 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Equilateral Crease (Posted on 2013-04-13) 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:
1. The length of EF in terms of x.
2. The area of each of the triangles BED, DEF and DFC in terms of x.

 No Solution Yet Submitted by K Sengupta Rating: 3.0000 (1 votes) Comments: ( Back to comment list | You must be logged in to post comments.) Solution with explicit values | Comment 2 of 4 | 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
 Posted by xdog on 2013-04-16 13:58:26 Please log in:

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