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A 7 m wide path is to be constructed all around, and outside of a circular field diameter 112 m . Find the cost of constructing the path at \( ₹ 50 \) per sq. m . The area enclosed by a circular plot is 616 sq . m . Find the diameter of the plot. Also find the cost of fencing it at the rate of \( ₹ 7 \) per metre. A rectangular garden is 120 m long and 90 m wide. In the middle of the garden, there is circular tank of radius 21 m . Find the cost of turfing the remaining portion at the rate of \( ₹ 4.50 \) per square metre. The areas of two circles are in the ratio \( 16: 25 \). Find the ratio of their circumferences. A circular park of radius 20 m has a road 5 m wide running around it. Find the cost of metaling

Ask by Hill Coleman. in India
Jan 26,2025

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Answer

**Summary of Solutions:** 1. **Path Construction:** - **Diameter of the field:** 112 m - **Radius of the field:** 56 m - **Radius of the larger circle (field + path):** 63 m - **Area of the path:** 2611.62 sq. m - **Cost of constructing the path:** ₹130,581 2. **Circular Plot Details:** - **Diameter of the plot:** 28 m - **Circumference for fencing:** 88 m - **Cost of fencing:** ₹616 3. **Rectangular Garden Turfing:** - **Remaining area to be turfed:** Approximately 9,414.06 sq. m - **Cost of turfing:** Approximately ₹42,363.27 4. **Circumference Ratio of Two Circles:** - **Ratio of circumferences:** 4 : 5 5. **Metaling Cost:** - **Area of the road:** 706.5 sq. m - **Cost calculation requires the metaling rate per sq. m.** **Note:** For the metaling cost in question 5, please provide the cost rate per square metre to determine the total cost.

