Q&A - P5 Math

[Jamesbond]

tianzhu:

Jamesbond:

Hi Tianzhu, Can u pl clear my doubt. Those 12 units are for bee bee alone or for bee bee and anna together?

Hi Jamesbond

You are correct, it's for Bee Bee and Anna altogether.

I am sorry for the error.

Best wishes

So is the answer 900? pl let me know. :nailbite:

[Jamesbond]

http://i46.tinypic.com/jgjyg0.jpg\">

Pl help...

[ChewingPencilLine]

Jamesbond:

http://i46.tinypic.com/jgjyg0.jpg\">

Pl help...

LOOK AT MATHIZZZFUN'S CORRECT ANSWER BELOW.

The drawing looks a bit odd... Why are there irregular unshaded patches?

Anyway, ignoring those, there are basically 3 shaded triangles within 3 rectangles (see attached picture below).

The top rectangle is 7cm by (7-3=4)cm.
Area = 7 x 4 = 28 cm square.
The bottom left rectangle is 3cm by (3+7=10)cm.
Area = 3 x 10 = 30 cm square
The last bottom right rectangle is 3cm by 5cm.
Area = 3 x 5 = 15 cm square

(If you have problem seeing why the rectangles are of the above sizes, feel free to ask.)

Therefore, to find the total area, find the area of the 3 triangles then sum them up!
1/2 x 28 = 14 cm square
1/2 x 30 = 15 cm square
1/2 x 15 = 7.5 cm square
14 + 15 + 7.5 = 36.5 cm square

http://i48.tinypic.com/dm79ec.png\">

[MathIzzzFun]

Jamesbond:

http://i46.tinypic.com/jgjyg0.jpg\">

Pl help...

http://i48.tinypic.com/x1loht.png\">

cheers.

[ChewingPencilLine]

MathIzzzFun:

Jamesbond:

http://i46.tinypic.com/jgjyg0.jpg\">

Pl help...

http://i48.tinypic.com/x1loht.png\">

cheers.

Ahh, I think my answer is slightly off because I presumed that the bottom right triangle is half of a 3cm by 5cm rectangle from looking at the diagram. Now that I think about it, that is not true (should be 2.8cm by 5cm, hmmm).

Thanks for submitting the right answer and catching me on my mistake ^_^.

[MathIzzzFun]

ChewingPencilLine:

MathIzzzFun:

[quote=\"Jamesbond\"]http://i46.tinypic.com/jgjyg0.jpg\">

Pl help...

http://i48.tinypic.com/x1loht.png\">

cheers.

Ahh, I think my answer is slightly off because I presumed that the bottom right triangle is half of a 3cm by 5cm rectangle from looking at the diagram. Now that I think about it, that is not true.

Thanks for submitting the right answer and catching me on my mistake ^_^.[/quote]the height of the rightmost triangle (inside square of side 5 cm) is = 5/12x7 cm = 2 11/12 cm

so, the shaded area cannot be obtained by adding the 3 triangles shown in your diagram.

You can add area of 3 triangles by cutting the shaded area into 3 triangles - add diagonal in square of side 7 cm from top-left to bottom right, add a line from top-right corner of square of side 3 cm to bottom-right corner of square 7 cm. The area of these 3 triangles are:
.. 1/2 x 3 x 3 cm2 = 4.5cm2
.. 1/2 x 4 x 7 cm2 = 14 cm2
.. 1/2 x 5 x 7 cm2 = 17.5cm2
total 36 cm2

cheers.

[cimman]

Jamesbond:

http://i46.tinypic.com/jgjyg0.jpg\">

Pl help...

Area problems can be categorized to heuristics as well. I call this heuristic \"Big minus Small\". The Big refers to the area of the overall shape. The small refers to the unshaded region. This heuristic can be used if both the overall shape and the unshaded region are regular shapes (triangle, square, rectangle) with known lengths. The overall shape is made up of 3 squares, so the total area can be easily calculated.
http://i50.tinypic.com/2cnvcp5.png\">

from here, it seems that we hit a snag. The top right portion of the diagram is not a regular unshaded region. However, it can be transformed into a regular triangle shape, if we extend the 5cm square to a rectangle, like so:
http://i49.tinypic.com/mkaog.png\">

from here on, we can start using the Big - Small heuristic:
http://i48.tinypic.com/2czwnz7.png\">
The total area of quadrilaterals A, B and C minus area J and K will give the red shaded region.
http://i50.tinypic.com/scuhy8.png\">

[cimman]

carol73:

thanks, btw, there are so many methods, how do you know which method to use for what type of questions


my girl is quite weak in choosing which method to use based on qn, ie when to use model, units method, guess and check and subsitution?

actually, there is a way to get around all the different methods. That way is through algebra. The reason why we have all the different methods is because children do not know how to use algebra. However, if your child is familiar with algebra, the relationships can be easily resolved without the use of different heuristics.

Having said that, the traditional way to using algebraic equation is not easy to teach, since most of the derivation of the equation is mental. Only the final equations are written down. The process of how the equations are derived are totally mental. I've developed an approach that guides students to the final equations through a step by step approach based on a visual analysis technique. All analysis are done on pen and paper, so the equation derivations can be easily followed by students. I do conduct workshops on this technique.
http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=43&t=32278&start=190

[Jamesbond]

Thanks for all your help.

[Jamesbond]

http://i50.tinypic.com/fcv5hc.jpg\">

Help needed.