More maths problem help please!

[annie987]

Hi
Please can anyone give a bit of help with these maths problems.
According to the maths teacher, there should be no need to use any decimal measurements nor simultaneous equations.
Thanks

Hi whole of the horizontal section area is 5x12 as given so that's 60 in total. Off that subtract the 20 and 16 given. The remainder is the area of the centre. Let me know if you can follow and agree and I'll continue

Yep with you so far

Gotcha...that leaves the centre part at 24, right?

How can the last one not be a decimal number in the “?” space? It’s the side length of two rectangles with a prime number area.
Sorry I’m no use, following out of interest!

Yep 👍🏻 and that’s where I stop!

Second one; you have a box which is 35cm^2 and known height 5, hence it's width must be 7

Total area of the bottom box is then 7 x height - and height is (40/105)*15. Which comes out to be 40.

So if the area of the centre rectangle (the part in the middle, common to both large rectangles) is 24, then the area of the rectangle running from top to bottom is 72 I think.

So each of the boxes top to bottom must share a factor that is their common width

The areas of the three boxes top to bottom are 21 (given) , 24 (just calculated) , 27 (given)

What factor is common to these, in other words which number in the times table do they all appear in?

Centre box 24, so top box height must be 521/24 and bottom box height must be 5 27/24, add those together with 5 for middle and it works out to height of 15.

I can't do it without using decimals

Yes but then I didn’t see how that could work with the height of the middle box being 5?

Second problem ? width is 35/5 = 7.

? height is 15*(40/105)

So area is 7 15 40 / 105 = 7 15 40 / (7*15) = 40

That was directed at Nooz.
RandomLondoner - you’ve just blown my mind! Think I need to sleep and look again in the morning!

Back to the first problem, since the widths are all the same (for the centre 3 boxes) the ratios of the heights must be the same as the ratios of the areas.

@StatisticallyChallenged

Second one; you have a box which is 35cm^2 and known height 5, hence it's width must be 7

Total area of the bottom box is then 7 x height - and height is (40/105)*15. Which comes out to be 40.

How is (40/105)*15 = 40 am I just not reading this right?

And in the second problem, since the 40 and 65 boxes have the same width, the areas must be proportional to height, the combined area is 105, the combined height is 15, so the height of the bottom box must be 15 * (40/105).

The second one is 40 because the height is the same and the width is 7 (can be worked out - 105/15, 35/5)

Ah, I was put off by the instruction no need to use decimal measurements in the first one. I read that as meaning there are no decimal measurements in the individual rectangles. There are - you just don’t need to work out what they are, in order to calculate the one missing length requested. I could have done it without the instructions, but not with
Following to see the answer for the third one when I get up tomorrow though

For the third problem, there are two ways to calculate the area of the whole shape. Let x be height.

A = 7x+8x-18 = 9x +17+ 19

Ignore "A=" and solve for x, x=9.

In the first one, the centre section is 24 as above. So the total area of the vertical section is 21+24+27 = 72. We know that the width of it is 24/5 (because of the centre section). So the length of the whole vertical section is 72 divided by 24/5 - 72*5/24 = 15.

In the second one, the total area of the left side is 105, so the width is 105/15 = 7. This means that the height of the bottom part is 40/7. The width of the right side is 35/5, so also 7. The area of the bottom right is then 7* (40/7) = 40.

I've got it the diagram is wrong!!

The height of the middle section is 4!

The person writing this knows the dimensions of the box that is are 20 has dimensions 4x5 and had written the 5 down by mistake.

If you redo the question with a height of 4 there it runs smoothly for whole numbers.

You're not mad, that's the problem here I promise you

Merryoldgoat you missed multiplying by 7 for the width.

Ah! Thanks!! I was worried I was going mad