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Wednesday, June 28 2017 @ 07:18 AM CDT
 Forum Index >  Test Paper Related >  Primary 6 Matters
 P6 Maths ! 2012 Prelims Red Swastika Paper 2 Q3,11
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By: geniuskids (offline)  Sunday, September 08 2013 @ 07:38 AM CDT (Read 890 times)  
geniuskids

hi, Pls help solve these Qs. Refer to upload!

Answer key

Q3) 68cm2
Q11) 2:3

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By: echeewh (offline)  Sunday, September 08 2013 @ 10:42 PM CDT  
echeewh

hey <geniuskids>,

Following are my worked solutions:


Q3

Refer to attached solution.

=========

Q11

Shaded Area : Unshaded Area
5 : 8

Area of Unshaded Semi-circle (S) : Area of Large Semi-circle (L)

S : L
4 : 9

(S / L) = (4 / 9)

Area S = (pi / 2) × r × r (where r=radius of S)
Area L = (pi / 2) × R × R (where R=radius of L)

(S / L) = (r × r) / (R × R) = (4 / 9)
(r / R) = [Square Root of (4 / 9)] = (2 / 3)

Hence, ratio r : R = 2 : 3

=========

Trust this helps.

Do let me know again if there are further clarifications.

Cheers,
Edward

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By: geniuskids (offline)  Monday, September 09 2013 @ 02:59 AM CDT  
geniuskids

Thank you Edward for the fast reply.

I understand Q3.

As for Q11, I don't understand the last second sentence. (r/R)={Square Root of (4/9)}=(2/3)

Why square root? And after that to assume its ratio 2:3???

Pls clarify.

Thank you

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