## Can anyone figure out the area of this hexagon?

(47 Posts)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!

I would do it as an inner rectangle and two triangles/a rectangle of the two sides.

Would that work?

How many squares from corner to corner horizontally?

That's what I would have thought but again, the answer given in the scheme doesn't seem to indicate that. May I ask what you think the "formula" for the area should be?

12, **Knockmedown**

I may be way out here but do you need to divide it into 2 right angled triangles and a rectangle in the centre?

The area of the two triangles added together into a square would be (y x y) and then the rectangle would be calculated by working out the hypotenuse of the right angled triangle using Pythagoras theorem and then multiplying it by x?

2x + 2y.

2(X x Y / 2)

Sorry, mine only works if it’s a square in the middle not a rectangle.

No sorry

2 (X x Y) /2

You’re working out the two trapeziums then adding them together

It's not a regular hexagon, so perhaps that's what has thrown you?

There is a central rectangle that is X * 8 and then there are two right angle triangles that make a square that would be y squared. So 8x + y^2. But I could be completely wrong...

If you say the side of the ‘rectangle’ is z

Would this work-

Y squared + y squared = z squared then the square root of that = z

You can work out the area of the rectangle

Then work out the area of the 2 triangles and add it all together?

It isnt a regular hexagon i.e. All sides equal length so I am not sure you can... I too would treat it as a rectangle and two triangles but can't see that you have enough info for the triangles, you only know the base length. Will watch to see if anyone remembers more geometry than I do!

Remember, this is Year 6 maths! None of the answers suggested so far are "right" (or even offered as options - **Caulk**'s came closest and was the closest to what my DD guessed).

I'll put you out of your misery soon...

There's an App called photomath which helps with working out of a problem . According to a friend with an older DC than mine.

Ah - if they are right angled triangles you can.

If horizontally it's 12 squares, and x is 4 squares, then the hexagon can be divided into 4 squares, each with sides of 4 squares. So the area would be 4x^2. Is that any where near close?

Primary teacher (ex). I loved teaching maths. It’s been a while tho...

Yes, I can see y is part of a right angle, and I did clock that the hexagon wasn't regular.

The answer on the mark scheme was 4x^2. Can anyone see that and explain why?

(And BTW, thanks very much to all for your replies so far.)

Area of the two triangles, plus the area on the rectangle in between.

Sorry **KnockMeDown**, crossed posts with yours - that's the answer!

Yesss!

Do I get a prize? And does my explanation make sense?

Oh you have y? And they are right angled triangles? Then Area (2 triangles) = area (1 square) = y^2 and area(rectangle) = 2x^2. Total volume = 2x^2 + y^2. Is that what they have?

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