Archive

Posts Tagged ‘Right Angled Triangles’

Concentric Circles Area Puzzle

June 27, 2015 6 comments

This morning I saw this post from Ed Southall (@solvemymaths):

image

And thought, that looks an interesting puzzle. I’ll have a little go. I think you should too, before reading any further…

Ok, so this is how I approached it. First I drew a sketch:

image

I assigned the arbitrary variables r and x to the radii of the larger and smaller circles respectively and used the fact that tangents are perpendicular to right angles, and the symmetry of isosceles triangles, to construct two right angled triangles.

I wrote an expression for the required area in r and x. Used Pythagoras’s Theorem to find an expression for x in terms or r, subbed it in and got the lovely answer of 25pi.

image

An interesting little puzzle, did you solve it the same way? I’d love to hear alternative solutions.

All you need is sine

June 2, 2015 12 comments

Today I was going through an M1 question with a year 13 student and was surprised to see the method he had used. The question involved finding an angle in a right angled triangle given the opposite and adjacent sides. The learner had used Pythagoras’s Theorem to find the hypotenuse then used the sine ratio to find the angle.

Puzzled I questioned further, thinking he may have instinctively found the hypotenuse without fully reading the question then having all 3 sides so going with the first. This turned out not to be the case:

“I know sine equals opposite over hypotenuse innit sir, I have trouble remembering the other ones so I just always use sine.”

This was extra interesting as earlier I had come across a markscheme which suggested the way to resolve a force at an angle of 30 degrees was to use Fsin30 for the vertical and Fsin60 for the horizontal! Further checking showed this learner did that too.

I wasn’t too sure what to make of it. It’s mathematically correct, so there’s no issue there. The learner has a grasp of the other ratios but is more confident with sine so I can see why he would default to that position, although I hope the extra time it takes isn’t an issue tomorrow. I can’t fathom, however, why the markscheme would show it this way in the first instance. (Not the only time a markscheme has confused me recently!)

What do you think? Have you got any quirky methods like this? Have any if your students? Do you have an idea why a markscheme would default to this position? I’d love to hear your response.

%d bloggers like this: