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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 $\pi = 3.14$)
Solution :
Given Variables & Initial Setup
To determine whether the sprinkler can cover the entire garden, we must analyze the spatial properties of both the garden and the sprinkler's reach. We are provided with the following parameters:
- Area of the circular flower garden, $A_g = 314 \text{ m}^2$
- Radius of the sprinkler's coverage, $r_s = 12 \text{ m}$
- Approximation for Pi, $\pi = 3.14$
Step 1: Determining the Radius of the Garden
The garden is circular. We can find its radius ($r_g$) by utilizing the standard formula for the area of a circle [Per the geometric theorem for circular area: $A = \pi r^2$].
Setting up the equation with the given area:
$A_g = \pi r_g^2$
Substituting the known values into the equation:
$314 = 3.14 \times r_g^2$
Isolating $r_g^2$ by dividing both sides by $3.14$ [By the Division Property of Equality]:
$r_g^2 = \frac{314}{3.14}$
$r_g^2 = 100$
Taking the principal square root of both sides to find the radius [Since distance must be a non-negative real number]:
$r_g = \sqrt{100} = 10 \text{ m}$
Step 2: Comparing the Radii
For a sprinkler located at the exact center of a circular garden to water the entire area, the radius of the sprinkler's coverage ($r_s$) must be greater than or equal to the radius of the garden ($r_g$).
$r_s = 12 \text{ m}$
$r_g = 10 \text{ m}$
Comparing the two values:
$12 \text{ m} > 10 \text{ m} \implies r_s > r_g$
Because the sprinkler throws water $2 \text{ m}$ beyond the boundary of the garden, it will successfully cover the entire garden.
Step 3: Verification via Area Comparison (Rigorous Proof)
Alternatively, we can verify this by calculating the total area the sprinkler can cover ($A_s$) and comparing it to the garden's area ($A_g$).
$A_s = \pi r_s^2$
$A_s = 3.14 \times (12)^2$
$A_s = 3.14 \times 144$
$A_s = 452.16 \text{ m}^2$
Since $452.16 \text{ m}^2 > 314 \text{ m}^2$, the area covered by the sprinkler strictly encompasses the area of the garden.
Visual Representation of Coverage
The following scale diagram illustrates the garden (green) and the sprinkler's coverage area (blue dashed line). The scale is $1 \text{ m} = 10 \text{ units}$.
Final Solution: Yes, the sprinkler will water the entire garden. The radius of the garden is $10 \text{ m}$, which is strictly less than the $12 \text{ m}$ radius covered by the sprinkler.
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
- 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.
- 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 $\pi = \frac{22}{7}$)
- 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? ($\pi = 3.14$)
- 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 $\pi = 3.14$)
CBSE Solutions for Class 9 Mathematics Coordinate Geometry
Chapters in CBSE - Class 9 Mathematics
Top Tutors who teach Coordinate Geometry
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