Area of an Unspecified Triangle
Date: 6/1/96 at 20:24:19 From: Anonymous Subject: Area of an unspecified triangle A square ABCD, with side "y", and an equilateral triangle DEF, with the same side "y", are in contact (at point "d") and their bases CD and DE are a continuous line segment. A line segment is drawn from point "b" of the square to point "e" of the triangle (it is the hypotenuse of the right-angled triangle BCE formed by the sides BC of the square and CE of the square plus the triangle DEF). There is a "space" between the side DA of the square and the side FD of the equilateral triangle. Another triangle is formed between the hypotenuse already drawn and this space (one of the angles of this new triangle is 30 degrees, the difference between 90, the supplementary angle of the square, and 60, the internal angle of the equilateral triangle). The question is: what is the area of this last triangle? b a xxxxxx f x x x x x x x <sides measuring "y" xxxxxxxxxxx c d e
Date: 6/1/96 at 20:56:22 From: Doctor Pete Subject: Re: Area of an unspecified triangle Draw the figure as follows: B __________ A F |-._ | /\ | -._ | / \ | -._|H / \ | |-./_ \ | | /G --._ \ |__________|/_________-\ C D E So you want the area of triangle DGH. As you pointed out, angle GDH is 30 degrees. But if CD = DE, then 2BC = CE, so BE intersects AD at the midpoint H. So DH = y/2. Also note that angle GHD = EBC. But tan(EBC) = 2/1 = 2. H |-._ |x -._ | -._ K |-----------/ G | h / | / | / | / | / tan(x) = 2 | / | / | / |30/ | / D |/ So we want h. Since GKD is a 30-60-90 right triangle, DK = sqrt(3)*h. So HK = y/2 - DK = y/2 - sqrt(3)*h. But tan(x) = h/HK = 2, so h/(y/2-sqrt(3)*h) = 2 ==> h = y - 2*sqrt(3)*h ==> (1+2*sqrt(3))h = y ==> h = y/(1+2*sqrt(3)) = (2*sqrt(3)-1)*y/11. So the area of triangle GHD = 1/2 * y/2 * h = (2*sqrt(3)-1)*y^2/44. -Doctor Pete, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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