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# Math Paper 32 CIE A level

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1. (Original post by TerraformingYou)
I just came here to ask, is a square a rhombus? Or are they "mutually exclusive" shapes? Because I proved ABCD, that is already KNOWN as a parallelogram, is a rhombus by proving that AC is perpendicular to BD. This is a valid method, right? Since the only two shapes that have their diagonals perpendicular are rhombuses and kites, right? And since ABCD is a parallelogram, then ABCD MUST be a rhombus, right?

People have been telling me that a square is not a rhombus and it's driving me crazy.. Even if it IS a square, all the question asks is to prove that ABCD is a rhombus, so even if ABCD IS a square, a square is a rhombus, right?

I'm going bonkers, I apologise..
Any shape with four sides is a parallelogram which includes square.

AC can not be perpendicular to BD. Though I didn't do it, rhombus actually has no 90 degree angles so we have to use dot product to show that the two direction vectors aren't equal to 0. I just proved the lengths are equal which should give one mark and finding OD would be another 1 or 2 marks.
2. (Original post by Devillain)
Yep. And I can remember 2.5 and 0.644 as well... Only thing being that I wrote 0.65 (I hate significant figures )
They asked us to show it to two decimal places so you are right. Rejoice
3. Question paper: http://justpastpapers.com/wp-content.../Scan20001.pdf

4. (Original post by Coloneltreavy)
Any shape with four sides is a parallelogram which includes square.

AC can not be perpendicular to BD. Though I didn't do it, rhombus actually has no 90 degree angles so we have to use dot product to show that the two direction vectors aren't equal to 0. I just proved the lengths are equal which should give one mark and finding OD would be another 1 or 2 marks.
AC and BD are the diagonals of the quadrilateral; proving that the diagonals in the parallelogram (you know what I'm talking about, right?) are equal makes it a rhombus, no?

EDIT: The diagonals I'm talking about don't mean the sides of the quadrilateral, but the two lines in the quadrilateral you can make through two opposite points in that quadrilateral..
5. (Original post by TerraformingYou)
AC and BD are the diagonals of the quadrilateral; proving that the diagonals in the parallelogram (you know what I'm talking about, right?) are equal makes it a rhombus, no?

EDIT: The diagonals I'm talking about don't mean the sides of the quadrilateral, but the two lines in the quadrilateral you can make through two opposite points in that quadrilateral..
Yea I get what you mean. But the diagonals can also prove it to be a square. What we need to do is to distinguish from a square.
6. (Original post by Coloneltreavy)
Yea I get what you mean. But the diagonals can also prove it to be a square. What we need to do is to distinguish from a square.
A square is a rhombus (I searched), and so it doesn't matter if it's a square or not.. And parallelograms cannot be kites..
7. what about statistics 72 (CIE) ? do you know any forum or page where I could find/discuss the answers?
8. (Original post by Rcnl)
what about statistics 72 (CIE) ? do you know any forum or page where I could find/discuss the answers?
http://www.thestudentroom.co.uk/foru...p?f=373&page=3

^you may be able to find a thread relevant to your needs over here. CIE ones are quite rare though.
9. I felt it was a bit harder, hoping for a mid 50 treshold. Do you guys remember you answers?
10. (Original post by Vuceee)
I felt it was a bit harder, hoping for a mid 50 treshold. Do you guys remember you answers?
it was actually a little harder. the real problem was, average/above-average kids could easily ace it if we had more time. only the math whizzes had a good exam, majority people i know, both good and bad at math, either had a so-so or a bad exam. a lot of kids could not even finish. so hoping the threshold is lowered, i don't think it will be above 60 cause I did June 2015 (32) and it was easier and had a threshold at 58 or sth for an A

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Updated: June 5, 2016
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