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Geometry has always been a piece of cake for me... but I'm [#permalink]

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21 Feb 2007, 15:17

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Geometry has always been a piece of cake for me... but I'm not so sure anymore after this one
Could pls. somebody explain...
I picked E, correct one is B.[/img]

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Triangele ADC.jpg [ 202.94 KiB | Viewed 773 times ]

Geometry has always been a piece of cake for me... but I'm not so sure anymore after this one Could pls. somebody explain... I picked E, correct one is B.[/img]

Between parallel lines 2 triangles having the same base have equal areas

B would suffice that axiom

Last edited by trivikram on 21 Feb 2007, 18:35, edited 1 time in total.

First of all, let's see what is the formula for area of triangle ADC. It is 1/2 * (base * height).

Base for triangle ADC is AC and height is BE. So Area of triangle ADC = 1/2 * (base * height) = 1/2 * (AC * BE). Well, the value for (AC * BE) is given which is 24. You got the area by simply putting the values of AC * BE in the formula. Area of triangle ADC is 12.

(Two ways to determine the height: either you draw a perpendicular line let's say DF, where F is the point on line l2 and now we got a square EBDF since lines l1 and l2 are parallel OR just take BE as the base...again DF and BE are same as we see that these two are the sides of the square EBDF, hence DF = BE)

I wish I could explain it by using some diagram or figures, then it would have been much easier for you to understand.

By the way, please write if you were able to understand it or not.

Thanks.

Last edited by Summer3 on 21 Feb 2007, 18:50, edited 3 times in total.

Between parallel lines 2 triangles wharing the same base have equal areas

B would suffice that axiom

Thanks, it is really a good piece of information you shared. Even I did not know this earlier. By using it, one can save time while solving the sums of such kind.

This is an easy one. The thing that bothered me is the notation. On the diagram it looks like we are looking for the area of the angle, not the triangle.