kichu and sachu were on a morning walk on teaching a junction kichu moved exactly east and sachu moved exactly north wards sachu was at a  speed of 30metres per minute than kichu.after 10 minutes the shortest distance between them was 1.5kilometer.draw a rough figure showing the position of kichu and sachu and the junction.calculate the distance Covered by each.find respective speed

Let's break down the problem step by step:

1. **Understanding the Problem:**
   - Kichu and Sachu start from the same junction.
   - Kichu moves eastward, and Sachu moves northward.
   - Sachu's speed is 30 meters per minute faster than Kichu's.
   - After 10 minutes, the shortest distance between them is 1.5 kilometers (1500 meters).

2. **Drawing a Rough Figure:**
   - Draw a junction as a point.
   - Draw a horizontal line towards the right to represent Kichu's path (East).
   - Draw a vertical line upwards to represent Sachu's path (North).
   - After 10 minutes, Kichu and Sachu's positions form a right-angled triangle with the junction.

3. **Let’s Calculate:**
   - Let Kichu's speed be \( v \) meters per minute.
   - Sachu’s speed will then be \( v + 30 \) meters per minute.
   - In 10 minutes:
     - Distance covered by Kichu: \( 10v \) meters.
     - Distance covered by Sachu: \( 10(v + 30) \) meters.

4. **Using the Pythagorean Theorem:**
   - The shortest distance between them is the hypotenuse of the right-angled triangle.
   - So, we have:
   \[
   \text{(Kichu's distance)}^2 + \text{(Sachu's distance)}^2 = (\text{Shortest distance})^2
   \]
   \[
   (10v)^2 + \left[10(v + 30)\right]^2 = 1500^2
   \]
   \[
   100v^2 + 100(v + 30)^2 = 2250000
   \]
   \[
   v^2 + (v + 30)^2 = 22500
   \]
   Expanding and simplifying the equation:
   \[
   v^2 + (v^2 + 60v + 900) = 22500
   \]
   \[
   2v^2 + 60v + 900 = 22500
   \]
   \[
   2v^2 + 60v - 21600 = 0
   \]
   Dividing by 2:
   \[
   v^2 + 30v - 10800 = 0
   \]

   - This is a quadratic equation in the form \( ax^2 + bx + c = 0 \).

5. **Solving the Quadratic Equation:**
   \[
   v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   \]
   Here, \( a = 1 \), \( b = 30 \), and \( c = -10800 \):
   \[
   v = \frac{-30 \pm \sqrt{30^2 + 43200}}{2}
   \]
   \[
   v = \frac{-30 \pm \sqrt{900 + 43200}}{2}
   \]
   \[
   v = \frac{-30 \pm \sqrt{44100}}{2}
   \]
   \[
   v = \frac{-30 \pm 210}{2}
   \]
   So, \( v \) can be:
   - \( v = \frac{180}{2} = 90 \) meters per minute (positive value makes sense).

6. **Distance Covered:**
   - Kichu's distance: \( 10 \times 90 = 900 \) meters.
   - Sachu's speed: \( 90 + 30 = 120 \) meters per minute.
   - Sachu's distance: \( 10 \times 120 = 1200 \) meters.

So, Kichu covered 900 meters, Sachu covered 1200 meters, and their respective speeds are 90 m/min and 120 m/min.

### Rough Figure:
- **Junction:** Point O.
- **Kichu's Position (after 10 minutes):** Point A on the right (east), 900 meters away from O.
- **Sachu's Position (after 10 minutes):** Point B upward (north), 1200 meters away from O.
- **Distance between Kichu and Sachu:** 1.5 km (1500 meters), which is the hypotenuse of triangle OAB.

Would you like to see the detailed figure, or do you have any questions?

### Related Questions:
1. What would happen if Sachu moved westward instead of northward?
2. How would the problem change if the shortest distance were given as 2 km?
3. Can we solve this problem using trigonometry instead of the Pythagorean theorem?
4. What if Kichu's speed was 30 meters per minute faster than Sachu's?
5. How would you solve this problem if only the time difference between them was given?

### Tip:
When solving problems involving distance, time, and speed, using the Pythagorean theorem is helpful for determining the shortest distance in right-angle scenarios.

Explore how to solve the problem of Kichu and Sachu's morning walk, calculating their distances covered and speeds. After 10 minutes, they are 1.5 kilometers apart. Includes step-by-step solution using the Pythagorean theorem and quadratic equation. Suitable for advanced high school students.

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