Solution

Sure, let's solve each of the problems step by step. --- ### **1. Cost of Constructing a 7 m Wide Path Around a Circular Field** **Given:** - Diameter of the circular field = 112 m - Radius of the field, \( r_1 = \frac{112}{2} = 56 \) m - Width of the path = 7 m - Radius of the larger circle (field + path), \( r_2 = 56 + 7 = 63 \) m - Cost of constructing the path = ₹50 per sq. m **Steps:** 1. **Calculate the area of the larger circle (field + path):** \[ \text{Area}_{\text{larger}} = \pi r_2^2 = \pi \times 63^2 = \pi \times 3969 \approx 3.14 \times 3969 = 12480.66 \text{ sq. m} \] 2. **Calculate the area of the field:** \[ \text{Area}_{\text{field}} = \pi r_1^2 = \pi \times 56^2 = \pi \times 3136 \approx 3.14 \times 3136 = 9869.04 \text{ sq. m} \] 3. **Determine the area of the path:** \[ \text{Area}_{\text{path}} = \text{Area}_{\text{larger}} - \text{Area}_{\text{field}} = 12480.66 - 9869.04 = 2611.62 \text{ sq. m} \] 4. **Calculate the cost of constructing the path:** \[ \text{Cost} = \text{Area}_{\text{path}} \times \text{Cost per sq. m} = 2611.62 \times 50 = ₹130,581 \] **Answer:** The cost of constructing the 7 m wide path is **₹130,581**. --- ### **2. Diameter of a Circular Plot with Area 616 sq. m and Cost of Fencing** **Given:** - Area of the circular plot = 616 sq. m - Cost of fencing = ₹7 per metre **Steps:** 1. **Use the area formula to find the radius:** \[ \text{Area} = \pi r^2 \Rightarrow 616 = \pi r^2 \] Assuming \( \pi = \frac{22}{7} \), \[ 616 = \frac{22}{7} r^2 \Rightarrow r^2 = \frac{616 \times 7}{22} = 196 \Rightarrow r = \sqrt{196} = 14 \text{ m} \] 2. **Find the diameter:** \[ \text{Diameter} = 2r = 2 \times 14 = 28 \text{ m} \] 3. **Calculate the circumference for fencing:** \[ \text{Circumference} = 2\pi r = 2 \times \frac{22}{7} \times 14 = 88 \text{ m} \] 4. **Determine the cost of fencing:** \[ \text{Cost} = 88 \times 7 = ₹616 \] **Answer:** - **Diameter of the plot:** 28 m - **Cost of fencing:** ₹616 --- ### **3. Cost of Turfing the Remaining Portion of a Rectangular Garden with a Circular Tank** **Given:** - Length of the garden = 120 m - Width of the garden = 90 m - Radius of the circular tank = 21 m - Cost of turfing = ₹4.50 per square metre **Steps:** 1. **Calculate the area of the rectangular garden:** \[ \text{Area}_{\text{garden}} = \text{Length} \times \text{Width} = 120 \times 90 = 10,800 \text{ sq. m} \] 2. **Calculate the area of the circular tank:** \[ \text{Area}_{\text{tank}} = \pi r^2 = \pi \times 21^2 = \pi \times 441 \approx 3.14 \times 441 = 1385.94 \text{ sq. m} \] 3. **Determine the remaining area to be turfed:** \[ \text{Remaining Area} = 10,800 - 1,385.94 \approx 9,414.06 \text{ sq. m} \] 4. **Calculate the cost of turfing:** \[ \text{Cost} = 9,414.06 \times 4.50 \approx ₹42,363.27 \] **Answer:** The cost of turfing the remaining portion of the garden is approximately **₹42,363.27**. --- ### **4. Ratio of Circumferences of Two Circles with Areas in the Ratio 16:25** **Given:** - Ratio of areas of two circles = 16 : 25 **Steps:** 1. **Let the areas be \( A_1 = 16x \) and \( A_2 = 25x \).** 2. **Since \( A = \pi r^2 \),** \[ \frac{A_1}{A_2} = \frac{16x}{25x} = \frac{16}{25} = \frac{\pi r_1^2}{\pi r_2^2} \Rightarrow \frac{r_1^2}{r_2^2} = \frac{16}{25} \] Taking square roots, \[ \frac{r_1}{r_2} = \frac{4}{5} \] 3. **Circumference is \( C = 2\pi r \), so the ratio of circumferences is:** \[ \frac{C_1}{C_2} = \frac{2\pi r_1}{2\pi r_2} = \frac{r_1}{r_2} = \frac{4}{5} \] **Answer:** The ratio of their circumferences is **4 : 5**. --- ### **5. Cost of Metaling a Road Around a Circular Park** **Given:** - Radius of the circular park = 20 m - Width of the surrounding road = 5 m - **Note:** The cost rate for metaling is not provided in the question. **Steps:** 1. **Calculate the radius of the larger circle (park + road):** \[ r_2 = 20 + 5 = 25 \text{ m} \] 2. **Calculate the area of the larger circle:** \[ \text{Area}_{\text{larger}} = \pi r_2^2 = \pi \times 25^2 = \pi \times 625 \approx 3.14 \times 625 = 1962.5 \text{ sq. m} \] 3. **Calculate the area of the park:** \[ \text{Area}_{\text{park}} = \pi r_1^2 = \pi \times 20^2 = \pi \times 400 \approx 3.14 \times 400 = 1256 \text{ sq. m} \] 4. **Determine the area of the road:** \[ \text{Area}_{\text{road}} = \text{Area}_{\text{larger}} - \text{Area}_{\text{park}} = 1962.5 - 1256 = 706.5 \text{ sq. m} \] 5. **Calculate the cost of metaling:** - **Missing Information:** The cost per square metre for metaling is not provided. To determine the total cost, you need the rate (e.g., ₹X per sq. m). **Formula:** \[ \text{Cost} = \text{Area}_{\text{road}} \times \text{Cost per sq. m} \] **Answer:** To calculate the cost of metaling the road around the park, we need the cost rate per square metre. If you provide the rate, the total cost can be determined using the formula: \[ \text{Cost} = 706.5 \times \text{(Cost per sq. m)} \] --- If you have the cost rate for metaling the road in question 5, please provide it, and I can help you complete the calculation!

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To calculate the cost of the circular path, first determine the total area of the circular field with the path. The radius of the field is 56 m (diameter of 112 m) and with the path, it becomes 63 m. The area of the larger circle is \( π \times (63^2) \), and the area of the circular field is \( π \times (56^2) \). The difference gives the area of the path, which you can then multiply by \( ₹ 50 \) to find the total cost. For the circular plot with an area of 616 sq. m, the formula for the area \( A \) of a circle is \( A = π \times r^2 \). By rearranging the formula to solve for the radius \( r \), we find the radius to be \( 7 \) m, hence the diameter is \( 14 \) m. To calculate the fencing cost, multiply the circumference \( 2πr \) by \( ₹ 7 \). Now, if you’re turfing the rectangular garden, you would find the area of the rectangular space and subtract the area of the circular tank. The remaining area is multiplied by \( ₹ 4.50 \) per square meter for turfing. Lastly, when comparing the circumferences of two circles with areas in a \( 16: 25 \) ratio, remember that the circumference ratio will be the square root of the area ratio, resulting in a ratio of \( 4: 5 \). Oh, and don't forget that for the circular park, you'll first find the area of the larger circle (radius 25 m) and subtract the area of the original circle (radius 20 m) to estimate the costs involved in metaling the road around it!

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