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

It’s a nasty one that is!

Ahh that is interesting - that could help.
Thanks all

I got the same answer as ds2019 through simultaneous equations- 13

If the height of the sides is y, 8y = 27 + x, 11y = 42 + x, therefore x = 8y - 27 and x = 11y - 42. 8y - 27 = 11y - 42, so 11y - 8y = 42 - 27, so 3y = 15. Therefore y = 5.
Substituting back in, 8 x 5 = 27 + x = 40, so x = 13. Check by putting it in the other calculation 11 x 5 = 42 + 13 which is correct.
The key is not to worry about working out the length of sides of x (which would be decimal) but focus on the total area of x.

My answer agrees with @ds2019 -
I called the height of the whole rectangle z and the area of the middle rectangle y.
So the left hand rectangle added to the central rectangle gives
42 + y = 11z
And the right hand rectangle added to the central rectangle gives
27 + y = 8z.
Subtracting the second equation from the first gives
15 = 3z. So z = 5.
Using the second equation, replacing z with 5 gives
27 + y = 8 x 5
27 + y = 40
So y must be 13.

Will now peruse with my glass of red for a quick trick that doesn't involve knowledge of simultaneous equations!

I think that it's simpler than that.
I might be wrong but this could be the answer?!

ds2019 is right

I did it by attacking ? directly without trying to solve any of the lengths.
If a is the height of the total rectangle then see photo.

But rosegold’s approach is simpler still.

Here

I think that they're trying to throw you by giving the length as it's not required. They only want the area and the two small are the same length as the larger rectangle, so, their areas added up should be equal.

Yep it is simultaneous equations but this can get introduced in a problem solving type way like this before being formally taught.

One method is the middle section is
equal to 8 x width - 27 which is also equal to 11 x width - 42 so therefore the width must be 5 (can be figured out by trial and error or by logic that 3 x width must be equal to 15)
If the width is 5 then the middle section is 13 (8x5-27)

Bright kids can solve these sorts of problems before being taught a formal method for simultaneous equations.

It's 18

The problem can be solved using areas. If you imagine 2 separate rectangles, one of 11cm wide and one of 8 cm wide and put them one on top of the other you are left with a rectangle 3cm wide. Both rectangles include the unknown section, so this cancels out, then area of 3cm rectangle will be 42-27=15. This tells us that the height of rectangle is 5cm.
Using this height we can then calculate the width of the 2 known rectangles. 42/5=8.4. By subtracting this from 11 we get 2.6 for the unknown width. As a check we can do the same for the other rectangle: 27/5=5.4, 8-5.4=2.6 also.
We now know the height and width of unknown rectangle, so area is 5x2.6=13.
Would be easier to explain with a diagram but can't figure out how to attach it!

TheQueen’sCousin - I can’t see anything that tells you that the 2 smaller rectangles are equal to the larger. Yes, they look like they are, but diagrams like this are not necessarily to scale.

You can find the height by dividing the area by the length. And you can do it in two ways, using the rectangle of length 11 or the rectangle of length 8, and they must both give the same answer. So you have:

(42 + ?)/11 = (27 + ?)/8

Rearrange using the usual techniques and ? = 13.

I used the same method as @ds2019 (before reading the full thread) and came out with the same answer of 13

I think that many of us have calculated that the unknown area is 13.

After doing it myself I gave this problem to my, "oh so smart he could cut cheese," PhD student son and he came up with the same answer and couldn't find a method that didn't use simultaneous equations.

Perhaps the teacher who set it has plans to teach simultaneous equations when your son returns to school?
Bloody love algebra! Thanks OP

I get 13 for the middle area, and 2.6 for the smaller unknown and 5 for the width but the method is very complex for Year 8

and couldn't find a method that didn't use simultaneous equations. Mine didn't explicitly use differential equations. By calculating the height in two different ways and saying they must give the same answer, you are bypassing the "eliminate one unknown" step of a simultaneous equation method.

Thanks for this. I got 13 too. Has he got any more homework as I've enjoyed this!

13 is correct. Your ds could try to work it out by trial and error (with some constraints) if he hasn’t done simultaneous equations yet (I have no clue what they have covered by y8)

Where Y is height of rectangle

11 * y = 42 + ?

8 * y = 27 + ?

? = 8y - 27

11y = 42 + 8y - 27
11y= 15 + 8y
3y=15
Y=5, ? = 13