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christoph
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the four-sided figure, ABCD,is inscribed in a semi-circle as Updated on: 24 Feb 2005, 10:19
Originally posted by christoph on 23 Feb 2005, 15:33.
Last edited by christoph on 24 Feb 2005, 10:19, edited 2 times in total.
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the four-sided figure, ABCD,is inscribed in a semi-circle as shown above. if AB=BC=CD, then the area of the triangle "E" is what fraction of the area of ABCD ?

a 1/2
b 4/7
c 5/8
d 3/4
e 11/18



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the four-sided figure, ABCD,is inscribed in a semi-circle as 24 Feb 2005, 00:14
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shown above - guess I need another pair of eyes :)
the figure please !
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BG
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the four-sided figure, ABCD,is inscribed in a semi-circle as 24 Feb 2005, 05:36
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the answer is 1/2 in case that triangle E is NOT triangle ABD
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the four-sided figure, ABCD,is inscribed in a semi-circle as 24 Feb 2005, 06:10
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takes too much time, but it's 1/2
abcd=2[2+SQRT(4-h^2)]*h/2
E = 2+SQRT(4-h^2)]*h/2
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the four-sided figure, ABCD,is inscribed in a semi-circle as 24 Feb 2005, 16:27
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ANyone could you explain the steps please......Thanks.
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MA
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the four-sided figure, ABCD,is inscribed in a semi-circle as 24 Feb 2005, 19:52
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it is A.

Lets denote the Point in bet A and D as O. now, extent the line BC to P and draw a perpendicular line from D to P. now we have rectangle BPDO and triangle CDP. with these figures, we can prove that triangle ABO=triangle CDO.

now,
trangle E=1/2 of rectangle BPDO.
trangle E=1/2 of quadilatrel ABCD (becasue triangle ABO=triangle CDO)

pls do correct, if any.
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the four-sided figure, ABCD,is inscribed in a semi-circle as 25 Feb 2005, 01:02
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when we image to cut off the left triangle AB? and stick it on the other side CD?, we will see that the triangle E is half of the rectangle ABCD.
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the four-sided figure, ABCD,is inscribed in a semi-circle as 25 Feb 2005, 01:11
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Quote:
MA
PostPosted: Thu Feb 24, 2005 6:52 pm Post subject: Re: PS - 44
it is A.

Lets denote the Point in bet A and D as O. now, extent the line BC to P and draw a perpendicular line from D to P. now we have rectangle BPDO and triangle CDP. with these figures, we can prove that triangle ABO=triangle CDO.

now,
trangle E=1/2 of rectangle BPDO.
trangle E=1/2 of quadilatrel ABCD (becasue triangle ABO=triangle CDO)

pls do correct, if any.
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that was nice ![/quote]
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the four-sided figure, ABCD,is inscribed in a semi-circle as 25 Feb 2005, 03:29
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christoph
when we image to cut off the left triangle AB? and stick it on the other side CD?, we will see that the triangle E is half of the rectangle ABCD.
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this is the fastest way, good intuition, thank you



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