1. The diagram shows two paths drawn inside a rectangular field \( \color{black} 50m \) long and \( \color{black} 35m \) wide. The width of each path is 5 m. Find the area of the shaded portion.
Solution
\[ \begin{align*} \\ (Length) \,EF & = 50m \\ (Breadth) \,FG & = 5m \\ \\ \text{Area of the } & \text{path EFGH} \\ &= length \times breadth \\ &= 50m \times 5m \\ &= \color{green} 250m^2 \\ \\(Length) \,IJ & = 5m \\ (Breadth) \,JK & = 35m \\ \\ \text{Area of the } & \text{path IJKL} \\ &= length \times breadth \\ &= 5m \times 35m \\ &= \color{green} 175m^2 \\ \\ \text{Let STUV be comman } & \text{area of both the paths} \\ (Side) ST &= 5m \\ \\\text{Area of the } & \text{square STUV} \\ &= side \times side \\ &= 5m \times 5m \\ &= \color{green} 25m^2 \\ \\ \text{Area of the shaded portions } &= (\text{Area of EFGH}) + (\text{Area of IJKL}) - (\text{Area of STUV}) \\ &= 250m^2 + 175m^2 - 25m^2 \\ &= 425 - 25 \\ &= \color{green} 400m^2 \\ \\ \end{align*} \]
Answer Area of the shaded portion \( = \color{red} 400m^2 \)
2. An oblong garden measures \( \color{black} 60m \) by \( \color{black} 55m \). From the centre of each side, a path \( \color{black} 2 \) metres wide goes across to the centre of the opposite side. Find the area of the paths.
Solution
\[ \begin{align*} \\ \text{Let EFGH and } & \text{IJKL be two paths.} \\ \\(Length) \,EF & = 60m \\ (Breadth) \,FG & = 2m \\ \\ \text{Area of the } & \text{path EFGH} \\ &= length \times breadth \\ &= 60m \times 2m \\ &= \color{green} 120m^2 \\ \\ (Length) \,IJ & = 2m \\ (Breadth) \,JK & = 55m \\ \\ \text{Area of the } & \text{path IJKL} \\ &= length \times breadth \\ &= 2m \times 55m \\ &= \color{green} 110m^2 \\ \\ \text{Let STUV be comman } & \text{area of both the paths} \\ (Side) ST &= 2m \\ \\ \text{Area of the } & \text{square STUV} \\ &= side \times side \\ &= 2m \times 2m \\ &= \color{green} 4m^2 \\ \\ \text{Area of the paths} &= (\text{Area of EFGH}) + (\text{Area of IJKL}) - (\text{Area of STUV}) \\ &= 120m^2 + 110m^2 - 4m^2 \\ &= 230 - 4 \\ &= \color{green} 226m^2 \\ \\ \end{align*} \]
Answer Area of the paths \( = \color{red} 226m^2 \)
3. A rectangular plot of land is \( \color{black} 300m \) long and \( \color{black} 250m \) broad. It has two roads, each \( \color{black} 3m \) wide running midway within it one parallel to the length and the other parallel to the breadth. Find the area of the roads. Also calculate the cost of constructing the roads at \(\color{black} \text{₹} 50\) per square metre.
Solution
\[ \begin{align*} \\ \text{Given } \\ \text{Length of plot } & = 300m \\ \text{Breadth of plot } & = 250m \\ \\ \text{Let EFGH and } & \text{IJKL be two roads.} \\ \\(Length) \,EF & = 300m \\ (Breadth) \,FG & = 3m \\ \\ \text{Area of } & \text{road EFGH} \\ &= length \times breadth \\ &= 300m \times 3m \\ &= \color{green} 900m^2 \\ \\ (Length) \,IJ & = 3m \\ (Breadth) \,JK & = 250m \\ \\ \text{Area of } & \text{road IJKL} \\ &= length \times breadth \\ &= 3m \times 250m \\ &= \color{green} 750m^2 \\ \\ \text{Let STUV be comman } & \text{area of both the roads} \\ (Side) ST &= 3m \\ \\ \text{Area of the } & \text{square STUV} \\ &= side \times side \\ &= 3m \times 3m \\ &= \color{green} 9m^2 \\ \\ \text{Area of the roads} &= (\text{Area of EFGH}) + (\text{Area of IJKL}) - (\text{Area of STUV}) \\ &= 900m^2 + 750m^2 - 9m^2 \\ &= 1650 - 9 \\ &= \color{green} 1641m^2 \\ \\ \text{Cost of constructing the roads} &= \text{Area of the roads} \times \text{Rate per } \text{m}^2 \\ &= 1641 \times 50 \\ &= \text{₹} 82050 \\ \end{align*} \]
Answer Area of the roads \( = \color{red} 1641m^2 \), Cost of constructing the roads \( = \color{red} \text{₹} 82050 \)
4. A rectangular piece of ground is \( \color{black} 45m \) long and \( \color{black} 30m \) wide. It has two roads each \( \color{black} 1m \) wide, running midway within it, one parallel to the length and the other parallel to the breadth. Find the area of the roads. Also calculate the area of the remaining ground.
Solution
\[ \begin{align*} \\ \text{Let ABCD be } & \text{rectangular ground} \\ (Length) \, AB & = 45m \\ (Breadth) \, BC & = 30m \\ \\\text{Area of } & \text{ground ABCD} \\ &= length \times breadth \\ &= 45m \times 30m \\ &= \color{green} 1350m^2 \\ \\\text{Let EFGH and } & \text{IJKL be two roads.} \\ \\(Length) \,EF & = 45m \\ (Breadth) \,FG & = 1m \\ \\ \text{Area of } & \text{road EFGH} \\ &= length \times breadth \\ &= 45m \times 1m \\ &= \color{green} 45m^2 \\ \\ (Length) \,IJ & = 1m \\ (Breadth) \,JK & = 30m \\ \\ \text{Area of } & \text{road IJKL} \\ &= length \times breadth \\ &= 1m \times 30m \\ &= \color{green} 30m^2 \\ \\ \text{Let STUV be comman } & \text{area of both the roads} \\ (Side) ST &= 3m \\ \\ \text{Area of the } & \text{square STUV} \\ &= side \times side \\ &= 1m \times 1m \\ &= \color{green} 1m^2 \\ \\ \text{Area of roads} &= (\text{Area of EFGH}) + (\text{Area of IJKL}) - (\text{Area of STUV}) \\ &= 45m^2 + 30m^2 - 1m^2 \\ &= 75 - 1 \\ &= \color{green} 74m^2 \\ \\ \text{Area of the remaining road} &= (\text{Area of ABCD}) - (\text{Area of raods}) \\ &= 1350 - 74 \\ &= \color{green} 1276m^2 \\ \\ \end{align*} \]
Answer Area of the roads \( = \color{red} 74m^2 \), Area of the remaining road \( = \color{red} 1276m^2 \)