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Geometry Problem Set D?

+1 vote
The strongest rectangular beam that can be cut from a circular log is one having a cross section in which the diagonal joining two vertices is trisected by perpendicular segments dropped from the other vertices.
If AB=a, BC=b, CE=x, and DE=y show that b/a =√2/1
The image is of a rectangle inside a circle, with each vertex touching the edge of the circle. The upper left vertex is D, the upper right vertex is C, the lower left vertex is A, and the lower right vertex is B. The diagonal extends from D to B. The first perpendicular segment extends from C to point E on the diagonal. The second perpendicular segment extends from A to point F on the diagonal.
asked Feb 28, 2013 in GEOMETRY by angel12 Scholar

1 Answer

+1 vote

Using Pythagorean theorem
From triangle BCD
BD2 = BC2 + CD2
(3y)2 = b2 + a2
9y2 = b2 + a2---------------> (1)
From triangle CED
CD2 = CE2 + DE2
a2 = x2 + y2 ------------------> (2)
From triangle CEB
BC2 = CE2 + EB2
b2 = x2 + (2y)2 ----------------> (3)
b2 = x2 + y2 + 3y2
b2 = a2 + 3y2
b2 - a2 = 3y2
substitute 3y2 =b2 - a2 in equation 1
3 (b2 - a2)= b2 + a2
3b2 - 3a2 = b2 + a2
2b2 = 4a2
b2 / a2= 4 /2 = 2/1
Take square root both sides
b/a = √2/1


answered Feb 28, 2013 by John Lyn Pupil

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