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 Forum Index >  Miscellaneous >  Question, Feedback and Comments 2010 NANYANG PRELIM PAPER 2 Q4 Q13
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 By: jlopez (offline)  Saturday, August 20 2011 @ 09:17 AM CDT (Read 1337 times)
jlopez

Hi, appreciate if someone could help me solve this 2 problem.

Newbie

Registered: 03/22/11
Posts: 3

 By: jo sarah (offline)  Monday, July 16 2012 @ 11:11 AM CDT
jo sarah

Hi,

For no. 4:
(4k/3) - 3 + 8k + 10 - 7k = (7k/3) + 7
Thus, if k = 6, the value is (7 x 6 / 3) + 7 = 14 + 7 = 21

No. 13:
If you know Pythagoras' Theorem,
the longest side of the right-angled triangle is sq root (12 sq + 16 sq) = 20 cm
Thus, the Perimeter required is 4 x 20 = 80 cm

If you don't know Pythagoras' Theorem,
then area of 1 triangle = (1/2) x 12 x 16 = 96 cm^2
area of the centre square = 16 cm^2
Total area = 96 x 4 + 16 = 400 cm^2
So that a side of the large square = 20 cm
and its Perimeter = 8 x 20 = 80 cm

Cheerio!

Regular Member

Registered: 03/20/12
Posts: 111

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