##
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 = *½*bh*.

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 = *½*bh*
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 = *½*bh*.

##
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 ½

*b* will be

(*x* + *b*) divided by ½

*b*, 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

denotes the smallest
integer greater than or equal to

*a*.

In the right triangle with
base *x*,

and thus

Therefore

As a check, in the example used above to illustrate the
process,

*b* = 2,

*h* = 6 and tan

*Î¸* = -3/2

.
The diagram confirms this, each of the parts H, F, A, D, G
and I
constituting one strip.

##
References

Densmore, D (ed., 2002). *Euclid’s Elements: all thirteen books
complete in one volume: the Thomas L. Heath translation*. Santa Fe: Green
Lion Press.

Pritchard, C (2016): *A Square Peg in a Round Hole: Adventures in the Mathematics of Area*.
Leicester: The Mathematical Association.