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 Forum Index >  Test Paper Related >  Primary 6 Matters
 Raffles Girls 2011 SA2
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By: Jane Lim (offline)  Wednesday, June 06 2012 @ 12:31 AM CDT (Read 3830 times)  
Jane Lim

Hi there,
Could anyone please help me with the following 2 questions.

Many thanks in advance!!

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By: echeewh (offline)  Wednesday, June 06 2012 @ 05:34 AM CDT  
echeewh

Hello Jane,

For Q14(b), pls refer to attachment. The worked solution is on the left side (not the working /method on the right).

==========================================

Q17.

Refer to re-modelled diagram in paper as per attached.

Given ODBE is a trapezium, angle ODB = 45°.

In Square Y, draw a diagonal labelled OF ( angle EOF = 45° ), where OF // DB. I.e. ODBF is a parallelogram and OF=DB=11.75cm.

In Square X, draw a diagonal labelled OG, where OG=OF=11.75cm (OG, OF are radii of semicircle with centre O). Angle FOG is a right-angled triangle (with angle FOG = 90°)

Area of Triangle FOG = (11.75 x 11.75) / 2 = (1 / 2) x Rectangle EFGH

(a)

Area (X + Y) = Area of Rectangle EFGH = 2 x Area of Triangle FOG
= 2 x [(11.75 x 11.75) / 2] = 11.75 x 11.75 = 138.06 cm2 (nearest 2 dec places)

(b)

Circumference of Semicircle = (pi / 2) x d = (pi / 2) x (11.75 x 2) = 11.75 pi

Perimeter of S,T = Circumference of Semicircle + (4 x r) , where r = radius of semicircle
= 11.75 pi + (4 x 11.75) = 83.91 cm (nearest 2 dec places)

=========================================

Trust the above helps.

Cheers,
Edward

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By: mlteo (offline)  Tuesday, June 12 2012 @ 11:19 PM CDT  
mlteo

Quote by: echeewh


(b)

Circumference of Semicircle = (pi / 2) x d = (pi / 2) x (11.75 x 2) = 11.75 pi

Perimeter of S,T = Circumference of Semicircle + (4 x r) , where r = radius of semicircle
= 11.75 pi + (4 x 11.75) = 83.91 cm (nearest 2 dec places)

=========================================

Hi Edward

Shouldn't the answer for (b) be 83.93 ?

(11.75 x pi) + (4 x 11.74) = 36.92857 + 47 = 83.92857, which rounded to 2 decimal places is
83.93.

Thanks

Regards
Teo

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By: echeewh (offline)  Wednesday, June 13 2012 @ 03:05 AM CDT  
echeewh

Quote by: mlteo

Quote by: echeewh


Hi Edward

Shouldn't the answer for (b) be 83.93 ?

(11.75 x pi) + (4 x 11.74) = 36.92857 + 47 = 83.92857, which rounded to 2 decimal places is
83.93.

Thanks

Regards
Teo



Hi Teo,

If you refer to the question, you should use the calculator's pi since question does not say to use 22/7 or 3.14.

In your case, you are using 22/7 , correct?? Your answer will be correct if the question explicitly says to use 22/7.

Hope this clarifies.

Thank you & best rgds.
Edward

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By: parent603 (offline)  Wednesday, June 13 2012 @ 04:10 AM CDT  
parent603

Hi Edward,

Referring to Q14(b), your answer is correct. However I do not understand why the difference between the shaded and unshaded area = 4 * rectangle instead of 2 * rectangle. Pls enlighten.

Thanks for the solution. Big Grin


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By: echeewh (offline)  Wednesday, June 13 2012 @ 08:04 PM CDT  
echeewh

Quote by: parent603

Hi Edward,

Referring to Q14(b), your answer is correct. However I do not understand why the difference between the shaded and unshaded area = 4 * rectangle instead of 2 * rectangle. Pls enlighten.

Thanks for the solution. Big Grin




Hello parent603,

If I understand you correctly, you were asking why Shaded - Unshaded is not 2 x Rectangle. Correct??

Yes it is .. and I checked I have written in that attachment as 2 x Rectangle (3).

Did I see it wrongly or have I misinterpreted your query??

Do let me know pls. Thanks.

Kind regards,
Edward

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By: mlteo (offline)  Thursday, June 14 2012 @ 02:42 AM CDT  
mlteo

Oops. You are right, I used 22/7 as pi.

Thanks.

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By: parent603 (offline)  Thursday, June 14 2012 @ 10:22 PM CDT  
parent603

Quote by: echeewh

Quote by: parent603

Hi Edward,

Referring to Q14(b), your answer is correct. However I do not understand why the difference between the shaded and unshaded area = 4 * rectangle instead of 2 * rectangle. Pls enlighten.

Thanks for the solution. Big Grin




Hello parent603,

If I understand you correctly, you were asking why Shaded - Unshaded is not 2 x Rectangle. Correct??

Yes it is .. and I checked I have written in that attachment as 2 x Rectangle (3).

Did I see it wrongly or have I misinterpreted your query??

Do let me know pls. Thanks.

Kind regards,
Edward



Hello Edward,

From your attachment, there are 2 rectangles ie 1.4 * 2.5 = 2 * 1.4 * 2.5.

I am wondering why the solution is based on 4 * 1.4 * 2.5. I do understand that your soution is correct. I am curious to know why.

Thanks, Question Big Grin

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By: Angch (offline)  Thursday, June 21 2012 @ 09:56 AM CDT  
Angch

hi., anyone able to help problem sum Q16 ?
Thanks.

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By: echeewh (offline)  Friday, June 22 2012 @ 12:12 AM CDT  
echeewh

Quote by: parent603

Quote by: echeewh

Quote by: parent603

Hi Edward,

Referring to Q14(b), your answer is correct. However I do not understand why the difference between the shaded and unshaded area = 4 * rectangle instead of 2 * rectangle. Pls enlighten.

Thanks for the solution. Big Grin




Hello parent603,

If I understand you correctly, you were asking why Shaded - Unshaded is not 2 x Rectangle. Correct??

Yes it is .. and I checked I have written in that attachment as 2 x Rectangle (3).

Did I see it wrongly or have I misinterpreted your query??

Do let me know pls. Thanks.

Kind regards,
Edward



Hello Edward,

From your attachment, there are 2 rectangles ie 1.4 * 2.5 = 2 * 1.4 * 2.5.

I am wondering why the solution is based on 4 * 1.4 * 2.5. I do understand that your soution is correct. I am curious to know why.

Thanks, Question Big Grin



Hello parent603,

Sorry i missed your clarification posted here earlier.

My solution is based on:

Shaded - Unshaded = 2 x Rectangle (3)

If u see carefully , this Rectangle (3) is 2 x (given small rectangle of 2.5 x 1.4) and will have length of 2.5 and breadth of 2.8.

I believe your term 'the solution' is referring to the Answerkey, which is 4 x (small rectangle, 2.5 x 1.4)

Hence, if you were to calculate carefully, both will give you the same results.

If you still have doubts on this, pls write me an email and i will do my best to address your concerns.

Thankyou once again for clarifying..

Cheers,
Edward

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