1. A rectangular garden \( \color{black} 30m \) long and \(\color{black} 25m \) wide has a path \( \color{black} 2.5m \) wide running all around inside it. Find the area of the path.
Solution
Let ABCD be the rectangular garden and EFGH be the internal boundries of the path.
\[ \begin{align*} \\ (Length) \,AB & = 30m \\ (Breadth) \,BC & = 25m \\ \\ \text{Area of the } & \text{rectangle ABCD} \\ &= length \times breadth \\ &= 30m \times 25m \\ &= \color{green} 750m^2 \\ \\(Length) \,EF &= 30 - (2.5 + 2.5) \\ &= 30 - 5 \\ &= 25m \\ \\ (Breadth) \,FG &= 25 - (2.5 + 2.5) \\ & = 25 - 5 \\ & = 20m \\ \\\text{Area of the } & \text{rectangle EFGH} \\ &= length \times breadth \\ &= 25m \times 20m \\ &= \color{green} 500m^2 \\ \\ \text{Area of the path } &= (\text{Area of ABCD}) - (\text{Area of EFGH}) \\ &= 750m^2 - 500m^2 \\ &= \color{green} 250m^2 \\ \end{align*} \]
Answer Area of the path \( = \color{red} 250m^2 \)
2. A path \( \color{black} 1 \, \text{m} \) wide is built inside a square park of side \( \color{black} 30 \, \text{m} \) along its sides. Find the area of the path. Calculate the cost of constructing the path at the rate of \( \color{black} \text{₹} 70 \text{ per m}^2 \).
Solution
Let ABCD be the square park and EFGH be the internal boundaries of the path.
\[ \begin{align*} \\ (Side ) \, AB & = 30 \, \text{m} \\ \\ \text{Area of the } & \text{square park ABCD} \\ &= \text{side} \times \text{side} \\ &= 30 \, \text{m} \times 30 \, \text{m} \\ &= \color{green} 900 \, \text{m}^2 \\ \\ (Side) \, EF &= 30 - (1 + 1) \\ &= 30 - 2 \\ &= 28 \, \text{m} \\ \\ \text{Area of the } & \text{inner square EFGH} \\ &= \text{side} \times \text{side} \\ &= 28 \, \text{m} \times 28 \, \text{m} \\ &= \color{green} 784 \, \text{m}^2 \\ \\ \text{Area of the path } &= (\text{Area of square ABCD}) - (\text{Area of square EFGH}) \\ &= 900 \, \text{m}^2 - 784 \, \text{m}^2 \\ &= \color{green} 116 \, \text{m}^2 \\ \end{align*} \]
\[ \begin{align*} \text{Cost of constructing the path} &= \text{Area of the path} \times \text{Rate per } \text{m}^2 \\ &= 116 \times 70 \\ &= \text{₹} 8120 \\ \end{align*} \]
Answer Area of the path \( = \color{red} 116 \, \text{m}^2 \), Cost of constructing the path \( = \color{red}\text{₹} 8120 \)
3. A table cover \( \color{black} 5m \times 3m\), is spread on a meeting table. If \( \color{black} 25 \, \text{cm} \) of the table cover is hanging all around the table, find the area of the table top.
Solution
Let ABCD be the dimensions of the table cover and EFGH be the dimensions of the table top.
\[ \begin{align*} \\ \text{Convert } &\text{cm to m } \\ 1cm &= \frac{1}{100}m \\ \\ 25cm &= \frac{25}{100}m \\ \\ 25cm & = 0.25m \\ \\ (Length) \,EF &= 5 - (0.25 + 0.25) \\ &= 5 - 0.5 \\ &= 4.5m \\ \\ (Breadth) \,FG &= 3 - (0.25 + 0.25) \\ & = 3 - 0.5 \\ & = 2.5 m \\ \\\text{Area of the } & \text{table top EFGH} \\ &= \text{length} \times \text{breadth} \\ &= 4.5 \, \text{m} \times 2.5 \, \text{m} \\ &= \color{green} 11.25 m^2 \\ \\ \end{align*} \]
Answer Area of the table top \( = \color{red} 11.25m^2 \)
4. A garden is \( \color{black} 120m \) long and \( \color{black} 85m \) wide. It has an inside path of uniform width \( \color{black} 3.5m \) all around it. The remaining part of the garden is covered by grass. Find the cost of covering the garden by grass at \( \color{black} 50 \, paise \) per square metre.
