The problem describes a piece of land consisting of a rectangle and a semi-circle. We need to find: (i) The perimeter of the whole land. (ii) The area of the semi-circular sector. (iii) The simplest ratio of the area of the rectangular part to the area of the semi-circular sector. (iv) Sketch a triangle inside the rectangle such that the area of the triangle is 1/8 of the rectangle's area, with one side along AD and the other side along AB.

1. Problem Description

The problem describes a piece of land consisting of a rectangle and a semi-circle. We need to find:

(i) The perimeter of the whole land.

(ii) The area of the semi-circular sector.

(iii) The simplest ratio of the area of the rectangular part to the area of the semi-circular sector.

(iv) Sketch a triangle inside the rectangle such that the area of the triangle is 1/8 of the rectangle's area, with one side along AD and the other side along AB.

2. Solution Steps

(i) Perimeter of the whole land:

The perimeter of the rectangular part is $2(40) + 2(21) = 80 + 42 = 122$ cm.

However, one side of the rectangle coincides with the diameter of the semi-circle, so it is not part of the total perimeter. The length of that side is 21 cm. Thus, the rectangular part contributes $40 + 21 + 40 = 101$ cm to the total perimeter.

The semi-circle has a radius $r = 21$ cm.

The length of the arc of the semi-circle is half the circumference of a full circle: $\frac{1}{2} (2\pi r) = \pi r = 21\pi$ cm.

The total perimeter is then $101 + 21\pi$ cm.

Using the approximation $\pi \approx \frac{22}{7}$, the perimeter is approximately $101 + 21 \times \frac{22}{7} = 101 + 3 \times 22 = 101 + 66 = 167$ cm.

(ii) Area of the sector:

The sector is a semi-circle with radius $r = 21$ cm.

The area of a full circle is given by $A = \pi r^2$.

The area of the semi-circle is half of the area of the circle: $\frac{1}{2} \pi r^2$.

Area of the sector = $\frac{1}{2} \pi (21)^2 = \frac{1}{2} \pi (441) = \frac{441}{2} \pi$ cm$^2$.

Using $\pi \approx \frac{22}{7}$, Area $= \frac{441}{2} \times \frac{22}{7} = \frac{63}{1} \times \frac{11}{1} = 693$ cm$^2$.

(iii) Ratio of the areas:

Area of the rectangle is $40 \times 21 = 840$ cm$^2$.

Area of the sector is $\frac{441}{2} \pi$ cm$^2$, which is approximately $693$ cm$^2$.

The ratio of the areas of the rectangle to the sector is:

$\frac{840}{\frac{441}{2}\pi} = \frac{840}{\frac{441}{2} \times \frac{22}{7}} = \frac{840}{693} = \frac{120}{99} = \frac{40}{33}$.

The ratio is $40:33$.

(iv) Area of rectangle is $40 \times 21 = 840$.

Area of the triangle is $\frac{1}{8} \times 840 = 105$.

Let the length of the triangle along AB be $x$. Then the area of the triangle is $\frac{1}{2} \times 21 \times x = 105$.

$21x = 210$, so $x = 10$.

The triangle has vertices at A, a point 10 cm along AB, and a point on AD.

3. Final Answer

(i) $167$ cm (approximately)

(ii) $693$ cm$^2$ (approximately)

(iii) $40:33$

(iv) The triangle has vertices A, point 10 cm along AB, and point D. The triangle is formed by the coordinates (0,0), (10,0), and (0,21).