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

I got 13 but did it a long winded way with many simultaneous equations!

Oh just realised it was the same as ds2019. I got x=5 and y=2.6 multiply them together and you get 13

The height is constant, and the middle rectangle is shared, so the difference between known areas (42 - 27 = 15) is due to the extra 3cm (11-8) width. Height therefore must be 5 as 3x5=15.

Then (8x5)-27 = 13.

Loads of ways to solve it. Don’t need complicated maths or decimals.

BudgieHammockBananaSmuggle

This method has so many holes in it! The difference between the two areas is irrelevant and nothing suggests that there is a 3cm difference in the widths.

@mrsBtheparker it’s a bit hard to explain my method so maybe you didn’t understand, but it is completely accurate! The 3cm difference is 11-8.

@BudgieHammockBananaSmuggle

But how do you know the width of the middle section is equal to 11-8 from that diagram? It may be, but it doesn’t have to be.

@BudgieHammockBananaSmuggler -
The only thing that you can surmise for definite is that the total width of the whole shape is 11 + 8 - x, where x is the unknown rectangle's width. That does not make x = 11 - 8.

Budgie is right actually and rather elegant
width of LHS+ Width of middle =11
Width of RHS + width of middle = 8
Therefore you know that the LHS is 3cm longer than the RHS.
So if you imagine putting the RHS rectangle on top of the LHS then the non-overlapping rectangle created is 3cm wide and 15cm squared in area, hence 5cm high.

@YesThisIsMe - So if you imagine putting the RHS rectangle on top of the LHS then the non-overlapping rectangle created is 3cm wide and 15cm squared in area, hence 5cm high.
I agree with that statement but that does not mean that the width of the unknown rectangle is 3cm.

Diagram -

She’s not talking about the “?” rectangle, she’s talking about a hypothetical rectangle formed by the difference between LHS and RHS, and using that to establish the height. Once she knows that the height is 5, then she knows that the size of (RHS + ?) is 40 so she can subtract RHS to give 13.

@YesThisIsMe - Got you now. Thanks for clarification.

Yes - that is an elegant solution @BudgieHammockBananaSmuggler

Thank you for explaining my method better than I could! Indeed the 3cm doesn’t relate to the middle rectangle, just a hypothetical rectangle. I think I used logic more than maths.