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Updated on: 30 Sep 2014, 00:38
Originally posted by vaivish1723 on 25 Jan 2010, 10:05.
Last edited by Bunuel on 30 Sep 2014, 00:38, edited 1 time in total.
Last edited by Bunuel on 30 Sep 2014, 00:38, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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43% (02:57) correct
57%
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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?
A. 1/3
B. \(\frac{\sqrt{2}}{4}\)
C. 1/2
D. \(\frac{\sqrt{2}}{2}\)
E. 3/4
OPEN DISCUSSION OF THIS QUESTION IS HERE: in-the-figure-each-side-of-square-abcd-has-length-1-the-length-of-li-54152.html
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25 Jan 2010, 14:24
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If someone can explain this, that'll be great. I keep on getting (rad 2) over 4, which is not any of the answers.
26 Jan 2010, 03:43
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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?
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.
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.
