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Sunday, October 22 2017 @ 05:50 PM CDT
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By: acyk (offline)  Saturday, June 18 2011 @ 05:52 AM CDT (Read 1785 times)  
acyk

Need help on question 9 of Paper 2 Ai Tong School 2010, viz

filemgmt_data/files/P5%20Maths%20SA2%202010%20Aitong.pdf


Thanks

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By: athamms (offline)  Sunday, June 19 2011 @ 11:42 AM CDT  
athamms


You have to break down the sides AB & AD into its components, assuming you are dropping a perpendicular line from the point Q
to the sides to indicate each of the triangle height. However one triangle's height can be another triangle base component

Let's go...
AB ==> a1 + a2
AD ==> b1 + b2

So Area AQB =1/2 x (a1 +a2) x b1 = 24
Area of DQC = 1/2 x (a1 + a2) x b2 = 45
Sum up the above 2
Therefore 1/2 x (a1 +a2) x (b1 + b2) = 69
(a1 +a2) x (b1 + b2) = 138 ----(1)

Area of rectangle ABCD = (a1 +a2) x (b1 + b2) = 24 + 51 + 45 + shaded area = 120 + shaded area ----(2)

Compare (1) & (2) ==> 138 = 120 + shaded area
Therefore, shaded area = 18 cm squared

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