Q&A - PSLE Math

[atutor2001]

Tang:

Vanilla Cake:

[quote=\"ck123\"]Need help with this question

http://i53.tinypic.com/2jfjg38.png[/IMG] x.png[/img]
Answer given : 43.73 sqcm

http://i56.tinypic.com/25uhzpy.png\">

Hi,

My answer is:

(144 - 24 pi) cm2.

As I do not have a calculator with me, please use your calculator and compute the answer yourself.



.....
6 x 3 = 18

30/360 x pi x 6 x 6 = 3 pi

(18 - 3 pi) x 2 = 36 - 6 pi

(108 - 18 pi) + (36 - 6 pi) = (144 - 24 pi) cm2[/quote]Hi Tang

I think there is one more triangle that needs to be subtracted from the rectangle 6 x 3 = 18.

Area of that triangle = (3x27^1/2)/2

So the area of that little small shaded part = [18-3pi-(3x27^1/2)/2] x 2

[Belle2011]

Thank-you very much tianzhu.

sigh… I also don’t know how to solve RGPS Q14b.

[tianzhu]

Belle2011:

Thank-you very much tianzhu.

sigh... I also don't know how to solve RGPS Q14b.

Hi Belle2011

You’re welcome.

I don’t have a set of 2011 test papers.

Please post the question if you need clarification.

Best wishes

[Belle2011]

tianzhu:

Belle2011:

Thank-you very much tianzhu.

sigh... I also don't know how to solve RGPS Q14b.

Hi Belle2011

You’re welcome.

I don’t have a set of 2011 test papers.

Please post the question if you need clarification.

Best wishes

http://i52.tinypic.com/wvskmd.jpg\">
Thank-you Mr Andrew Lee.

[atutor2001]

Andrew Lee:

Can help me solve Q14b:


http://i52.tinypic.com/wvskmd.jpg\">

Following the labeling in your diagram

Area of Quadrant = a+b+c+d

Shaded Area = (a+b+c+d) + c + d + (b+d) + (a+d)
= 2a+2b+2c+4d

Unshaded Area = b + (b+c) + a + (a+c)
=2a+2b+2c

Shaded - Unshaded = 4d = 2.5x1.4x4 = 14

[htn]

Hi, anyone tried Pei Chun 2011 prelim paper (paper 2) question 14, can teach me how to solve ?


Tia

[Andrew Lee]

Belle2011:

tianzhu:

[quote=\"Belle2011\"]Thank-you very much tianzhu.

sigh... I also don't know how to solve RGPS Q14b.

Hi Belle2011

You’re welcome.

I don’t have a set of 2011 test papers.

Please post the question if you need clarification.

Best wishes

http://i52.tinypic.com/wvskmd.jpg\">
Thank-you Mr Andrew Lee.[/quote]You are welcome, I also cant solve this question too

[tianzhu]

Hi Belle


I did a quick check for Q14b

Is the answer 14?

Best wishes

[Belle2011]

atutor2001:

Andrew Lee:

Can help me solve Q14b:


http://i52.tinypic.com/wvskmd.jpg\">

Following the labeling in your diagram

Area of Quadrant = a+b+c+d

Shaded Area = (a+b+c+d) + c + d + (b+d) + (a+d)
= 2a+2b+2c+4d

Unshaded Area = b + (b+c) + a + (a+c)
=2a+2b+2c

Shaded - Unshaded = 4d = 2.5x1.4x4 = 14

I understand area of quadrant but unsure how do you get Shaded Area and Unshaded Area.

[Belle2011]

tianzhu:

Hi Belle

I did a quick check for Q14b
Is the answer 14?
Best wishes

Dear tianzhu,
Yes, this answer is the same as atutor.
Thank-you.