Monday, 17 February 2020

Fun with numbers - right-angle triangles and side length ratios.

Maths problem - you are told that one side of a 'right triangle' (as the Americans call them), is a certain fraction (1/n) of the total perimeter (the length of all three sides of the triangle added together). You have to work out the length of the other sides and/or the area.

For example... you are told that one side (the height) is one-sixth of the total perimeter.

Write down "n = 6" to get started.

The term "n" is hardly used from here on in, what comes up all the time is "n-1" so we might as well treat this as a separate variable, 'modified n', for which I use capital "N".

So write down "N = 5".

The relative length of the height is simply N x 2* 'units'**.

* Note: override rule: if "N" is an odd number, see separate section below. But I think it's easier to just remember one rule which works whether "N" is odd or even.

** Note: 'units' means relative length, not absolute length.

So write down height = N x 2 = 10 units.

The total length of the other two sides = height (10) x N (5) = 50 units.

If this were an isosceles triangle, the other two sides would simply be half that, 25 units each.

In a right triangle, the hypotenuse is longer than the base, and in these problems, the hypotenuse is simply 2 units longer than the base. That's the magic here. I'll have to have a think about why, but for now, just accept that that's how this works.

(UPDATE, I've had a think and explained it (to myself, at least) and it turns out to be pretty non-magical at all. The height is always 2N units and the difference between base and hypotenuse is always 2 units because that's how you set up the equations. Still a handy trick, should you ever need it.)

So take the side length from the theoretical isosceles triangle (25 units), add 1 for hypotenuse (= 26 units) and deduct 1 for base (= 24 units).

That gives you height 10 units, base 24 units, hypotenuse 26 units, total perimeter 60 units, which you can simplify to 5-12-13. If you apply the override rule, you would have got 5-12-13 straightaway.
Override rule example, same facts as above:

n = 6, N = 5, so "N" is odd.

Height = N = 5 units.

Total length of other two sides = height x N (or N^2, if you are that way inclined) = 25 units.

In an isosceles triangle, the other two sides would be 12.5 each. This is a right triangle, so add and deduct 0.5 to find length of hypotenuse and base.

Answer = 5 - 12 - 13.

Remember, this only works if "N" is odd; the rule that height = N x 2 always works, whether "N" is odd or even!
You might be told, or able to work out, that the total perimeter is (say) 72 centimetres (or 90 yards), or whatever, you divide that by the number of units (= 1.2 centimetres/unit or = 1.5 yards/unit) and multiply up again.

So sides would be height 12 cm; base 28.8 cm, hypotenuse 31.2 cm (or 15 yards; 36 yards; 39 yards).
Area is simply base x height x 1/2.

So using above examples, area would be 12 x 28.8 x 1/2 = 172.8 sq cm (or 270 sq yards).
Presh Talwalkar, at the end of this video, reckons you can short-circuit calculating the area if you know the perimeter and "n".

The formula, if you can remember it, is...

(perimeter ^2) x (N-1) ÷ (4N^2 + 4N).

So if n = 6, N = 5 and perimeter is 72 centimetres (or 90 yards), the area is...

72 x 72 x 4 ÷ (100 + 20) = 172.8 sq cm.

90 x 90 x 4 ÷ (100 + 20) = 270 sq yards.
Here endeth.


James Higham said...

Wot, before breakfast?

Mark Wadsworth said...

JH I couldn't sleep very well, so I did the calcs in my head and then looked for a pattern.

Bayard said...

"The relative length of the height is simply N x 2* 'units'**."

I can't see from where you derive that.

Mark Wadsworth said...

B, quadratic equations, which I did in my head. I might do a follow up post on this.