= 3 \ \text{m} \\ h &=21 \ \text{m}\\ \color{green}\text{Inner C.S.A} &= \color{green}2\pi r h \\ &= 2 \times \frac{22}{\cancel7_1}\times 3 \times \cancel{21}^3 \\ &= 44 \times 9 \\ &= 396 \ \text{m}^2 \\ \\ \text{Cost} &= 396 \times 9.50 \\ &= \text{₹ } 3762 \end{aligned} \]Answer Cost of plastering \(= \color{red}{\text{₹ }3762}\)
23. The outer diameter of a metallic cylindrical tube is 14 cm and the thickness of the tube is 1 cm. Find the weight of 2 m long tube, if the density of the metal is 7 gm/cm3.
Solution
\[ \begin{aligned} D &= 14 \ \text{cm} \\ \text{outer Radius }(R) &= 7 \ \text{cm} \\ thickness & = 1 \ cm \\ \text{inner radius }(r) &= 7 - 1 \implies 6 \ \text{cm} \\ h &= 2 \ m \implies 200 \ \text{cm}\\ \\ \color{green}\text{Volume of metal} &= \color{green}\pi h(R^2-r^2) \\[6pt] &= \frac{22}{7}\times 200 \times (7^2 - 6^2)\\[6pt] &= \frac{22}{7}\times 200 \times (49-36)\\[6pt] &= \frac{22}{7}\times 200 \times 13 \\[6pt] Volume & = \frac{57200}{7} \ \text{cm}^3\\[6pt] \text{Weight} &= \text{Volume} \times \text{density} \\[6pt] &= \frac{57200}{\cancel7_1}\times \cancel7^1 \\[6pt] &= 57200 \ \text{gm} \\[6pt] & \implies 57.2 \ \text{kg} \end{aligned} \]Answer Weight of the tube \(=\ \color{red}{57.2 \ \text{kg}}\)
24. The length of the roller is 40 cm and its diameter is 21 cm. It takes 300 complete revolutions to move once to level the floor of a room. Find the area of the floor of the room in m2.
Solution
\[ \begin{aligned} \text{Diameter} (d) &= 21 \,\text{cm} \\[6pt] \text{Radius} (r) &= \frac{21}{2} \,\text{cm} \\[6pt] \text{Height }(h) &= 40 \,\text{cm}\\[6pt] \color{green}\text{C.S.A} &= \color{green}2\pi r h \\[6pt] & = \cancel 2 \times \frac{22}{\cancel7_1} \times \frac{\cancel{21}^3}{\cancel2} \times 40 \\[6pt] & = 66 \times 40 \\[6pt] \text{C.S.A} & = 2640 \ cm^2 \\[6pt]Revolutions &= 300 \\ & = 300 \times 2640 \\ \text{Area of room} & = 792000 \ cm^2 \\[6pt] 1 \ cm^2 & = \frac{1}{10000} \ m^2 \\[6pt] & = \frac{792000}{10000} \ m^2 \\[6pt] &= 79.2\,\text{m}^2 \end{aligned} \]
Answer Area of the floor \(=\ \color{red}{79.2 \ \text{m}^2}\)
25. The Curved Surface Area of the cylinder is 1848 cm2. If the circumference of the base is 132 cm, then find the height and volume of the cylinder.
Solution
\[ \begin{aligned} \text{Circumference } & = 132 \\ 2\pi r & = 132 \\[6pt] 2 \times \frac{22}{7} \times r & = 132 \\ r & = \frac{\cancel{132}^3 \times 7}{\cancel{44}_1} \\r & = 21 \ \text{cm}\\[6pt] \color{green}\text{C.S.A} &= 1848 \\ \color{green}2\pi r h & = 1848 \\ 132 \times h & = 1848 \\[6pt] h & = \frac{1848}{132} \\[6pt] h &= 14 \ \text{cm}\\[8pt] \color{green}\text{Volume} &= \color{green}\pi r^2 h \\[6pt] & = \frac{22}{\cancel7_1}\times 21^2 \times \cancel{14}^2 \\[6pt] &= 22 \times 441 \times 2 \\ &= 22 \times 882 \\ &= 19404 \ \text{cm}^3 \end{aligned} \]Answer Height \(= \color{red}{14 \ \text{cm}}\), Volume \(= \color{red}{19404 \ \text{cm}^3}\)
