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Updated on: 26 Nov 2018, 06:39
Originally posted by singh_amit19 on 19 Oct 2007, 04:37.
Last edited by Bunuel on 26 Nov 2018, 06:39, edited 7 times in total.
Last edited by Bunuel on 26 Nov 2018, 06:39, edited 7 times in total.
Renamed the topic, edited the question and added the OA.
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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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
Attachment:
squareabcd.jpg [ 17.9 KiB | Viewed 264331 times ]
Solving this question helps.
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25 Jan 2012, 10:53
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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
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 [ 12.11 KiB | Viewed 268139 times ]
30 Jul 2016, 00:07
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singh_amit19
My advise is not to use any fancy triangle properties or weird geometry.
This is a 2D geometry question and the shape of the object is a KITE.
This question can be solved under 20 seconds, The trick is use obscure properties of shapes to your advantage ; in this case the area of a KITE.
If you see the question properly it is asking us to find the area of half KITE and then subtract the area of a Half square
Mathematically
Area = 1/2*area of KITE-1/2* area of square
Area = 1/2 (area of KITE-area of square)
Area of the square is super easy .. its 1; half of it will be = 1/2
For finding the area of the KITE you will need its two diagonal. Smaller one is \(\sqrt{2}\). Bigger one is \(1+\sqrt{2}\)
FORMULA FOR KITE AREA= \(\frac{1}{2}\)*(diagonal1 * diagonal 2)
Solve it and you will get \(\frac{\sqrt{2}}{4}\)
