A rectangle ABCD has a length of 50 m and a width of 25 m - Junior Cycle Mathematics - Question 2 - 2017

To find the perimeter of a rectangle, we use the formula:

$\text{Perimeter} = 2 \times (\text{Length} + \text{Breadth})$

Substituting the values: $\text{Perimeter} = 2 \times (50 \text{ m} + 25 \text{ m}) = 2 \times 75 \text{ m} = 150 \text{ m}$

Thus, the length of the perimeter of rectangle ABCD is 150 m.

To calculate the area of the shaded region AEFD, we can divide it into a rectangle and a triangle.

1. Calculate the area of rectangle AEBF:

   • The dimensions are 10 m by 25 m, hence: $\text{Area}_{AEBF} = 10 \text{ m} \times 25 \text{ m} = 250 \text{ m}^2$

2. Calculate the area of triangle EFD:

   • The base (EF) is 20 m (length from E to F) and the height is 10 m. The area of the triangle is given by: $\text{Area}_{EFD} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 20 \text{ m} \times 10 \text{ m} = 100 \text{ m}^2$

3. Total area of shaded region AEFD: $\text{Area}_{AEFD} = \text{Area}_{AEBF} + \text{Area}_{EFD} = 250 \text{ m}^2 + 100 \text{ m}^2 = 350 \text{ m}^2$

Thus, the area of the shaded region AEFD is 350 m².

In triangle EFD, we apply the Pythagorean theorem:

$c^2 = a^2 + b^2$

Let:

• |EF| be side c,
• The vertical distance (height) ED be a = 20 m,
• The horizontal distance (width) DF be b = 25 m.

Therefore, $|EF|^2 = 20^2 + 25^2 = 400 + 625 = 1025$

Thus, taking the square root: $|EF| = \sqrt{1025} \approx 32.02 \text{ m}$

Rounded to the nearest metre, |EF| = 32 m.

To find the perimeter of the shaded region AEFD, we add the lengths of all its sides:

1. Lengths of AE = 10 m,
2. Lengths EF = 32 m (found above),
3. Lengths FD = 10 m,
4. Lengths AD = 25 m.

Calculating the total perimeter: $\text{Perimeter}_{AEFD} = 10 + 32 + 10 + 25 = 77 \text{ m}$

Thus, the length of the perimeter of the shaded region AEFD is 77 m.