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maths Q ]
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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
其他解答:
*****************************************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^^
maths Q ]
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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: ... 顯示更多 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/最佳解答:
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
其他解答:
*****************************************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^^
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