26. The Curved Surface Area of a cylindrical rod is 132 cm2. Find its length, if the radius is 0.35 cm.
Solution
\[ \begin{aligned} r&=0.35 \\ \color{green}\text{C.S.A} &= 132 \\ \color{green}2\pi r h &= 132 \\ 2 \times \frac{22}{7}\times 0.35 \times h &= 132\\[6pt] \frac{22}{\cancel7_1}\times \cancel{0.7}^{ \ 0.1} \times h &= 132\\[6pt] 22 \times 0.1 \times h &= 132\\[6pt] 2.2 \times h &= 132\\[6pt] h &= \frac{132 \times 10}{2.2 \times 10} \\[6pt] h &= \frac{\cancel{1320}^{ \ 60}}{\cancel{22}_1} \\[6pt] h & = 60 \ \text{cm} \end{aligned} \]Answer Length of the rod \(= \color{red}{60 \ \text{cm}}\)
27. The area of the base of a right circular cylinder is 616 cm2 and its height is 2.5 cm. Find the Curved Surface Area of the cylinder.
Solution
\[ \begin{aligned} \text{Base area } &= 616 \ cm^2 \\ \pi r^2 &= 616 \\ \frac{22}{7} \times r^2 &= 616 \\[6pt] r^2 &=\frac{\cancel{616}^{28} \times 7}{\cancel{22}_1} \\[6pt] r^2 &= 28 \times 7 \\[6pt] r^2 &= 196 \\[6pt] r &= \sqrt{196} \\[6pt] r &= 14 \ \text{cm}\\[6pt] h &= 2.5 \ \text{cm}\\[6pt] \color{green}\text{C.S.A} &= \color{green} 2\pi r h \\ &= 2 \times \frac{22}{\cancel7_{\color{green}1}}\times \cancel{14}^{\color{green}2}\times 2.5 \\ &= 2 \times 22 \times 2 \times 2.5 \\ &= 44 \times 5 \\ &= 220 \ \text{cm}^2 \end{aligned} \]Answer Curved Surface Area \(= \color{red}{220 \ \text{cm}^2}\)
28. A rectangular sheet of paper is rolled along its length to form a cylinder. The sheet is 44 cm long and 20 cm wide. Find the Total Surface Area of the cylinder.
Solution
\[ \begin{aligned} \text{Circumference } &= 44 \\ 2\pi r &= 44 \\[6pt] 2 \times \frac{22}{7} \times r & = 44 \\[6pt] \frac{44}{7} \times r & = 44 \\[6pt] r & = \frac{\cancel{44} \times 7}{\cancel{44}} \\[6pt] r &= 7\ \text{cm}\\ h &= 20 \ \text{cm}\\[4pt] \color{green}\text{T.S.A} &= \color{green}2\pi r(h+r)\\[6pt] &= 2\times \frac{22}{\cancel7}\times \cancel7 \times (20+7)\\[6pt] &= 44 \times 27 \\ &= 1188 \ \text{cm}^2 \end{aligned} \]Answer Total Surface Area \(=\ \color{red}{1188 \ \text{cm}^2}\)
29. The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. Calculate the ratio of their Curved Surface Areas.
Solution
Let the radii of the two cylinders be 2x and 3x and their respective heights 5y and 3y.
\[ \begin{align*} \text{CSA}_1 : \text{CSA}_2 &= \frac{\text{CSA}_1}{\text{CSA}_2} \\[6pt] &= \frac{2\pi \ r_1 \ h_1}{2\pi \ r_2 \ h_2} \\[6pt] &= \frac{\cancel{2\pi} \times 2 \cancel x \times 5 \cancel y}{\cancel{2\pi} \times 3 \cancel x \times 3 \cancel y} \\[6pt] &= \frac{10}{9} \\[6pt] \text{CSA}_1 : \text{CSA}_2 &= 10 : 9 \end{align*} \]
Answer The ratio of their curved surface areas is \(= \color{red} 10:9 \)
