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 By: geniuskids (offline)  Saturday, May 05 2012 @ 11:47 AM CDT (Read 1443 times)
geniuskids

Can someone pls help?

Kindly refer to test paper for figure info-did not upload it.

Q13) In the figure below, ABF and ADF are triangles. DE is half the length BG. Given that the area Triangle ABF is 63cm2 and Triangle ACF is 9cm2, find the area of the shaded figure.

Many thanks!

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Registered: 11/12/11
Posts: 169

 By: echeewh (offline)  Saturday, May 05 2012 @ 02:38 PM CDT
echeewh

Hi there again

Following pls find worked solution:

Shaded ABC = 63 - 9 = 54 cm2.

Given that height DE is 1/2 the height BG and the formula for Area of Triangle = 1/2 x Base x Height, the Area of Triangle is proportional to the Height since both triangles ABF and ADF are having a common base , AF.

So we have ...

Area of Triangle ADF = 1/2 x Area of Triangle ABF
= 63 / 2 = 31.5 cm2

Shaded CDF = 31.5 - 9 = 22.5 cm2

Shaded ABC + Shaded CDF = 54 + 22.5 = 76.5 cm2

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

Trust the above helps.

Cheers,
Edward

Active Member

Registered: 04/21/11
Posts: 625

 By: geniuskids (offline)  Saturday, May 05 2012 @ 10:40 PM CDT
geniuskids

Thank you for your help. Still confused with ' Area of Triangle is proportional to the height ?

Active Member

Registered: 11/12/11
Posts: 169

 By: echeewh (offline)  Sunday, May 06 2012 @ 11:11 PM CDT
echeewh

Hi there,

Hope the following explanation helps.

Area of Triangle = 1/2 x Base x Height

Since both triangles ABF and ADF have a common base AF, when you calculate the Area for each triangle, the area is dependent on the respective perpendicular heights. As such, given the perpendicular height DE ( for ADF ) is 1/2 that of BG ( for ABF ), the Area of ADF = 1/2 x Area of ABF.

This is the principle behind the Proportion method.

Lets hope you can understand this in a better light.

Cheers,
Edward

Active Member

Registered: 04/21/11
Posts: 625

 By: geniuskids (offline)  Sunday, May 06 2012 @ 11:37 PM CDT
geniuskids

yes thanks, finally got the concept. thanks much

Active Member

Registered: 11/12/11
Posts: 169

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