A paper has the precise shape of a triangle which is denoted by ABC, with AB = BC = CA = x (say) and, D is a point on BC.
Vertex A is joined onto D forming the crease EF - where E is on AB and F is on AC.
Given that DF is perpendicular to BC, determine:
- The length of EF in terms of x.
- The area of each of the triangles BED, DEF and DFC in terms of x.
CDF is a 30-60-90 right triangle. Set CF=2y. Then FD=y*sqrt3. But FD=FA=x-2y. Solve and find y=(2-sqrt3) * x.
From E drop a perpendicular to BC at G. Set EB=2z. Then for right triangle EGD, EG=z*sqrt3, DG=x-y-z, ED=x-2z. This gives z=(-1+sqrt3)/4 * x.
Now the bases and altitudes of the three triangles and the coordinates of F and E are determined which is enough information to calculate areas and distance.
Area BED=(-3+2sqrt3)/4 * x^2 = 0.1160254037 * x^2
Area DEF=(27-15sqrt3)/8 * x^2 = 0.1274047358 * x^2
Area DFC=(-12+7sqrt3)/2 * x^2 = 0.0621778264 * x^2
Distance FE=sqrt(14-8sqrt3) * 3/2 * x = 0.5684060729 * x
**Thanks to Charlie for his GS-derived values which served as a good check.
Edited on April 17, 2013, 9:43 am
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Posted by xdog
on 2013-04-16 13:58:26 |
Solution with explicit values
