Can anyone figure out the area of this hexagon?

[mommybunny]

I'm trying to help DD10 with maths homework and she has to state the correct formula for the area of this shape:

Although I don't need the answer (just the formula for figuring it out), you can't see the graph lines but assume that the line x is 4 squares long, and from the top x to the bottom there are 8 squares.

The answer given by the book is not what I'd have thought, and I was wondering where I went wrong.

Thanks!

Sort of, but I'm struggling a little with 4 equal squares.

I see. You divide the central rectangle into two squares of area x^2 then the two edges into triangles which are each half of one of those squares so four in total.

Can I ask for a diagram ;-) ?

The y is obviously meant to be a red herring.

The triangle look right angled. Squish the two triangles together to get a square. It's area is y^2.

The area of the rectangle is x times whatever you get from pythagoros - sqrt of 2y^2

So y2 + 2xy2

Y^2(1+2x)

I had to look at this for a while, and I'm a secondary maths teacher! Athough I'm on holiday and may have had some wine.
Draw two vertical lines from the top corners to the bottom ones, the a horizontal line from the left-hand corner to the right-hand one. Each square has an area of x^2. There are two whole ones, and four half squares (making them two whole squares).
Add them together. [proud emoticon] !

Ah.

So would it be fair to guess that it would be unfair to put such a question on an 11+maths exam?

Snap. 4x^2. Passes year 6 maths

An exam if they've taught the thinking is fair enough, a good exam has some challenging questions doesn't it? Bear in mind we haven't thought about geometry for many many years, as opposed to weekly classes.

I do get it now, and timeisnotaline's diagram has helped.

Many thanks again, everyone.

A rectangle and two triangles.

My DD isn't terribly strong in maths, while my maths isn't bad, but I'm thinking that if I'm struggling to get it she stands no chance - better to focus on (and increase the number of) the areas we can get really, really comfortable with and accept that for a question like that, she may just have to make a blind guess.

Just divide it into triangles - there are 4. A=4x180

Ignore me I’m doing angles not area - doh!

That is not yr 6 maths - I teach yr 6 maths and at an absolute push it might class as greater depth/mastery work.
Here are the relevant objectives for he end of yr 6 for this area -

I can calculate the area of parallelograms and triangles

I can investigate relationships between area and perimeter e.g. shapes with the same area can have different perimeters and vice versa
I can recognise when it is possible to use formulae to calculate area

Not sure if anyone is still with me here, but I spent a lot of time thinking about this last night and was wondering if there was any reason the answer couldn't have been 2x2 + y2? 2x2 would cover the internal "squares" and y2 the "square" made by adding the 2 right triangles together?

Yes it could be. From Pythagoras, y2=2x2.

It could also be 2y^2.

Why not try drawing out timeisnotaline's diagram on graph paper, cut out the triangles then let your dd see how they can be put into squares of different sizes?

Oh wow JustRichmal, you have officially blown my mind with Pythagorean explanations. I had never seen Pythagoras like that before (though maybe I should have?)

Respect. . And many thanks.

The answer I guessed above wasn't one of the choices, but the one that others, clearly much cleverer than I, guessed (4x^2) was.

This has been a fascinating discussion. While I think I could seriously get into solving these sorts of problems for my own intellectual enrichment, at the moment I have to teach a DD enough maths to pass an exam. If she will miss at most one or two marks not getting this right, when it has taken some clearly brilliant mathematical minds a fair bit of time, then I think our time is better spent - frankly - on making sure she has percentages totally nailed.

2 trapeziums - area of a trapezium is ((top+base)/2)xheight - we have 2 of them

so area is (top+base)xheight of the trapezium... top =4=x, base =12 =3x, height =4=x

So we have (x+3x) multiplied by x =4x^2

Seems odd that it mixed asking for a formula, not an answer, with some real values for sizes, I would have assumed the formula had to be all in x and y, and not assume the proportions that happen to be on the graph paper....