DAV Class 8 Maths Chapter 14 Worksheet 3
Mensuration Worksheet 3
1. Find the length of the side of a cube whose total surface area measures \( 600 \, \text{cm}^2. \)
Solution
\[ \begin{align*} \text{Let side of cube} & = l \\ \text{Total surface area of the cube} &= 600 \, cm^2 \\ \color{green} \text{Total surface area of the cube} &= \color{green} 6\,l^2 \\ 6\,l^2 &= 600 \\ l^2 &= \frac{\cancel{600}^{100}}{\cancel{6}_1} \\ l^2 &= 100 \\ l &= \sqrt{100} \\ l &= 10 \, cm \\ \end{align*} \]
Answer Side of a cube \( = \color{red} 10 \, cm\)
2. The length, breadth, and height of a room are 5 m, 4 m, and 3 m respectively. Find the cost of white washing the inner walls and ceiling at the rate of ₹ 50 per square metre.
Solution
\[ \begin{align*} \text{Length} &= 5 \, \text{m} \\ \text{Breadth} &= 4 \, \text{m} \\ \text{Height} &= 3 \, \text{m} \\ \\ & \text{Total area to be white washed} \\ &= \color{green} \text{Area of 4 walls} + \text{Ceiling area} \\ &= [2 \times (l + b) \times h] + (l \times b) \\ &= [2 \times (5 + 4) \times 3] + (5 \times 4) \\ &= [6 \times 9] + 20 \\ &= 54 + 20 \\ Area &= 74 \, \text{m}^2 \\ \\ & \text{Total cost of white washing} \\ &= 74 \times 50 \\ Cost&= \text{₹ }3700 \end{align*} \]
Answer The cost of white washing the inner walls and ceiling \( = \color{red} \text{₹ }3700 \).
3. Find the total surface area of a closed cardboard box of length 0.5 m, breadth 25 cm, and height 15 cm.
Solution
\[ \begin{align*} \text{Length} &= 0.5 \, \text{m} \implies 50 \, \text{cm} \\ \text{Breadth} &= 25 \, \text{cm} \\ \text{Height} &= 15 \, \text{cm} \\ \\ \color{green} \text{Total surface area} &= \color{green} 2(lb + bh + hl) \\ &= 2(50 \times 25 + 25 \times 15 + 15 \times 50) \\ &= 2(1250 + 375 + 750) \\ &= 2 \times 2375 \\ \text{Total surface area} &= 4750 \, \text{cm}^2 \end{align*} \]
Answer The total surface area of the box \( = \color{red} 4750 \, \text{cm}^2 \).
4. You are given two boxes. Which box will need more paper to cover the whole box?
Solution
\[ \begin{align*} \\ &\color{blue} \text{First box calculation}\\ l &= 50 \, \text{cm} \\ b &= 40 \, \text{cm} \\ h &= 30 \, \text{cm} \\ & \color{green} \text{Total surface area of cuboid} \\ &= \color{green} 2(lb + bh + hl) \\ &= 2(50 \times 40 + 40 \times 30 + 30 \times 50) \\ &= 2(2000 + 1200 + 1500) \\ &= 2 \times 4700 \\ &= 9400 \, \text{cm}^2 \\ \\ &\color{blue} \text{Second box calculation}\\ l &= 40 \, \text{cm} \\ & \color{green} \text{Total surface area of cube} \\ &= \color{green} 6(l)^2 \\ &= 6 \times (40)^2 \\ &= 6 \times 1600 \\ &= 9600 \, \text{cm}^2 \\ \\ \end{align*} \]
Answer \( \color{red} \text{Second box } \)will need more paper to cover the whole box.
5. The dimensions of an oil tin are \( 26 \, cm \times 26 \, cm \times 45 \, cm \). Find the area of tin sheet required to make 20 such tins.
Solution
\[ \begin{align*} \text{Length} &= 26 \, \text{cm} \\ \text{Breadth} &= 26 \, \text{cm} \\ \text{Height} &= 45 \, \text{cm} \\ \\ & \color{green} \text{Total surface area of one tin} \\ & = \color{green} 2(lb + bh + hl) \\ &= 2(26 \times 26 + 26 \times 45 + 45 \times 26) \\ &= 2(676 + 1170 + 1170) \\ &= 2 \times 3016 \\ &= 6032 \, \text{cm}^2 \\ \\ & \text{Total area for 20 tins} \\ & = 20 \times (\text{Total surface area of one tin}) \\ &= 20 \times 6032 \\ &= 120640 \, \text{cm}^2 \\ \\ \implies 1 \, \text{m}^2 &= 10000 \, \text{cm}^2 \\ &= 12.064 \, \text{m}^2 \, \end{align*} \]
Answer The total area of tin sheet required for 20 tins is \( = \color{red} 120640 \, \text{cm}^2 \text{ (or) } 12.064 \, \text{m}^2 \).
6. A swimming pool is 20 m in length, 15 m in breadth, 4 m in depth. Find the cost of cementing its floor and walls at ₹ 35 per \( m^2 \).
Solution
\[ \begin{align*} \text{Length} &= 20 \, \text{m} \\ \text{Breadth} &= 15 \, \text{m} \\ \text{Depth} &= 4 \, \text{m} \\ \\ & \text{Total area to be cemented} \\ &= \color{green} \text{Area of 4 walls} + \text{Floor area} \\ &= [2 \times (l + b) \times h] + (l \times b) \\ &= [2 \times (20 + 15) \times 4] + (20 \times 15) \\ &= [2 \times 35 \times 4] + 300 \\ &= [70 \times 4] + 300 \\ &= 280 + 300 \\ Area &= 580 \, \text{m}^2 \\ \\ &\text{Total Cost of cementing} \\ &= 580 \times 35 \\ Cost &= \text{₹ } 20300 \end{align*} \]
Answer The cost of cementing the floor and walls \( = \color{red} \text{₹ } 20300 \).
7. A cubical box with lid has a length of 30 cm. Find the cost of painting the inside and outside of the box at ₹ \( 5.50 \text{ per } m^2 \).
Solution
\[ \begin{align*} \text{Length} &= 30 \, \text{cm} \\ & = 0.3 \, \text{m} \\ \\ &\text{Total surface area of the cube (inside + outside)} \\ &= 2 \times 6 \times l^2 \\ &= 2 \times 6 \times (0.3)^2 \\ &= 2 \times 6 \times 0.09 \\ &= 6 \times 0.18 \\ &= 1.08 \, \text{m}^2 \\ \\ &\text{Cost of painting} \\ &= 1.08 \times 5.50 \\ &= \text{₹ } 5.94 \end{align*} \]
Answer The cost of painting the inside and outside of the box \( = \color{red} \text{₹ } 5.94 \).
8. Two cubes of side 4 cm are fixed together. Find the total surface area of the new solid formed.
Solution
\[ \begin{align*} \text{Side of each cube} & = 4 \, \text{cm} \\ \text{When two cubes } & \text{are joined, new solid (cuboid) is formed} \\ Length &= 8 \, \text{cm} \\ Breadth &= 4 \, \text{cm} \\ Height &= 4 \, \text{cm} \\ & \color{green} \text{Total surface area of cuboid} \\ & = \color{green} 2(lb + bh + hl) \\ &= 2(8 \times 4 + 4 \times 4 + 4 \times 8) \\ &= 2(32 + 16 + 32) \\ &= 2 \times 80 \\ &= 160 \, \text{cm}^2 \\ \end{align*} \]
Answer The total surface area of the new solid \( = \color{red} 160 \, \text{cm}^2 \).