This is the maths Q that i don't know...
In the figure, ABCD is a rectangle of dimensions 6 cm *8 cm. E and F are the mid-pts of BC and AD respectively. BD intersects EF at G.
1, Prove that ABGF is a right- angled trapezium.
2. Find the area of trapezium.ABGF.
Pls help.. THX..
Diagram: http://www.flickr.com/photos/51470019@N07/4732632919/
maths Q ]
2010-06-26 6:57 am
回答 (2)
2010-06-26 8:53 am
✔ 最佳答案
As follows:圖片參考：http://i707.photobucket.com/albums/ww74/stevieg90/02-95.jpg
2010-06-26 00:53:29 補充：
http://i707.photobucket.com/albums/ww74/stevieg90/02-95.jpg
2010-06-27 3:49 am
*****************************************Solution to Q1******************************
Since ABCD is a rectangle, angle A must be 90 degree.
Also, F and E are mid points on AD and BC respectively and AD = BC (by properties of rectangle). So, AF = BE and angle AFE = 90 degree.
As angle A = angle B = 90 degree (from rectangular property) and AF = BE (proved above), EF // AB. Hence, GF // AB.
Combining the three results above, ABGF is a right-angled trapezium.
***************************************************************************************************************************************Solution to Q2********************************
Note that AF = 8/2 = 4 cm because F is the mid point of AD. Using the mid-point theorem, FG = 6/2 = 3 cm. Thus, the area of trapezium ABGF = (1/2)*(upperbase + lower base )*(height) = (1/2)*(FG+AB)*(AF) = (1/2)*(3+6)*(4) = 18 cm2.
********************************************************************************************
Hope this helps^^
Since ABCD is a rectangle, angle A must be 90 degree.
Also, F and E are mid points on AD and BC respectively and AD = BC (by properties of rectangle). So, AF = BE and angle AFE = 90 degree.
As angle A = angle B = 90 degree (from rectangular property) and AF = BE (proved above), EF // AB. Hence, GF // AB.
Combining the three results above, ABGF is a right-angled trapezium.
***************************************************************************************************************************************Solution to Q2********************************
Note that AF = 8/2 = 4 cm because F is the mid point of AD. Using the mid-point theorem, FG = 6/2 = 3 cm. Thus, the area of trapezium ABGF = (1/2)*(upperbase + lower base )*(height) = (1/2)*(FG+AB)*(AF) = (1/2)*(3+6)*(4) = 18 cm2.
********************************************************************************************
Hope this helps^^
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