## Problem:

A rectangular swimming pool, with a width of 6.1 m and the
cross section as shown in the diagram below, is to be lined with
epoxy paint. If a minimum dry coat thickness of 3.0 mm is
required and the shrinkage upon drying is 25%, how many litres
of paint are needed?

## Solution:

The pool has two trapezoidal sides as shown above. (Called
Area 1).

The pool water on the top is flat, so each end wall will be
rectangular.

Area 2 = "deep end" end wall which will be 2.1 x 6.1
m

Area 3 = "shallow end" which will be 0.6 x 6.1 m

Area 4 = the bottom of the pool. If you look down at it from
above the pool you will see it is a rectangle which is 6.1 m
wide. We will get its length by creating a right angle triangle
in the diagram above. The left side of the triangle will have
length 2.1-0.6 = 1.5 m

The Interior Surface Area will consist of:

(Area 1 x 2) + (Area 2) + (Area 3) + (Area 4)

Area 1 = trapezoid:

Area 2 = Rectangle:

Area 3 = Rectangle:

Find the length of the angled bottom wall of the pool by using
the Pythagorean Theorem.

Area 4 = Rectangle:

Total Area:

Imagine laying all the pieces out flat. You want these covered
uniformly with a very thin layer of epoxy paint. In this case

Thus the volume of paint required before shrinkage
allotment.

Volume=Area x thickness:

Since the paint shrinks by 25% of what you start with,
then 75% of the total required = 341.64L

Total Paint Required: