Tutor MathsGuru: Ask me for your burning Maths questions!

[peapot]

Hi everyone!


I need to post here a P1 question which was from my DD recent test paper.

Please help to solve.

Jason uses some sticks to form some squares as shown below. How many such squares can he make if he has 24 sticks?

[/img][/list]

[peapot]

sorry, I miss out the picture as I can’t seems to upload it. Exasperating!!!

[acehkr3009]

YLH88:

iFruit:

[quote=\"YLH88\"]
If you join 2 of the same isosceles triangles together, as shown in the diagram below, it will become a square.

Not necessarily. It will become a rhombus but not necessarily a square.

The perpendicular line shown in the picture bisects AB. So in the right angle triangle, base = 6, hypotenuse = 10. So height = 8 (sqrt of 100-36)

So area of the triangle is 1/2 x 12 x 8 = 48 cm²

Hi Dharma, iFruit,

Thanks for pointing out my mistake and I have already deleted my solution posted earlier. But the kids don't learn Pythagoras theorem in primary school ....

acehkr3009,
May I know the source of this question ? is it for primary level ?[/quote]Hi everyone,

I saw this question in P5 test paper 2 Section A Question 5 MCQ, from Advanced Learners Final Exam paper (grey colour loose plastic pack in Popular). 4 options here-(1) 40 cm sq (2) 48 cm sq (3) 50 cm sq (4) 60 cm sq & Ans is 48 cm sq. But dont know how to explain to my son if using Pythagoras theorem.

My solution to my son....I told him that the height can never be longer than slanted side which is 10cm. With this, use guess & check to do 6 (half of 12cm) x 9 or 8 or 7... until he gets one of the answer in the MCQ, which is 48 cm sq. I think that will be the simplest option in the context of Primary level.

Thanks all for trying..

[atutor2001]

acehkr3009:

..


My solution to my son....I told him that the height can never be longer than slanted side which is 10cm. With this, use guess & check to do 6 (half of 12cm) x 9 or 8 or 7... until he gets one of the answer in the MCQ, which is 48 cm sq. I think that will be the simplest option in the context of Primary level.

Thanks all for trying..

I believe your approach is the same as what setter of the question has in mind. To find the area without knowledge of Pythagoras theorem is equivalent to the discovery of Pythagoras theorem. Was googling and found this site very interesting on how Pythagoras theorem can be proven.

http://www.1728.com/pytproof.htm

[acehkr3009]

atutor2001:

acehkr3009:

..


My solution to my son....I told him that the height can never be longer than slanted side which is 10cm. With this, use guess & check to do 6 (half of 12cm) x 9 or 8 or 7... until he gets one of the answer in the MCQ, which is 48 cm sq. I think that will be the simplest option in the context of Primary level.

Thanks all for trying..

I believe your approach is the same as what setter of the question has in mind. To find the area without knowledge of Pythagoras theorem is equivalent to the discovery of Pythagoras theorem. Was googling and found this site very interesting on how Pythagoras theorem can be proven.

http://www.1728.com/pytproof.htm

Thanks for the link...great visual in showing Pythagoras theorem instead just remembering the formula = More fun in understanding maths....

[iFruit]

atutor2001:

acehkr3009:

..


My solution to my son....I told him that the height can never be longer than slanted side which is 10cm. With this, use guess & check to do 6 (half of 12cm) x 9 or 8 or 7... until he gets one of the answer in the MCQ, which is 48 cm sq. I think that will be the simplest option in the context of Primary level.

Thanks all for trying..

I believe your approach is the same as what setter of the question has in mind. To find the area without knowledge of Pythagoras theorem is equivalent to the discovery of Pythagoras theorem. Was googling and found this site very interesting on how Pythagoras theorem can be proven.

http://www.1728.com/pytproof.htm

I'm not too sure about that. Surely the P4/P5s know decimals/fractions and that lengths need not be whole numbers, no?

IMHO, it is better not to teach this sort of heuristics as they might become a habit to the detriment of conceptual understanding.

I think the question was mistakenly set out of syllabus. It is probably best to introduce Pythagoras theorem at this point (they will love it) or just skip the problem.

Just my 2 cents (you can throw away )

[wkong]

Hi All,


Need help for the following:

Based on the diagram I have attached, what is the maximum amount of Figure A that can be cut out of the rectangle?


http://postimage.org/image/1eyvrals4/

[iFruit]

wkong:

Hi All,


Need help for the following:

Based on the diagram I have attached, what is the maximum amount of Figure A that can be cut out of the rectangle?


http://postimage.org/image/1eyvrals4/

If you join two of the triangles, it becomes a 4x3 cm rectangle.

Maximum number 4x3 rectangles that can fit in the big rectangle = (28/3) x (16/4) = 36 (3cm along length wise, 4cm along breadth wise)

So max triangles that can be cut out = 36x2 = 72.

PS: if you cut the other way, you will end up with 70.

[wkong]

Hi iFruit,


Thanks for getting the answer. My kid got the answer 72 too, but another classmate of his say is 74. So which one is correct? Confused. :?

[iFruit]

wkong:

Hi iFruit,


Thanks for getting the answer. My kid got the answer 72 too, but another classmate of his say is 74. So which one is correct? Confused. :?

Hi wkong,

I think 74 is arrived at by rounding off (area of rectangle / area of triangle) = (28x16/6). But it is wrong as you can't cut more than 72 triangles as the remaining area can't be cut into required triangles.

HTH