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 By: keenlearner (offline)  Thursday, December 20 2012 @ 02:32 PM CST (Read 1370 times)
keenlearner

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 shaded area of the figure. (Ans: 76.5cm2)

Junior

Registered: 11/08/12
Posts: 22

 By: echeewh (offline)  Friday, December 21 2012 @ 11:52 PM CST
echeewh

Hey keenlearner,

Following pls find my worked solution:

Area of Triangle = (1/2) × B × H

Given base of both triangles ABF, ADF are common (same = AF) and DE is half the length BG (where BG is perpendicular height for triangle ABF), we can apply the proportion (ratio) method as follows;

Since B is constant,

Area of Triangle is proportional to H (Height)

Given height of triangle ABF is twice that of triangle ADF (since DE is half BG),

Area of Triangle ABF = 2 × Area of Triangle ADF

Area of Triangle ADF = 63 ÷ 2 = 31.5 cm2

Hence, Shaded Area of figure = (Area of Triangle ABF) + (Area of Triangle ADF) - (2 × Area of Triangle ACF)

= 63 + 31.5 - 18 = 76.5 cm2

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

Trust this helps.

Do let me know again if there's further clarification.

Cheers,
Edward

Active Member

Registered: 04/21/11
Posts: 627

 By: keenlearner (offline)  Wednesday, December 26 2012 @ 05:53 AM CST
keenlearner

Hi Edward,
Sincere thanks to providing the work solutions to many of my math questions, you are really super!
Have a very Happy New Year! :-)

Junior

Registered: 11/08/12
Posts: 22

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