
By: geniuskids (offline) Saturday, May 05 2012 @ 11:47 AM CDT (Read 1423 times)



geniuskids 
Can someone pls help?
Kindly refer to test paper for figure infodid 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!

Active Member
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: 623





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: 623





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



