 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 | Comment 3 of 4 | The Triangle ABC let AF=h, AE=i, CD=j then

h =root3 / ( 2+root3 )x  = (0.4641016x)

i = ( 1.5 + 0.5root3 )/(2+root3 ) x  = (0.6339746x)

j = 1/ ( 2+root3 ) x  = (0.2679492x)

If perpendicular from D to EF meets EF at G let DG= k & FG=l  then

k = ( 1.5 + 0.5root3 )/(2+root3  ) /root2   x  = (0.4482877x)

l = root3 /( 2+ root3 )*(root6-root2 )/4  x  = (0.1201183x)

Area of BED, DEF & DFC are

1/2*(x-i)*i , 1/2*k*(l+k), 1/2*h*j respectively

Or 0.1160254x2, 0.1274047x2, 0.0621778x2 respectively

And length EF = l+k = 0.568406x

Edited on April 22, 2013, 2:40 am
 Posted by Salil on 2013-04-22 02:25:27 Please log in:
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