Bunuel wrote:
What is the ratio of the area of quadrilateral BCDE to the area of triangle ABE?
(1) BC = 1⁄2*AC
(2) Angle CBE = 135°
Solution: In the above diagram, \(\angle CAD=180-(90+45)=45^o \) and \(\angle ABE=180-(90+45)=45^o\)
Attachment:
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We can see that \(\triangle ACD\) and \(\triangle ABE\) are similar by AAA condition
This means \(\frac{Area of \triangle ACD}{Area of \triangle ABE} = (\frac{AC}{AB})^2=(\frac{CD}{BE})^2=(\frac{AD}{AE})^2\)
We are asked the \(\frac{Area of BCDE}{Area of \triangle ABE}\) which can easily be found after getting \(\frac{Area of \triangle ACD}{Area of \triangle ABE}\)
Statement 1: BC = 1⁄2*AC
According to this statement, \(\frac{BC}{AC}=\frac{1}{2}\) or \(\frac{AC}{BC}=\frac{2}{1}\) or \(\frac{AC}{AB}=\frac{2}{1}\)
This means \(\frac{Area of \triangle ACD}{Area of \triangle ABE} = (\frac{AC}{AB})^2=(\frac{2}{1})^2=\frac{4}{1}\)
Thus, we can ultimately say \(\frac{Area of BCDE}{Area of \triangle ABE}=\frac{4-1}{1}=\frac{3}{1}\)
So,
statement 1 alone sufficient and we can eliminate options B, C and EStatement 2: Angle CBE = 135°
This statement is not only insufficient but also redundant because from the given diagram itself we can infer that \(\angle CBE=360-(90+90+45)=135\)
Attachment:
similar5.png [ 6.11 KiB | Viewed 1453 times ]
Hence the right answer is
Option A