I thought BC is the hypotenuse. Thus, by pythagoras thm, CO = square root of [1^2 - (sq root 2/2)^2]?
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
Note that as BE=DE the triangles BCE and CDE are congruent (all three sides are equal) --> \(\angle{BCE}=\angle{DCE}\). So if we extend 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 of BCE) = (The area of BOE) - (The area of 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 \(BCE=BOE-BOC=\frac{\sqrt{2}+1}{4}-\frac{1}{4}=\frac{\sqrt{2}}{4}\)
Answer: B.
Attachment:
Square ABCD.JPG