30. The Curved Surface Area of a cylinder is 1320 cm2 and its base has diameter 21 cm. Find the height of the cylinder.
Solution
\[ \begin{align*} \text{Diameter} (d) &= 21 \, \text{cm} \\[6pt] \text{Radius} (r) &= \frac{21}{2} \, \text{cm} \\[6pt] \color{green} \text{Curved Surface Area} &= 1320 \, \text{cm}^2 \\[6pt] \color{green} 2\pi rh & = 1320 \\[6pt] \cancel 2 \times \frac{22}{\cancel7_1} \times \frac{\cancel{21}^3}{\cancel 2} \times h & = 1320 \\[6pt] 66 \times h & = 1320 \\[6pt] h &= \frac{1320}{66} \\[6pt] h &= 20 \, \text{cm} \end{align*} \]
Answer Height of the cylinder \(= \color{red}{20 \ \text{cm}}\)
31. The height of a right circular cylinder is 10.5 cm. If three times the sum of the areas of its two circular faces is twice the area of the Curved Surface Area, then find the radius of its base.
Solution
\[ \begin{aligned} h &= 10.5 \ \text{cm} \\[4pt] \text{Areas of two base} &= 2 \pi r^2 \\ CSA &= 2 \pi rh \\ \\ \color{green} 3 \times\underbrace{(2\pi r^2)}_{\text{two bases}} & = \color{green} 2 \times \underbrace{(2\pi r h)}_{\text{C.S.A}} \\[4pt] 6\pi r^2 & = 4\pi r h \\[4pt] 6 \times \pi \times r \times r & = 4 \times \pi \times r \times 10.5 \\[6pt] r & = \frac{\cancel{42}^7 \times \cancel \pi \times \cancel r}{\cancel6_1 \times \cancel \pi \times \cancel r } \\[6pt] r & = 7 \ cm \end{aligned} \]Answer Radius of base \(= \color{red}{7 \ \text{cm}}\)
32. Twenty one cylindrical pillars of the Parliament House are to be cleaned. If the diameter of each pillar is 0.50 m and height is 4 m, then find the cost of cleaning them at the rate of ₹ 2.50 per sq.m.
Solution
\[ \begin{aligned} d &= 0.50 \ \text{m} \\ r &= 0.25 \ \text{m} \\ h &= 4 \ \text{m}\\[4pt] \color{green}\text{C.S.A of 1 pillar} &= \color{green}2\pi r h \\[6pt] &= 2 \times \frac{22}{7} \times 0.25 \times 4 \\[6pt] &= \frac{44}{7} \times 1 \\[6pt] \text{C.S.A of 1 pillar} &= \frac{44}{7}\ \text{m}^2 \\[6pt] \text{CSA of 21 pillars} &= \frac{44}{\cancel7_1} \times \cancel{21}^3 \\[6pt] &= 3 \times 44 \\[6pt] &= 132 \ \text{m}^2 \\[6pt] \text{Total Cost} &= 132 \times 2.50 \\ &= \color{red}{\text{₹ }330} \end{aligned} \]Answer Cost of cleaning 21 pillars \(= \color{red}{\text{₹ }330}\)
33. The sum of the radius of the base and height of a solid cylinder is 37 m. If the Total Surface Area of the solid cylinder is 1628 m2, then find the circumference of its base.
Solution
\[ \begin{aligned} r+h &= 37 \ \text{m}\\ \color{green}\text{T.S.A} & = 1628 \ m^2 \\[4pt] \color{green}2\pi r(h+r) & = 1628 \\[4pt] 2\pi r \times 37 &= 1628 \\[6pt] 2\pi r &= \frac{1628}{37} \\[6pt] 2\pi r &= 44 \\[6pt] \text{Circumference } (2\pi r )&= \color{green}{44 \ \text{m}} \end{aligned} \]Answer Circumference of base \(= \color{red}{44 \ \text{m}}\)
34. A cylindrical vessel without lid has to be tin-coated on its both sides. If the radius of the base is 70 cm and its height is 1.4 m, then calculate the cost of tin-coating at the rate of ₹ 3.50 per 1000 cm2.
