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In the figure, each side of square ABCD has length 1, the [#permalink]
25 Jan 2010, 09:05

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In the figure, each side of square ABCD has length 1, the length of line Segment CE is 1, and the length of line segment BE is equal to the length Of line segment DE. What is the area of the triangular region BCE?

In the figure, each side of square ABCD has length 1, the length of line Segment CE is 1, and the length of line segment BE is equal to the length of line segment DE. What is the area of the triangular region BCE?

This proble can be solved in many ways. One of the approaches:

Note that as BE=DE the triangles BCE and CDE are congruent (all three sides are equal) --> \angle{BCE}=\angle{DCE}. So if we continue the line segment CE it will meet with the crosspoint of the diagonals, let's call it point O. Also note that EO will be perpendicular to BD, as diagonals in square make a right angle.

The area BCE=BOE-BOC.

Area of BOC is one fourth of the square's =\frac{1}{4}.

Area BOE, \frac{1}{2}BO*EO. BO=\frac{\sqrt{2}}{2}, half of the diagonal of a square. EO=CE+CO=1+\frac{\sqrt{2}}{2}=\frac{2+\sqrt{2}}{2}, CO is also half of the diagonal of a square. So AreaBOE=\frac{1}{2}BO*EO=\frac{1}{2}*\frac{\sqrt{2}}{2}*\frac{2+\sqrt{2}}{2}=\frac{\sqrt{2}+1}{4}.

Area BCE=BOE-BOC=\frac{\sqrt{2}+1}{4}-\frac{1}{4}=\frac{\sqrt{2}}{4}

None of the answer choices shown is correct. _________________

in triangle BCE , draw a perpendicular to base BE which meets BE at Z

Now in Triangle BCZ , BC = 1 Angle EBC = 45/2

=> BZ = cos(45/2) CZ = sin(45/2)

Area of triangle BCE = \frac{1}{2}* BE * CZ = \frac{1}{2}* 2*cos(45/2) * Sin(45/2) =\frac{1}{2}* sin45 using -> sin(2A) = 2 * SinA * cosA =\frac{1}{2\sqrt{2}}

which should be 2^ (-3/2)

I think you have wrongly written your ans.. Please check if D is 2^ (-3/2) _________________

Basically adding a picture for more clarity... original one is throwing bit off...! OB=OC=(sq. rt 2)/2 rest is simplification... agree with the answers above of sqrt2/4.

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