Trapezium Four
Trapezium Four printable sheet
You may wish to use GeoGebra to explore this problem.
What makes a shape a trapezium?
The diagonals of a trapezium divide it into four parts.

Can you create a trapezium where two of those parts are equal in area?

Can you create a trapezium where three of those parts are equal in area?

Can you create a trapezium where all four parts are equal in area?
In each case can you explain how much freedom you have?
Or, if any of those challenges are impossible, say why you think that.
Well done to Dominic from St Paul's School, Tom from Bristol Grammar School, and Margaret in Cornwall who in different ways pointed out that :
When you put together the purple and blue triangles this larger triangle must have the same area as the larger triangle made from orange and blue because they share the same base and have the same height, therefore must match for area. Then as blue is common to both these triangles the bits that are purple and orange must also be equal in area.
Well done Margaret for pointing out that yellow and blue are similar triangles :
So if, looking at the two parallel side lengths, we said that the longer side is $k$ times the shorter then the blue triangle has $k^2$ times the area of yellow because the blue height is $k$ times bigger than the yellow height.
Looking along a diagonal of the trapezium and comparing yellow with purple they have the same "height away from the diagonal" but purple's "base along that diagonal" is $k$ times as big as yellow's base (those similar triangles again) and so purple's area will be $k$ times as big as the area of yellow.
Comparing all four areas : purple and orange will always be equal, yellow will always be smaller than them while blue will always be larger.
The ratio of areas for yellow, purple, orange and blue is $1 : k : k : k^2$.
Well done to Tom and Dominic for pointing out that the only time anything happens other than just two areas equal is on the occasion that the parallel sides are the same length, the diagonals then intersect at their midpoints, the trapezium at that moment becomes a parallelogram, and the four colours occupy equal areas.
Why do this problem?
This problem has been created to draw attention to area relationships within any trapezium.
Possible approach:
Encourage students to play with the form and extract (propose and support with reasoning) as many properties as possible.
Students may need to draw a variety of trapezia until they have a mental map of the situation. This will give them confidence and help them see the generality within the problem.
Key questions:

Have you made a number of different trapezia and found the area of each of the four triangles? (Dynamic geometry software may be useful here)

What did you find?

What are the cases to be considered in the problem? [all four areas different, two matching, three matching, and all four equal] How would the trapezium have to be to make each case occur?
Possible support:
Students who are not ready for this challenge without preliminary activity might explore a figure made from two parallel lines of length 3cm and 5cm, at a perpendicular distance of 4cm from each other. Lines are drawn between the left end of the 3cm line and the right end of the 5cm, and similarly with the other ends, to create two similar triangles. Students can then explore the ratio between lengths in the figure.
 What can you say about these two triangles?
 What can you say about their areas?
 What can you say about line ratios in this figure?
 How much freedom to move does this figure possess, if the conditions involving the lengths 3cm, 4cm and 5cm are maintained? And how are the answers above altered by that movement?
Possible extension:
What is the ratio of the four triangle areas? start with some specific lengths for the parallel sides and the distance between them if that helps.