1. A field is in the shape of a rectangle whose length is 130 m and breadth is 95 m respectively. Find the cost of fencing the field at Rs 20 per metre.
\[ \begin{align*} \text{Length} & = 130 \, \text{m} \\ \text{Breadth} & = 95 \, \text{m} \\ \text{Perimeter of the } & \text{rectangle (P)} \\ & = 2 \times (\text{length} + \text{breadth}) \\ & = 2 \times (130 \, \text{m} + 95 \, \text{m}) \\ & = 2 \times 225 \, \text{m} \\ & = 450 \, \text{m} \\ \\ \text{Cost of fencing one metre} & = Rs.\, 20 \\ \text{Total cost} & = \text{Perimeter} \times \text{Cost per metre} \\ & = 450 \, \text{m} \times Rs.\, 20 \\ & = Rs.\, 9000 \\ \\ \text{Answer: The cost of fencing } & \text{ the field is Rs. 9000.} \end{align*} \]
2. Akshay goes three times around a rectangular field whose length and breadth are 25 m and 19 m respectively. How much distance does he cover?
\[ \begin{align*} \text{Length} & = 25 \, \text{m} \\ \text{Breadth} & = 19 \, \text{m} \\ \text{Perimeter} & = 2 \times (\text{length} + \text{breadth}) \\ & = 2 \times (25 \, \text{m} + 19 \, \text{m}) \\ & = 2 \times 44 \, \text{m} \\ & = 88 \, \text{m} \\ \\ \text{Total distance } & \text{covered by Akshay} \\ & = 3 \times \text{Perimeter} \\ & = 3 \times 88 \, \text{m} \\ & = 264 \, \text{m} \\ \text{Answer:} & \quad \text{Akshay covers 264 m.}\\ \end{align*} \]
3. The length of one side of a square field is 25 m. Find the cost of fencing at the rate of Rs 18 per metre.
\[ \begin{align*} \text{Side of the square field} & = 25 \, \text{m} \\ \text{Perimeter of the square field (P)} & = 4 \times \text{Side} \\ & = 4 \times 25 \, \text{m} \\ & = 100 \, \text{m} \\ \\ \text{Cost of fencing one metre} & = Rs.\, 18 \\ \text{Total cost} & = \text{Perimeter} \times \text{Cost per meter} \\ & = 100 \, \text{m} \times Rs.\, 18 \\ & = Rs.\, 1800 \\ \\ \text{Answer: The cost of fencing } & \text{ the square field is Rs. 1800.} \end{align*} \]
4. Raju ran around a rectangular garden of length 30 m and breadth 15 m. Manu ran around a square field of side 35 m. Who covered more distance and by how much?
\[ \begin{align*} \text{Distance covered } & \text{by Raju} \\ \text{Length} & = 30 \, \text{m} \\ \text{Breadth} & = 15 \, \text{m} \\ \text{Perimeter of rectangle} & = 2 \times (\text{Length} + \text{Breadth}) \\ & = 2 \times (30 \, \text{m} + 15 \, \text{m}) \\ & = 2 \times 45 \, \text{m} \\ & = 90 \, \text{m} \\ \\ \text{Distance covered } & \text{by Manu} \\ \text{Side} & = 35 \, \text{m} \\ \text{Perimeter of square} & = 4 \times \text{Side} \\ & = 4 \times 35 \, \text{m} \\ & = 140 \, \text{m} \\ \\ \text{Manu covered } & \text{ more distance.} \\ \text{Difference} & = 140 \, \text{m} - 90 \, \text{m} \\ & = 50 \, \text{m} \\ \text{Answer: Manu covered } & \text{ 50 m more distance than Raju.} \end{align*} \]
5. Vineet wants to polish the floor of a room which is 140 cm long and 120 cm wide. The charges for polishing the floor is Rs 52 per square metre. What will be the cost of polishing the floor of the room?
\[ \begin{align*} \text{Length} & = 140 \, \text{cm} \\ \text{Width} & = 120 \, \text{cm} \\ \\ \text{Area of } & \text{the rectangle } \\ \text{Area} & = \text{Length} \times \text{Width} \\ & = 140 \, \text{cm} \times 120 \, \text{cm} \\ & = 16800 \, \text{cm}^2 \\ \\ \text{Convert } & {cm}^2 \, \text{to} \, {m}^2\\ 100 \, \text{cm} & = 1 \, \text{m}\\ 1 \, \text{cm} & = \frac{1}{100} \, \text{m}\\ 1 \, \text{cm}^2 & = \frac{1}{10000} \, \text{m}^2\\ \\ \text{Area in m}^2 & = \frac{16800}{10000} \, \text{m}^2 \\ & = \frac{168\cancelto{}{00}}{100\cancelto{}{00}} \, \text{m}^2 \\ & = \frac{168}{100} \, \text{m}^2 \\ & = 1.68 \, \text{m}^2 \\ \\ \text{Cost per square metre} & = Rs.\, 52 \\ \text{Total cost} & = \text{Area in m}^2 \times \text{Cost} \\ & = 1.68 \, \text{m}^2 \times Rs.\, 52 \\ & = Rs.\, 87.36 \\ \\ \text{Answer: The cost of polishing } & \text{the floor is Rs. 87.36} \end{align*} \]
6. A carpet is 6 m long and 4.5 m wide. Find the area of the carpet. Also find the cost of the carpet if it costs Rs 45 per square metre.
\[ \begin{align*} \text{Length} & = 6 \, \text{m} \\ \text{Width} & = 4.5 \, \text{m} \\ \text{Area} & = \text{Length} \times \text{Width} \\ & = 6 \, \text{m} \times 4.5 \, \text{m} \\ & = 27 \, \text{m}^2 \\ \\ \text{Cost per square meter} & = Rs.\, 45 \\ \text{Total cost} & = \text{Area} \times \text{Cost} \\ & = 27 \, \text{m}^2 \times Rs.\, 45 \\ & = Rs.\, 1215 \\ \\ \text{Answer: The area of the carpet is } & \text{ 27 m}^2 \text{ and the cost is Rs. 1215.} \end{align*} \]
7. A square park is to be watered. If one side of the park is 4.2 m, find the area to be watered.
\[ \begin{align*} \text{Side} & = 4.2 \, \text{m} \\ \text{Area of} & \text{the Square}\\ \text{Area} & = \text{Side} \times \text{Side} \\ & = 4.2 \times 4.2 \\ & = 17.64 \, \text{m}^2 \\ \text{Answer: The area } & \text{to be watered is 17.64 m}^2. \end{align*} \]
8. Which playground has bigger area-one that measures 50 m by 40 m or the other that measures 75 m by 20 m?
\[ \begin{align*} \text{Area of }& \text{the First playground} & \\ \text{Length} & = 50 \, \text{m} \\ \text{Breadth} & = 40 \, \text{m} \\ \text{Area} & = \text{Length} \times \text{Breadth} \\ & = 50 \, \text{m} \times 40 \, \text{m} \\ & = 2000 \, \text{m}^2 \\ \\ \text{Area of }& \text{the Second playground} & \\ \text{Length} & = 75 \, \text{m} \\ \text{Breadth} & = 20 \, \text{m} \\ \text{Area} & = \text{Length} \times \text{Breadth} \\ & = 75 \, \text{m} \times 20 \, \text{m} \\ & = 1500 \, \text{m}^2 \\ \\ \text{Answer: First } & \text{playground has bigger area}. \end{align*} \]