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#1 2006-02-24 20:57:18

naturewild
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Need help for a triangle

ABCD is a trapezoid in which AB is parallel to DC and AB = 9, DC = 15. The perpendicular distance between AB and CD is 4. Sides DA and CB are extended and meet at point E. If F and G are midpoints of Da and CB respectively, determine the area of triangle FGE.

#2 2006-02-24 23:34:30

gnitsuk
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Re: Need help for a triangle

Area (FGE) = Area (CDE) - AREA(CDFG)

Now length FG = 12cm which is 9 + half the difference between 9 and 15 (As FG is halfway down AD and BC).

So area of trapezoid CDFG is half sum of parallel sides times height  = [(15 + 12) / 2] * 2 = 27 sq cm

Now we need area of CDE. We can easilty see that a formula for the width of this triangle as a function of the height from the bottom of the triangle must be linear, in fact it must be W = 15 - (3/2)*H.

**
This is constructed from realising that the foumula must give 15 when height from bottom (H) = 0, and 9 when H = 4 (also 12 when H = 2). So we can imagine a graph of H (x-axis) versus W (y-axis) and we must have a straight line passing through (4,1) and (2,0.5) and we can therefore find its equation easily as y = x/4 so our formula is W = 15 - (H/4)*(15 - 9) = 15 - (3/2)*H (as above)
**

So rearrange this to make H the subject and insert W = 0 (as it does at the top of the triangle) and we get H = 10. So the height of CDE is 10 and the base is 15 therefore area CDE is 0.5 * 10 * 15 = 75 sq cm.

Therefore area of FGE = 75 - 27 = 48 sq cm

#3 2006-02-25 03:42:23

John E. Franklin
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Re: Need help for a triangle

Yes, I also get 48. 
But I did it maybe differently.
Make base of final triangle FG.
So area is 12H/2, where H is
2 + 6.
The 2 is half of the 4 distance given.
The 6 is 150% of the 4 distance given
because DC shrinks to AB and then
shinks to point E.  The perpendicular
heights reduce by same ratio as the
bases.  So (15-9):4 equals 9:6, so that
last 6 is the 150% bigger than the 4 in the
last ratio.  You need to draw it to see it.
So gnitsuk is correct!


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