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primary 6 | Maths | Geometry
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Sanjana
Sanjana

primary 6 chevron_right Maths chevron_right Geometry chevron_right Singapore

Pls help, Thanks in advance!

Date Posted: 4 months ago
Views: 22
Eric Nicholas K
Eric Nicholas K
4 months ago
This has to do with counting because each of the four parts (top left, top right, bottom left and bottom right) have the same number of sides.

The easiest way is to let the top side rhombus be 3 cm in length and the bottom side rhombus be 6 cm in length.
Eric Nicholas K
Eric Nicholas K
4 months ago
Altogether, the total length is 36 cm.

If you are a good observant, every one short line on top can be paired with one long line at the bottom to combine and form a length 9 cm.

There are four pairs of short-long 9 cm combos, so the total length is 36 cm.

With algebra it’s easier, but I won’t use it since you probably have not learnt algebra in detail yet.
Sanjana
Sanjana
4 months ago
Ok Understood now, Thank You so much
J
J
4 months ago
There are actually 8 shorter sides of equal length on the upper portion of the figure and 8 longer sides on the lower portion of the figure.

XY is made up of 1 shorter side and 1 lower side. And XY is 9 cm.

Since there are 8 shorter sides and 8 longer sides in total for the perimeter, it should be 9 cm x 8 = 72 cm and not 36 cm.


Furthermore, the lengths of the individual longer and shorter side are not given. We cannot assume as 6cm and 3cm as that is mathematically incorrect. You might incur a deduction of marks.

The purpose of such type of questions is to get students to realise that individual lengths need not always be given/known for solving questions. Many students tend to be too reliant on them and end up getting stuck on such questions.
Sanjana
Sanjana
4 months ago
Thanks for the clarification. Yes, the answer is 72cm.
Eric Nicholas K
Eric Nicholas K
4 months ago
Oops, I miscounted
Eric Nicholas K
Eric Nicholas K
4 months ago
From the looks of the question, it appears to be an mcq question though, so letting one length be 3 cm makes computation easier for students. The fact that the question can be asked without a length is testament to the fact that the diagram works for all lengths between 0 cm and 9 cm.

The proper way to do this is still as J described. We assume a length is unknown because we must show that the figure has the same perimeter regardless of length.
J
J
4 months ago
I see part a) and part b) printed on the back of the paper, so I would infer this is either a Paper 1 Booklet B question or Paper 2 short answer question. Probably not long enough to be a problem sum.

Actually assigning lengths in this makes computation more tedious and time consuming. The working would be as such :

9 = 6 + 3
(or other combinations like 4 + 5, 2 + 7, 3.5 + 6.5 etc)

3 x 8 = 24

6 x 8 = 48

24 + 48 = 72

This takes up 4 steps and relies on assumption of lengths.


The correct method, which tests students to recognise that the lengths should be paired together, would just be :

9 x 8 = 72, which is much faster.


What I see very often is that students try to assign lengths for this questions and then try to add them up together. They would try many different combinations of lengths. Then they will struggle in their attempts to justify why the lengths are such.

This wastes time and they end up realising that it will still amount to the same final answer if they had just paired them up in the first place.

They may also have the misconception that since it 'works' for this question, it will work for other questions as well. Students will repeat this, only to find that they got the next questions incorrect.



Sanjana , for further learning, you might want to take a look for a similar question under the 'unanswered' category by a person with username DS 3 days ago.

The student posted a question on a figure made with a wire bent into 4 equilateral triangles. He/she tried to assign lengths until he got the one that matched the answer but could not prove why they were so and asked for help.

I provided an explanation there in the comments. Hope it can enhance your learning.
Eric Nicholas K
Eric Nicholas K
4 months ago
That’s true, but often more than not, if I am unable to prove something, I assign one number to see that the statement holds for a specific number before continuing to find the actual proof. I guess it’s just a matter of preference by solvers.

Sanjana, follow what J has noted. In proof questions, when you proceed to higher levels, you will need to prove these type of questions using reasoning or algebraic and geometric manipulation.
J
J
4 months ago
Yes, I do that too to try to get some hint or clue on how to get a proof.

As a side working on rough paper it would be ok. What's most important is that the steps and working that are actually used on the answer sheet/space is conceptually and mathematically correct.