2012 DSE Maths paper 2 Q17

Join AF.
Since AD : DG = 1 : 1
So area of triangle DAF = area of triangle DFG = 3 cm^2.
Since DF : FC = 3 : 4
So area of triangle DAF : area of triangle FAC = 3 : 4
So area of triangle FAC = 4 x area of triangle DAF/3 = 4 cm^2.
Area of triangle DAF + area of triangle FAC = 3 + 4 = 7 cm^2 = half the area of parallelgram ABCD, so area of parallelogram of ABCD = 7 x 2 = 14 cm^2. (B)

Hai... I have been working out on this question too... too bad... Ask me next time if there are any more math problem... ^_^

It may be other method.

Using the graph in
https://s.yimg.com/rk/HA00430218/o/779754713.png

Area of ΔGDF = (3ab sin ∠GDF)/2 = 3
∴ab sin∠GDF = 2
Area of ABCD = 2(7ab sin ∠GDF)/2 = 7(2) = 14 cm²

2015-04-10 01:13:52 補充：
係， 哈~

btw，岩岩回覆後又出現番 意見者 嘅英文名 -.-

我答遲了，我退到意見欄吧～

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加油！！！　還有五日！！！

https://s.yimg.com/rk/HA00430218/o/779754713.png
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2015-04-09 21:42:28 補充：
其實wy 網友的方法更佳，請學習！

2015-04-10 00:58:11 補充：
Thanks, HK~

"There" may be "an"other method.