TEACHING MATHS TEACHING

Kinsler Crosses

In his excellent book A Square Peg in a Round Hole: Adventures in the Mathematics of Area, Dr Chris Pritchard introduces an idea he calls ‘Kinsler’s Cross’ after the colleague who introduced it to him.

This article provides a detailed walkthrough of how Kinsler crosses can be applied to find the area of a triangle and thereby justify the traditional formula A = ½bh.

The cross ‘works’ as follows: inside the figure whose area is to be calculated, draw two perpendicular line segments, as shown in this diamond:

The idea is to find positions for the line segments such that their product is equal to the area of the figure: in other words, such that the area of the yellow product rectangle below is equal to the area of the diamond:

This obviously isn’t the right position in the case shown above: there’s more of the rectangle outside the diamond than there is unfilled area within it.

Crosses in triangles

For a triangle, you need to drop a perpendicular from one vertex to the opposite side (one of the triangle’s altitudes) to form one segment of the cross:

Here are a few possible positions for the second segment, with the product rectangles shown:

The first clearly gives an area that is too small, while the second is too large. It turns out that when the second segment joins the midpoints M and N of the two sides it touches, the area is correct.

This is easy to demonstrate by dissecting the rectangle and rotating some of the parts. First, split the rectangle into three parts as shown below:

Part A is completely inside the triangle, so it does not need to be moved. To show it has become fixed, I’ve coloured it pink. The two ‘exterior’ parts of the rectangle are now labelled B and C:

Next, rotate B through a half-turn about midpoint M:

Now rotate C about N in the same way:

The whole of the rectangle is now contained within the triangle.

Now, the second segment of the cross joins the midpoints of two sides of the triangle:

The smaller triangle is similar to the original and half the size, meaning that MN is half the base of the original triangle. The area of the product rectangle is thus ‘half the base’ times ‘the height’.

The mathematical meaning of A = ½bh is usually taken as ‘half of (the base × the height)’, which is justified by doubling the triangle to make a rectangle. To remind yourself how this works, use the altitude that formed the first segment of the cross to split a copy of the original triangle into two right triangles, as shown here:

Now rotate both of the right triangles through a half-turn:

Finally, match up the hypotenuses of the right triangles with the corresponding sides of the original triangle:

This forms a rectangle whose area is twice that of the original triangle.

The base of this rectangle is equal to the base of the triangle, and it has the same height as the triangle. Hence, ‘half of (the base × the height)’.

However, students will often quote the formula as ‘half the base’ times ‘the height’, and will even calculate with it in this sense. The Kinsler cross provides an appropriate justification for this perfectly reasonable parsing of A = ½bh.

Obtuse triangles

The Kinsler cross method can be used in any obtuse triangle in the straightforward manner illustrated above, by choosing the longest side of the triangle as base (this is the one opposite the obtuse angle, by Proposition I.19 of Euclid’s Elements). However, it’s also possible to demonstrate how the process works using the ‘horizontal’ base, as follows. The dissection is more complicated, but the result is very satisfying!

I start with this configuration:

The rule that the altitude has to pass through a vertex makes it impossible to see a cross at all, let alone a product rectangle; however, if I translate the altitude to the midpoint of the second segment, the situation becomes more obvious. Does it look as if the rectangle has the same area as the triangle?

The process starts by identifying that part of the rectangle that is currently within the triangle. This is the trapezium marked A below. Then the protruding portions of the rectangle are labelled B and C (they are both right-angled trapezia). The midpoints are labelled M and N, as before.

Next, rotate B and C through half-turns about N and M respectively:

The parts of B and C that now lie within the triangle are labelled D and F, respectively. This leaves two protruding parts: the right triangle E and trapezium G.

Now rotate E and G by half-turns about M and N, respectively:

G is now entirely contained within the triangle; E needs to be split into triangle H, which is inside the triangle, and right triangle I, which is outside.

Finally, rotate I through a half-turn about N:

All of the parts made by dissecting the rectangle are now inside the triangle. Just to show that the rectangle really can be recreated from the pieces, the second diagram below shows what they look like when reassembled:

This configuration of the constituent parts yields some interesting relationships. In the diagram below, line segments that are parallel to the sides of the triangle have been identified. Note the intersections of the produced segments with the lower vertices of the triangle. Also, segments that are parallel to each other are necessarily of the same length.

The number of times this dissection/rotation process has to be carried out will depend on the ratio of height to base of the rectangle, and also the size of the obtuse angle. To analyse this, it will be easier to think of the triangle being dissected to form the rectangle, rather than the other way around.

In the diagram above, the number of strips n into which the triangle will have to be divided to form a rectangle of width ½b will be (x + b) divided by ½b, rounded up to give a whole number, as there is no reason why this should be an integer. Note that the following lines use the ceiling function: the notation