Posted by Joel on October 12, 2002 at 12:51:35:
In Reply to: Re: Area problem (sorry, my other post was messed up because my angles were interpreted as HTML tags- I hope this one will be clearer) posted by Joel on October 11, 2002 at 17:14:03:
: : Huge triangle ABC has an area of 20,000,000 square feet.
: : D is on AB, E is on BC and F is on CA.
: : AD = 1/16th of AB, BE = 1/16th of BC, CF = 1/16th of CA.
: : Find area of triangle DEF.
: : Thank you.
: I will get you started.
: 1. Draw a picture. Obviously you could put A, B and C anyplace, but it will be easier to see my explanation if we are both looking at the same picture. So, I put A at the top, B at the right end of the base, and C at the left end of the base. Now put D, E, and F as described in your problem. Next, drop a line from A, perpendicular to BC, meeting BC at point G. Also drop a line from F perpendicular to BC, meeting BC at H.
: 2. Let's call the length of BC = a, AC = b, and AB = c. Let's call the height AG = h, and the height of the small triangle (FEC) FH = x.
: 3. Notice that the two triangles FEC and ABC share the same angle {< ACB}. What is the sine of that angle? sin{< ACB} = h/b = x/((1/16)b)
: Think about that & you will see that you now have enough information to calculate the area of triangle FEC.
: You can calculate the areas of triangles EDB and DFA by the same method.
: Then what?
: Got it?
h/b = x/(b/16)
x = h/16
and since CE = (15/16)a .....[a=BC in my drawing]
area of triangle FEC = (1/16)*(15/16)*20,000,000
The same applies to triangles EDB and DFA so the area of triangle DEF = 20,000,000*(1-(3/16)(15/16)) = 20,000,000*(211/256) = 16,484,375