Can anyone please help with this maths problem?

[annie987]

My son (year 8) is struggling with this maths problem.
Although some of the sides may be decimals there should be a trick to find the answer and avoid using the decimal sides.
I’m usually able to help him but I’m flummoxed.
Any help gratefully received. Thanks

Ok first thought is 42 and 27 both appear in the 3 times table

Do you take the 8cm length from the 11cm length to fund the short length of the unknown rectangle - 3cm.

That makes the long length of the 42cm2 rectangle 8cm which means the short length must be 6cm (8x6=42)

So the middle rectangle must be 3cmx6cm so 18cm2?

What’s the question?!

What time s ?

Ignore me that would make the last rectangle (square) 6x5. Which is 30cm2.

Hmm that's a tricky one

Yup.

11cm-8cm to give 3cm for the bottom of the middle shape.

Therefore the whole length is 11cm+ (8cm-3cm) = 16cm.

The bigger rectangle is 16-5-3 = 8cm.

As it is 42cmsqu, the upright must be 6cm.

So the smaller rectangle is 6x3cm squared.

Do you have any other info OP? Seems to be missing something

Bernadette what makes you want to do 11-8?

Is it just a case of setting up all the simultaneous equations and solving that way?

I would be dividing the area by sides to try and find the numbers. E.g if the middle length was 3, then 8-3=5. 27/5=5.4 but then 11-3=8 and 8 x 5.4=43.2 so isn't that.
Adjust slightly and try again.

The middle two smaller rectangles are the same size as the larger one so the middle one is 15cm/sq. I think

We want to find 2 values : y the overlapping length that creates the inner rectangle and x, the height of the rectangle. We can then multiply them to get the area of the inner rectangle.

You get two simultaneous equations for the two areas:
(11-x)*y=42 and
(8-x)*y = 27

This becomes
11y-xy=42
8y-xy=27

subtract these simultaneous equations and you get
3y=15
so y=5

Putting y back into the simultaneous eqn
we get
40-5x=27
so x=13/5=2.6
Then xy = 52.6=5*13/5 = 13

Has he covered simultaneous equations?

Because if you take the 8cm line away from the 11cm line it shows what the overlap of the two is.

But 6 x 8 isn't 42 or have I missed something?

Sorry buggered that post up. The answer is 15cm/sq.

Haven't done maths in years but I think it is simultaneous equations and ? = 13

ds2019 👏🏻👏🏻👏🏻

I’ve got a pen and paper out and now need to compare your answer to what I’ve scribbled, I definitely went wrong somewhere!!

I did wonder re simultaneous equations but they haven’t covered them yet.

Surely the height of the rectangles is 4.25 cm as the total area of the two RHS rectangles is 27+7sq cm and the length is 8cm, 34/8=4.25.
Once you know the height you can work out everything else!

I get the middle rectangle to be 21cm squared.
As follows: the 42cm squared must be 6 x 7 cm.
Either the 6cm or 7cm will be the side measurement of the large rectangle.
On the right, the 27cm rectangle, one of these sides will therefore be either 6 or 7cm.
7cm doesn't divide into 27, so it must be 6cm x 4.5cm, with the right hand side line being the 6cm. Then you deduct the 4.5cm from the 8cm noted on the top, which is 3.5cm x the 6cm = 21cm squared.
So

Call the width of all the rectangles a
Call the missing area X
Then you have
11a=43 + X
8a=27 + X
and solve using a simultaneous equation method

Sorry, ignore me, just realised I've done it wrong, the 11cm at the bottom doesn't work out

Oooh, I revisited my simultaneous equations from 1983 . It's 13 (5 x 2.6)

I don't think it means anything that they haven't done simultaneous equations yet. Mine is frequently getting work she hasn't done yet!

If x is the area of the small rectangle, I did
(42+x) / 11 = (27+x) / 8
8x +336 = 11x +297
8x+39 = 11x
39 = 3x
x=13