Solution
\[ \begin{aligned} r &= 70\ \text{cm} \\ h &=1.4\ \text{m} \implies 140\ \text{cm}\\[4pt] \color{green}{\text{Area to coat tin}} & = \begin{cases} \color{green} \text{CSA (inner + outer)} \\ \color{green}\quad \quad \quad + \\ \color{green}\text{base area (inner + outer)}\\ \end{cases}\\[6pt] & = \underbrace{2\times 2\pi r h}_{\text{inner+outer CSA}} + \underbrace{2\times \pi r^2}_{\text{inner+outer base}} \\[6pt] & = 2 \pi r(2h + r) \\[6pt] & = 2 \times \frac{22}{\cancel7_1} \times \cancel{70}^{10} \times (2 \times 140 + 70) \\[6pt] & = 44 \times 10 \times (280 + 70) \\[6pt] & = 440 \times 350 \\[6pt] &= 154000 \ \text{cm}^2 \\[6pt] \text{Cost} &= \cancel{154000}^{ \ 154} \times \frac{3.50}{\cancel{1000}_1} \\[6pt] & = 154 \times 3.50 \\ Cost &= \color{green}{\text{₹ }539} \end{aligned} \]Answer Cost of tin-coating \(= \color{red}{\text{₹ }539}\)
35. Two cubes have their volumes in the ratio \( 8 : 27 \). Find the ratio of their surface areas.
Solution
\[ \begin{aligned} \text{Let the volumes of cube be } & V_1 \text{ and } V_2 \\[4pt] \text{Let the sides of cube be } & s_1 \text{ and } s_2 \\[4pt] \color{green} \text{Volume of cube} &= \color{green} (s)^3 \\[4pt] \color{green}V_1 : V_2 &= 8 : 27 \\[6pt] \frac{ {V_1}}{ V_2} &= \frac{8}{27} \\[6pt] \frac{ {({s}_1)^3}}{ {({s}_2)^3}} &= \frac{8}{27} \\[6pt] \left(\frac{ {{s}_1}}{ {{s}_2}}\right)^3 &= \frac{8}{27} \\[6pt] \frac{ {{s}_1}}{ {{s}_2}} &= \sqrt[3]{\frac{8}{27}} \\[6pt] \frac{ {{s}_1}}{ {{s}_2}} &= \frac{2}{3} \\ \\ \color{green}\text{TSA of a cube} &= \color{green}6(\text{side})^2 \\[6pt] \frac{ {{TSA}_1}}{ {{TSA}_2}} &= \frac{\cancel6(2^2)}{\cancel6(3^2)} \\[6pt] &=\frac{4}{9} \\[4pt] &= 4 : 9 \end{aligned} \]Answer Ratio of their surface areas \(= \color{red}{4:9}\)
36. A school provides milk to the students daily in a cylindrical glasses of diameter 7 cm. If the glass is filled with milk up to height of 12cm, then find how many litres of milk is needed to serve 1600 students.
Solution
\[ \begin{aligned} d &= 7 \ \text{cm} \\[6pt] r &= \frac{7}{2} \text{cm} \\[6pt] h &= 12 \ \text{cm} \\[6pt] \color{green}\text{Volume of 1 glass} &= \color{green}\pi r^2 h \\[6pt] &= \frac{\cancel{22}^{11}}{\cancel7}\times \frac{\cancel7}{\cancel2_1} \times \frac{7}{\cancel2_1} \times \cancel{12}^6 \\[6pt] &= 11 \times 7 \times 6 \\[6pt] &= 77 \times 6 \\[6pt] &= 462 \ \text{cm}^3 \\[6pt] \text{Volume for }1600\text{ students} &= 462 \times 1600 \\ &= 739200 \ \text{cm}^3 \\[6pt] 1000 \ \text{cm}^3 &= 1 \ \text{L} \\[6pt] \text{Milk needed} &= \frac{739200}{1000} \\[6pt] &= 739.2 \ \text{L} \end{aligned} \]Answer Milk needed to serve 1600 students \(= \color{red}{739.2 \ \text{litres}}\)
37. Four times the area of a Curved Surface Area of a cylinder is equal to the six times the sum of the area of its base. If the height is 12 cm, then find its Curved Surface Area.
