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Q4:
A gardener wants to fence a circular garden of diameter $21$ m. Find the length of the rope he needs to purchase, if he makes $2$ rounds of fence. Also find the cost of the rope, if it costs ₹ $4$ per meter. (Take $\pi = \frac{22}{7}$)
A gardener wants to fence a circular garden of diameter $21$ m. Find the length of the rope he needs to purchase, if he makes $2$ rounds of fence. Also find the cost of the rope, if it costs ₹ $4$ per meter. (Take $\pi = \frac{22}{7}$)
Solution :
Step 1: Given Variables & Initial Setup
Let us define the geometric and financial parameters provided for the circular garden:
- Diameter of the circular garden, $d = 21 \text{ m}$
- Radius of the circular garden, $r = \frac{d}{2} = \frac{21}{2} \text{ m} = 10.5 \text{ m}$
- Number of rounds of fencing required, $n = 2$
- Cost of the rope per meter, $C_{\text{unit}} = \text{₹ } 4 \text{ per m}$
- Approximation for Pi, $\pi = \frac{22}{7}$
Step 2: Geometric Visualization
The following diagram illustrates the circular garden from a top-down perspective, including the diameter and the two concentric rounds of rope fencing required.
Step 3: Calculating the Circumference of the Garden
The length of rope required for a single round of fencing is exactly equal to the perimeter (circumference) of the circular garden. [Per the geometric definition of a circle's circumference, the boundary length is a function of its diameter].
The formula for the circumference $C$ is given by:
$C = \pi d$
Substituting the given values into the equation:
$C = \left(\frac{22}{7}\right) \times 21 \text{ m}$
By simplifying the fraction [dividing $21$ by $7$ yields $3$]:
$C = 22 \times 3 \text{ m} = 66 \text{ m}$
Thus, one complete round around the garden requires $66 \text{ m}$ of rope.
Step 4: Determining the Total Length of Rope Required
Since the gardener intends to make $2$ complete rounds of the fence, the total length of the rope ($L_{\text{total}}$) is the circumference multiplied by the number of rounds. [Total Length = $n \times C$].
$L_{\text{total}} = 2 \times 66 \text{ m}$
$L_{\text{total}} = 132 \text{ m}$
Step 5: Calculating the Total Cost of the Rope
The total financial cost is the product of the total length of the rope and the unit cost per meter. [Total Cost = $L_{\text{total}} \times C_{\text{unit}}$].
$\text{Total Cost} = 132 \text{ m} \times \text{₹ } 4 / \text{m}$
$\text{Total Cost} = \text{₹ } 528$
Final Solution: The total length of the rope required to fence the garden with 2 rounds is $132 \text{ m}$, and the total cost of purchasing the rope is ₹ $528$.
More Questions from Class 9 Mathematics Coordinate Geometry EXERCISE 9.2
- Q1(a): Find the circumference of the circles with the following radius: (Take $\pi = \frac{22}{7}$) (a) $14$ cm
- Q1(b): Find the circumference of the circles with the following radius: (Take $\pi = \frac{22}{7}$) (b) $28$ mm
- Q1(c): Find the circumference of the circles with the following radius: (Take $\pi = \frac{22}{7}$) (c) $21$ cm
- Q10: From a circular card sheet of radius $14$ cm, two circles of radius $3.5$ cm and a rectangle of length $3$ cm and breadth $1$cm are removed. (as shown in the adjoining figure). Find the area of the remaining sheet. (Take $\pi = \frac{22}{7}$)
- Q11: A circle of radius $2$ cm is cut out from a square piece of an aluminium sheet of side $6$ cm. What is the area of the left over aluminium sheet? (Take $\pi = 3.14$)
- Q12: The circumference of a circle is $31.4$ cm. Find the radius and the area of the circle? (Take $\pi = 3.14$)
- Q13: A circular flower bed is surrounded by a path $4$ m wide. The diameter of the flower bed is $66$ m. What is the area of this path? ($\pi = 3.14$)
- Q14: A circular flower garden has an area of $314$ m$^2$. A sprinkler at the centre of the garden can cover an area that has a radius of $12$ m. Will the sprinkler water the entire garden? (Take $\pi = 3.14$)
- Q15: Find the circumference of the inner and the outer circles, shown in the adjoining figure? (Take $\pi = 3.14$)
- Q16: How many times a wheel of radius $28$ cm must rotate to go $352$ m? (Take $\pi = \frac{22}{7}$)
- Q17: The minute hand of a circular clock is $15$ cm long. How far does the tip of the minute hand move in $1$ hour. (Take $\pi = 3.14$)
- Q2(a): Find the area of the following circles, given that: (a) radius = $14$ mm (Take $\pi = \frac{22}{7}$)
- Q2(b): Find the area of the following circles, given that: (b) diameter = $49$ m
- Q2(c): Find the area of the following circles, given that: (c) radius = $5$ cm
- Q3: If the circumference of a circular sheet is $154$ m, find its radius. Also find the area of the sheet. (Take $\pi = \frac{22}{7}$)
- Q5: From a circular sheet of radius $4$ cm, a circle of radius $3$ cm is removed. Find the area of the remaining sheet. (Take $\pi = 3.14$)
- Q6: Saima wants to put a lace on the edge of a circular table cover of diameter $1.5$ m. Find the length of the lace required and also find its cost if one meter of the lace costs ₹ $15$. (Take $\pi = 3.14$)
- Q7: Find the perimeter of the adjoining figure, which is a semicircle including its diameter.
- Q8: Find the cost of polishing a circular table-top of diameter $1.6$ m, if the rate of polishing is ₹ $15/m^2$. (Take $\pi = 3.14$)
- Q9: Shazli took a wire of length $44$ cm and bent it into the shape of a circle. Find the radius of that circle. Also find its area. If the same wire is bent into the shape of a square, what will be the length of each of its sides? Which figure encloses more area, the circle or the square? (Take $\pi = \frac{22}{7}$)
CBSE Solutions for Class 9 Mathematics Coordinate Geometry
Chapters in CBSE - Class 9 Mathematics
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