Solution
Let ABCD be the garden and EFGH be the inner part of the garden covered by grass. \[ \begin{align*} AB & = 120m \\ BC & = 85m \\ Width &= 3.5m \\ \\ \text{Area of the } & \text{grass EFGH} \\ (Length) \,EF &= 120 - (3.5 + 3.5) \\ &= 120 - 7 \\ &= 113m \\ \\ (Breadth) \,FG &= 85 - (3.5 + 3.5) \\ & = 85 - 7 \\ & = 78m \\ \\\text{Area}&= \text{length} \times \text{breadth} \\ &= 113m \times 78m \\ &= \color{green} 8814m^2 \\ \\ \end{align*} \]
\[ \begin{align*} \text{Cost of covering EFGH by grass} &= \text{Area } \times \text{Rate per } \text{m}^2 \\ &= 8814 \times 0.50 \\ &= \text{₹} 4407 \\ \end{align*} \]
Answer Cost of covering the garden by grass \( = \color{red}\text{₹} 4407 \)
5. Calculate the area of the shaded region in the diagram
Solution
Let ABCD be the outer rectange and EFGH be inner rectange.
\[ \begin{align*} (Length) \,AB & = 43m \\ (Breadth) \,BC & = 27m \\ \\ \text{Area of the } & \text{outer rectangle ABCD} \\ &= length \times breadth \\ &= 43m \times 27m \\ &= \color{green} 1161m^2 \\ \\(Length) \,EF &= 43 - (1.5 + 1.5) \\ &= 43 - 3 \\ &= 40m \\ \\ (Breadth) \,FG &= 27 - (1.5 + 1.5) \\ & = 27 - 3 \\ & = 24m \\ \\\text{Area of the } & \text{inner rectangle EFGH} \\ &= length \times breadth \\ &= 40m \times 24m \\ &= \color{green} 960m^2 \\ \\ \text{Area of the shaded region} &= (\text{Outer area ABCD}) - (\text{Inner area EFGH}) \\ &= 1161m^2 - 960m^2 \\ &= \color{green} 201m^2 \\ \end{align*} \]
Answer Area of the shaded region \( = \color{red} 201m^2 \)
6. A rectangular lawn measuring \( \color{black} 50m \) by \( \color{black} 36m \) is surrounded externally by a path which is \( \color{black} 2m \) wide. Find the area of the path. Also, calculate the cost of levelling of the path at \( \text{₹} 16 \text{ per m}^2 \).
Solution
\[ \begin{align*} \\ \text{Let EFGH be the } & \text{rectangular lawn} \\ (Length) \,EF &= 50m \\ (Breadth) \,FG &= 36m \\ \\ \text{Area of the } & \text{rectangular lawn EFGH} \\ &= \text{length} \times \text{breadth} \\ &= 50m \times 36m \\ &= \color{green} 1800m^2 \\ \\ \text{Let ABCD be the } & \text{outer boundries} \\(Length) \,AB & = 50 + 2 + 2 \\ & = 54m \\ (Breadth) \,BC & = 36 + 2 + 2 \\ & = 40m \\ \\\text{Area of the } & \text{outer rectangle ABCD} \\ &= \text{length} \times \text{breadth} \\ &= 54m \times 40m \\ &= \color{green} 2160m^2 \\ \\ \text{Area of the path } &= (\text{Outer area ABCD}) - (\text{Inner area EFGH}) \\ &= 2160m^2 - 1800m^2 \\ &= \color{green} 360m^2 \\ \end{align*} \]
\[ \begin{align*} \text{Cost of levelling the path} &= \text{Area of the path} \times \text{Rate per } \text{m}^2 \\ &= 360 \times 16 \\ &= \text{₹} 5760 \\ \end{align*} \]
Answer Area of the path \( = \color{red} 360m^2 \), Cost of levelling the path \( = \color{red}\text{₹} 5760 \)
7. A painting \( \color{black} 40 \, \text{cm} \) long and \( \color{black} 28 \, \text{cm} \) wide is painted on a cardboard such that there is a margin of \( \color{black} 2 \, \text{cm} \) along each of its sides. Find the total area of the margin.
