Problem Solving. Work Backwards

A good mathematics problem will not have an obvious solution. We must consider what information we have been given and what we already know, and how these things may relate to the problem. If we are systematic in our thinking this will often lead us to the solution.

If we don’t seem to be making progress, we can:

• Read the problem again;
• Reread the information given and focus on key information that could be useful;
• Use what we already now to reduce the scope of the problem;
• Record work done carefully, so it is easy to retrace steps and to verify or change the method.

Examples of working backwards to tackle a problem

First we will read the two examples and have a quick think about them and then we will look at how working backwards can help us with each one:

Angle Problem

The following diagram shows an isosceles triangle and a square drawn on a straight line. Find the size of angle A:

The following diagram is of a square with four semicircles drawn inside. What faction of the square is shaded?

Worked Solutions

Angle Problem

The following diagram shows an isosceles triangle and a square drawn on a straight line. Find the size of angle A:

Solution

We can start by adding in the angles that we know about in the square, marking each of them as 90 degrees. Then we can think about the isosceles triangle and the properties of it that we know about. Using this we can calculate that each of the base angles in this triangle must be 70 degrees. Once we have marked each of these angles on, it should be clearer to see what angle A is.

The following diagram is of a square with four semicircles drawn inside. What faction of the square is shaded?

Solution

We don’t have a formula for finding this specific area and we don’t know any measurements, so we will have to use letters to represent some of the unknown lengths.

Let’s start by looking at one semicircle, the radius of which we will call r:

Using the formula for the area of a circle, we know that the area of this semicircle is ½πr2.

But we don’t want the whole semicircle. We just want the petals. The “white” shape that we need to exclude has a strange shape, but we can tackle the problem instead by excluding a triangle within the semi-circle, which will leave us with “half-petals” from which we can easily find the area of the petals:

Because we have already named the radius of the semicircle as “r”, we can find the area of this “white” triangle, because it has base length 2r and height r, so its area is ½ x 2r x r = r2.

So the two red half petals on the diagram above have a combined area or ½πr2 – r2. There are four times these in the whole diagram, so the total area shaded is 4(½πr2 – r2).

The area of the rectangle is 2r x 2r = 4r2.

So the fraction we want is 4(½πr2 – r2) / 4r2. This simplifies to five ½π – 1, which is about 0.57.

The following questions can be solved in different ways. The worked solutions provided afterwards are based on the ways suggested above.

29 Questions of increasing difficulty

Worked Solutions to Questions