The rectangle ABCD represents a map of an area 30m X 60m
A mobile phone mast, M, is to be placed such that it is equidistant from A and B and 20m from point E, such that BE : EC = 1 : 2.

a/ Draw the map using a scale drawing of 1cm = 5cm
b/ Showing your construction lines clearly, find the shortest distance of M from D

Let's break down the problem step by step to solve it correctly:

### Part a: Drawing the map using a scale of 1 cm = 5 m

1. **Dimensions of the rectangle ABCD**:
   - Length of AB and CD: 60 m
   - Width of AD and BC: 30 m

   Using the scale of 1 cm = 5 m, we can convert the actual dimensions to scaled dimensions:
   - Scaled length of AB and CD: \( \frac{60}{5} = 12 \) cm
   - Scaled width of AD and BC: \( \frac{30}{5} = 6 \) cm

2. **Steps for the drawing**:
   - Draw a rectangle ABCD where:
     - AB = 12 cm
     - AD = 6 cm

### Part b: Finding the position of mast \( M \)

#### Step 1: Placing \( M \) equidistant from points \( A \) and \( B \)
- \( M \) must lie on the perpendicular bisector of line segment \( AB \), as it is equidistant from \( A \) and \( B \).
  - To draw this, construct the perpendicular bisector of \( AB \) by:
    1. Drawing an arc of equal radius from points \( A \) and \( B \), ensuring that the radius is more than half of the length of \( AB \).
    2. The intersection of these arcs above and below \( AB \) defines the perpendicular bisector.

#### Step 2: Finding point \( E \) such that \( BE : EC = 1 : 2 \)
- Divide line segment \( BC \) in the ratio \( 1:2 \).
  - The total length of \( BC \) is 30 m (scaled 6 cm), so:
    - \( BE = \frac{1}{3} \times BC = 10 \) m (scaled 2 cm)
    - \( EC = \frac{2}{3} \times BC = 20 \) m (scaled 4 cm)
  
  - Mark point \( E \) on \( BC \) such that \( BE = 10 \) m and \( EC = 20 \) m.

#### Step 3: Placing \( M \) 20 m from \( E \)
- \( M \) is 20 m from \( E \), which in the scale is \( 20/5 = 4 \) cm.
  - Draw a circle centered at \( E \) with a radius of 4 cm. \( M \) must lie on this circle as well as the perpendicular bisector of \( AB \).

#### Step 4: Intersection of the perpendicular bisector and the circle
- The intersection of the perpendicular bisector of \( AB \) and the circle centered at \( E \) will give the location of point \( M \).

### Part c: Finding the shortest distance from \( M \) to \( D \)
- The shortest distance from \( M \) to point \( D \) is the perpendicular distance from \( M \) to line \( AD \).

  **To find this distance:**
  - Draw a perpendicular from \( M \) to line \( AD \).
  - Measure this distance using the scale (1 cm = 5 m). This will give the shortest distance from \( M \) to \( D \).

Would you like me to clarify any steps, or do you have any questions?

### Follow-up Questions:
1. How would the construction change if \( M \) were equidistant from other points?
2. What if the ratio \( BE:EC \

The rectangle ABCD represents a map of an area 30m X 60m. A mobile phone mast, M, is to be placed such that it is equidistant from A and B and 20m from point E, such that BE : EC = 1 : 2. a/ Draw the map using a scale drawing of 1cm = 5m. b/ Showing your construction lines clearly, find the shortest distance of M from D.

Learn how to solve a geometry problem involving a 30m x 60m rectangle map, a mobile phone mast equidistant from points A and B, and 20m from point E. The solution includes steps to draw a scale map, use the perpendicular bisector, and find the shortest distance of M from point D. Suitable for students in Grades 9-11.

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