Solution
Let EFGH be the dimensions of the painting and ABCD be the dimensions of the cardboard.
\[ \begin{align*} \text{Let EFGH be the } & \text{painting area} \\ (Length) \,EF &= 40 \, \text{cm} \\ (Breadth) \,FG &= 28 \, \text{cm} \\ \\ \text{Area of the } & \text{painting EFGH} \\ &= \text{length} \times \text{breadth} \\ &= 40 \, \text{cm} \times 28 \, \text{cm} \\ &= \color{green} 1120 \, \text{cm}^2 \\ \\ \text{Let ABCD be the } & \text{cardboard area} \\ (Length) \,AB & = 40 + 2 + 2 \\ &= 44cm \\ (Breadth) \,BC & = 30 + 2 + 2 \\ &= 32cm \\ \\\text{Area of the } & \text{cardboard ABCD} \\ &= \text{length} \times \text{breadth} \\ &= 44 \, \text{cm} \times 32 \, \text{cm} \\ &= \color{green} 1408 \, \text{cm}^2 \\ \\ \text{Area of the margin } &= (\text{Area of cardboard ABCD}) - (\text{Area of painting EFGH}) \\ &= 1408 \, \text{cm}^2 - 1120 \, \text{cm}^2 \\ &= \color{green} 288 \, \text{cm}^2 \\ \end{align*} \]
Answer Total area of the margin \( = \color{red} 288 \, \text{cm}^2 \)
8. Each side of a square flower bed is \( \color{black} 3 \, \text{m} \, 80 \, \text{cm} \) long. It is extended by digging a strip \( \color{black} 50 \, \text{cm} \) wide all around it. Find the area of the enlarged flower bed and also the increase in the area of the flower bed.
Solution
Let EFGH be the square flower bed and ABCD be the enlarged flower bed.
\[ \begin{align*} \text{Let EFGH be the } & \text{square flower bed} \\(Side)EF &= 3 m \, \, 80 cm \\ &= 3.8m \\ \\\text{Area of } & \text{square EFGH} \\ &= side \times side \\ &= 3.8 m \times 3.8 m \\ &= \color{green} 14.44 m^2 \\ \\ \text{Let ABCD be the } & \text{enlarged flower bed} \\ 50cm &= 0.5m \\ (Side)AB &= 3.8 \, \text{m} + 0.5 \, \text{m} + 0.5 \, \text{m} \\ &= 3.8 \, \text{m} + 1 \, \text{m} \\ &= 4.8 \, \text{m} \\ \\ \text{Area of } & \text{square ABCD} \\ &= side \times side \\ &= 4.8 m \times 4.8 m \\ &= \color{green} 23.04 m^2 \\ \\ \text{Increase in area of flower bed } & = (\text{Area of enlarged flower bed}) - (\text{Area of original flower bed}) \\ &= 23.04 - 14.44 \\ &= 8.60 \\ &= \color{green} 8.6 m^2 \\ \end{align*} \]
Answer Area of the enlarged flower bed \( = \color{red} 23.04 m^2 \), Increase in the area of the flower bed \( = \color{red} 8.6 m^2 \)