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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 $\pi = 3.14$)
Solution :
Given Variables & Initial Setup
We are tasked with finding the total cost of polishing the top surface of a circular table. The following parameters are provided:
- Diameter of the circular table-top ($d$) = $1.6 \text{ m}$
- Rate of polishing ($R$) = $₹ 15 \text{ per m}^2$
- Constant approximation for $\pi$ = $3.14$
Step 1: Determining the Radius of the Circular Table-Top
The standard formula for the area of a circle requires the radius ($r$). The radius is defined as exactly half of the diameter [Per the fundamental geometric properties of a circle].
$r = \frac{d}{2}$
$r = \frac{1.6 \text{ m}}{2} = 0.8 \text{ m}$
Step 2: Calculating the Area of the Circular Region
Polishing is a surface-level operation, meaning we must calculate the two-dimensional area of the table-top. We apply the area formula for a circle [Derived from the integration of concentric circular rings, $A = \pi r^2$].
$A = \pi r^2$
Substituting the given values into the equation:
$A = 3.14 \times (0.8 \text{ m})^2$
First, square the radius:
$(0.8)^2 = 0.8 \times 0.8 = 0.64 \text{ m}^2$
Next, multiply by $\pi$:
$A = 3.14 \times 0.64 \text{ m}^2$
$A = 2.0096 \text{ m}^2$
The total area to be polished is exactly $2.0096 \text{ m}^2$.
Step 3: Computing the Total Cost of Polishing
The total cost ($C$) is the product of the total area and the rate of polishing per square meter. [Dimensional analysis confirms this operation: $\text{m}^2 \times \frac{₹}{\text{m}^2} = ₹$].
$C = \text{Area} \times \text{Rate}$
$C = 2.0096 \text{ m}^2 \times 15 \frac{₹}{\text{m}^2}$
Performing the multiplication:
$C = 30.144 \text{ ₹}$
In standard currency formatting, values are typically rounded to two decimal places (the nearest paisa). Therefore, $30.144$ rounds to $30.14$.
Final Solution: The total cost of polishing the circular table-top is ₹ 30.14.
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 $\pi = \frac{22}{7}$)
- 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 $\pi = \frac{22}{7}$)
- 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.
- 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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