ABC is a triangle with AB = 12, AC = 14 and BC = 16.
D is the midpoint of AB.
The vertex C is joined with D forming the crease EF with E on AC and F on BC.
Find the area of the quadrilateral BDEF
(In reply to Long Solution
I agree, it looks nasty:
The height of triangle ABC is sqrt( 14^2-3.5^2) = 7sqrt15/2, so that triangle's area is 42*sqrt15/2, A. Already nasty.
Turning to the kite DFCE, the larger pair of triangles (b) is bigger than the smaller pair (s) by the proportion 41/35.
Then 2b+ discarded (d) = 2(35/41)b +kept (k) = A/2.
The area d is easy: sqrt(32/5^2- 8/5^2)*3 = 24 sqrt15/5.
Now we have 2b+24 sqrt15/5 = 21*sqrt15/2, and b is the again nasty 57 sqrt15/20
The entire kept area, K, is s+b+k = 1/2A-s+b, = 21*sqrt15/2+ b- 35/41b = 267 sqrt15/20 - 399sqrt15/164, or (2238 sqrt(15))/205.
Posted by broll
on 2016-04-11 05:15:12