The problem describes a plot of land consisting of a trapezium ABCD and a semi-circle DCP. The trapezium is reserved for a supermarket, and the semi-circle for a children's park. The perimeter of the supermarket area (trapezium) is 52m. We are asked to find: (i) the radius of the children's park, (ii) the length of the arc DPC, (iii) the perimeter of the entire land, (iv) the area of the entire land, and (v) a point E on AB such that the area of triangle ADE is half the area of the trapezium. $\pi$ is given as $22/7$.

1. Problem Description

The problem describes a plot of land consisting of a trapezium ABCD and a semi-circle DCP. The trapezium is reserved for a supermarket, and the semi-circle for a children's park. The perimeter of the supermarket area (trapezium) is 52m. We are asked to find: (i) the radius of the children's park, (ii) the length of the arc DPC, (iii) the perimeter of the entire land, (iv) the area of the entire land, and (v) a point E on AB such that the area of triangle ADE is half the area of the trapezium. $\pi$ is given as $22/7$.

2. Solution Steps

(i) Finding the radius of the children's park.

The radius of the semi-circle is the same as the length of DC. The perimeter of the trapezium ABCD is given as 52m. The perimeter is $AD + DC + CB + BA$. We are given $AD = 8m$, $CB = 10m$, and $BA = 20m$. Thus, we have:

$8 + DC + 10 + 20 = 52$

$DC + 38 = 52$

$DC = 52 - 38$

$DC = 14m$

The radius of the children's park is $DC/2 = 14/2 = 7m$.

(ii) Finding the length of the arc DPC.

The length of an arc in a semicircle is given by $\pi r$, where $r$ is the radius. Here, $r = 7m$, and $\pi = 22/7$.

Arc length DPC = $\pi r = (22/7) \times 7 = 22m$.

(iii) Finding the perimeter of the land.

The perimeter of the entire land is given by $AD + AB + BC + \text{arc DPC} = 8 + 20 + 10 + 22 = 60m$.

(iv) Finding the area of the land.

The area of the land is the sum of the area of the trapezium and the area of the semi-circle.

The area of a trapezium is given by:

Area = $\frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$

Area of trapezium ABCD = $\frac{1}{2} \times (8 + 10) \times 20 = \frac{1}{2} \times 18 \times 20 = 180 m^2$.

The area of the semicircle is given by:

Area = $\frac{1}{2} \pi r^2 = \frac{1}{2} \times \frac{22}{7} \times 7^2 = \frac{1}{2} \times \frac{22}{7} \times 49 = \frac{1}{2} \times 22 \times 7 = 11 \times 7 = 77 m^2$.

The area of the entire land is $180 + 77 = 257 m^2$.

(v) Mark the point E on AB with measurements such that the area of triangle ADE is equal to half the area of the trapezium.

The area of trapezium ABCD is $180 m^2$. Half of this area is $90 m^2$.

The area of triangle ADE is $\frac{1}{2} \times AE \times AD = \frac{1}{2} \times AE \times 8 = 4 \times AE$.

We require $4 \times AE = 90$, so $AE = 90/4 = 22.5 m$.

Since AB is 20m, it is not possible to mark a point E on AB such that $AE = 22.5 m$. This means that there is an error in the problem statement or figure.

Since A is at the very left side and B is at the right side, AE has to be from A to E towards the right. AE can't be more than AB, meaning AE can't be greater than 20m.

Since AE is 22.5 m, we should locate E outside AB, which is impossible.

It's likely there's a misunderstanding of what needs to be half the trapezium area, it should be area of triangle ADE = half the trapezium = $180/2 = 90$. Then $1/2 * AE * AD = 90 => 1/2 * AE * 8 = 90 => 4AE = 90 => AE = 90/4 = 22.5$. But AE can be at most 20, so point E will be outside AB. The original problem statement might be asking area of triangle CDE equals half area of the trapezium ABCD.

3. Final Answer

(i) Radius of the children's park: 7m

(ii) Length of the arc DPC: 22m

(iii) Perimeter of the land: 60m

(iv) Area of the land: 257 $m^2$

(v) AE = 22.5m. Point E is not located on segment AB since AE is larger than length AB, thus this instruction does not work.