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Q3:

To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1m each. 100 flower pots have been placed at a distance of 1m from each other along AD, as shown in Fig. 7.12. Niharika runs $\frac{1}{4}$th the distance AD on the 2nd line and posts a green flag. Preet runs $\frac{1}{5}$th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?

Solution :

Given:

1. The school ground is rectangular, represented by a coordinate plane where the x-axis represents the lines (1 to 10) and the y-axis represents the distance along AD (in meters).

2. Total distance along AD = $100$ meters (since there are 100 flower pots placed 1m apart).

3. Niharika runs on the 2nd line ($x_1 = 2$) and covers $\frac{1}{4}$ of the distance AD.

4. Preet runs on the 8th line ($x_2 = 8$) and covers $\frac{1}{5}$ of the distance AD.

To Find:

1. The distance between the green flag (Niharika) and the red flag (Preet).

2. The coordinates where Rashmi should post her blue flag (the midpoint of the two flags).

A(0,0) B D(0,100) Green Flag (2, 25) Red Flag (8, 20)

Step 1: Determine the coordinates of the flags.

Let the position of the green flag be $G(x_1, y_1)$.

$x_1 = 2$

$y_1 = \frac{1}{4} \times 100 = 25$

So, $G = (2, 25)$.

Let the position of the red flag be $R(x_2, y_2)$.

$x_2 = 8$

$y_2 = \frac{1}{5} \times 100 = 20$

So, $R = (8, 20)$.

Step 2: Calculate the distance between the two flags.

Using the Distance Formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$d = \sqrt{(8 - 2)^2 + (20 - 25)^2}$

$d = \sqrt{(6)^2 + (-5)^2}$

$d = \sqrt{36 + 25}$

$d = \sqrt{61}$

$d \approx 7.81$ meters.

Step 3: Determine the position of the blue flag.

The blue flag is at the midpoint $M(x, y)$ of the line segment joining $G(2, 25)$ and $R(8, 20)$.

Using the Midpoint Formula: $M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$

$x = \frac{2 + 8}{2} = \frac{10}{2} = 5$

$y = \frac{25 + 20}{2} = \frac{45}{2} = 22.5$

Therefore, the blue flag should be posted on the 5th line at a distance of 22.5 meters from AD.

Final Answer: The distance between the two flags is $\sqrt{61}$ m (approx 7.81 m). Rashmi should post her blue flag at the coordinates (5, 22.5).


More Questions from Class 10 Mathematics Coordinate geometry EXERCISE 7.2


CBSE Solutions for Class 10 Mathematics Coordinate geometry


Chapters in CBSE - Class 10 Mathematics


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