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by chaitanyareddy » Fri Aug 20, 2010 11:22 pm
Hi as per my calculation the answer should be 18.

Here is my answer.

In the figure , BE is parallel to the base CD and also divides , AC into two equal halves. As this Line segment BE is parallel to the base and bisects AC , it will also Bisect AD at E.

As per the data given BC = AB = 3 , and AE = 4. Therefore ED should also be Equal to 4 since , E is the mid point for AD.

Now consider triangle ACD. AC= 6 , AD = 8 , CD = 10. There for AC^2 + AD ^2 = CD^2.

36 + 64 = 100.

Which means that ACD is a Right angle triangle , with right angle at angle CAD.

So you can draw the figure now something like this.


Image

So the area of the trapezium is the area of the triangle ACD - area of triangle ABE.

So , area of triangle ACD is 1/2*6*8 = 24.
area of triangle ABE is 1/2*3*4 = 6.

There fore area of the trapezium should be 24-6 = 18.
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by anantbhatia » Sat Aug 21, 2010 12:16 am
oops missed the 6,8,10 triplet. That makes it easier. I solved it in another way:-

the sides of the bigger triangle measure 6,8,10. So the area of that triangle using Hero's theorem is 24.

The inner triangle is half of the outer triangle by dimensions(and both the triangles are similar). So it's area must be 1/4 of the bigger triangle- 24/4=6.

bigger triangle - smaller triangle= 24-6=18.
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