Solution
\[ \begin{aligned} h &= 12 \ \text{cm} \\ \text{CSA} &= 2\pi r h \\ \text{Sum of base areas} &= 2\pi r^2 \\[6pt] \color{green}6\times(\text{sum of base areas}) & = \color{green}4\times \text{CSA} \\[6pt] 6(2\pi r^2) & = 4(2\pi r h) \\[6pt] 6 \times 2 \pi \times r \times r & = 4 \times 2 \pi \times r \times 12 \\[6pt] r & = \frac{\cancel{48}^{8} \times \cancel {2 \pi} \times \cancel r }{\cancel 6_1 \times \cancel {2 \pi} \times \cancel r } \\[6pt] r &= 8 \ \text{cm} \\[8pt] \color{green}\text{CSA} &= \color{green}2\pi r h \\[6pt] &= 2\times \frac{22}{7}\times 8\times 12 \\[6pt] &= \frac{4224}{7} \ \text{cm}^2 \\[6pt] &= 603.43 \ \text{cm}^2 \end{aligned} \]Answer Curved Surface Area \(= \color{red} 603.43\ \text{cm}^2 \)
38. The Volume of a cuboidal box is 48 cm3. If its height and length are 3 cm and 4 cm respectively, then find its breadth.
Solution
\[ \begin{aligned} h &= 3 \ \text{cm} \\ l &= 4 \ \text{cm} \\ b &= ? \\ \color{green}Volume &= 48 \ cm^3 \\[6pt] \color{green} l \times b \times h &= 48 \\[6pt] 3 \times b \times 4 &= 48 \\[6pt] 12 \times b &= 48 \\[6pt] b &= \frac{\cancel{48}^4}{\cancel{12}_1} \\[6pt] b &= 4 \ \text{cm} \end{aligned} \]Answer Breadth \(= \color{red}{4 \ \text{cm}}\)
39. The population of a village is 4000. They required 150 litres water per head per day. It has a tank which is 20 m long, 15 m broad and 6 m high. For how many days will the water of this tank last?
Solution
\[ \begin{aligned} l &= 20 \ \text{m} \\ b &= 15 \ \text{m} \\ h &= 6 \ \text{m} \\[4pt] \color{green}\text{Tank volume} &= \color{green}l \times b \times h \\ &= 20 \times 15 \times 6 \\ &= 1800 \ \text{m}^3 \\[6pt] 1 \ \text{m}^3 &= 1000 \ \text{L} \\ \text{Water in tank} &= 1800 \times 1000 \\ &= 1800000 \ \text{L} \\[6pt] \text{Water used in 1 day} &= 4000 \times 150 \\ &= 600000 \ L \\[6pt] \text{No. of Days} &= \frac{1800000}{600000} \\[6pt] &= 3 \ days \end{aligned} \]Answer Water will last \(=\ \color{red}{3 \ \text{days}}\)
40. The dimensions of an oil tin are 26 cm × 26 cm × 45 cm respectively. Find the area of the tin sheet required for making 20 such tins. If 1 square metre of the tin sheet costs ₹ 10, then find the cost of tin sheet used for the 20 tins.
Solution
\[ \begin{aligned} l &= 26 \ \text{cm} \\ b &= 26 \ \text{cm} \\ h &= 45 \ \text{cm} \\[4pt] \color{green}\text{T.S.A (1 tin)} &= \color{green}2(lb + bh + hl) \\ &= 2(26\times 26 + 26\times 45 + 45\times 26) \\ &= 2(676 + 1170 + 1170) \\ &= 2 \times 3016 \\ &= 6032 \ \text{cm}^2 \\[8pt] \text{Area for }20\text{ tins} &= 6032 \times 20 \\ &= 120640 \ \text{cm}^2 \\[6pt] 1 \ \text{m}^2 &= 10000 \ \text{cm}^2 \\[6pt] \text{Area in } \text{m}^2 &= \frac{120640}{10000} \\[6pt] &= 12.064 \ \text{m}^2 \\[6pt] \text{Cost} &= 12.064 \times 10 \\ &= \text{₹ }120.64 \end{aligned} \]Answer Required tin sheet \(= \color{red}{12.064 \ \text{m}^2}\), Cost of tin sheet used \(= \color{red}{\text{₹ }120.64}\)