Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 14"

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==Problem==
 
==Problem==
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[[Image:2006 CyMO-14.PNG|250px|right]]
[[Image:2006 CyMO-14.PNG|250px]]
 
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The rectangle <math>AB\Gamma \Delta</math> is a small garden divided to the rectangle <math>AZE\Delta</math> and to the square <math>ZB\Gamma E</math>, so that <math>AE=2\sqrt{5}m</math> and the shaded area of the triangle <math>\Delta BE</math> is <math>4m^2</math>. The area of the whole garden is
 
The rectangle <math>AB\Gamma \Delta</math> is a small garden divided to the rectangle <math>AZE\Delta</math> and to the square <math>ZB\Gamma E</math>, so that <math>AE=2\sqrt{5}m</math> and the shaded area of the triangle <math>\Delta BE</math> is <math>4m^2</math>. The area of the whole garden is
  
A. <math>24m^2</math>
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<math>\mathrm{(A)}\ 24\ \text{m}^2\qquad\mathrm{(B)}\ 20\ \text{m}^2\qquad\mathrm{(C)}\ 16\ \text{m}^2\qquad\mathrm{(D)}\ 32\ \text{m}^2\qquad\mathrm{(E)}\ 10\sqrt{5}\ \text{m}^2</math>
 
 
B. <math>20m^2</math>
 
 
 
C. <math>16m^2</math>
 
 
 
D. <math>32m^2</math>
 
 
 
E. <math>10\sqrt5m^2</math>
 
  
 
==Solution==
 
==Solution==

Revision as of 10:30, 27 April 2008

Problem

2006 CyMO-14.PNG

The rectangle $AB\Gamma \Delta$ is a small garden divided to the rectangle $AZE\Delta$ and to the square $ZB\Gamma E$, so that $AE=2\sqrt{5}m$ and the shaded area of the triangle $\Delta BE$ is $4m^2$. The area of the whole garden is

$\mathrm{(A)}\ 24\ \text{m}^2\qquad\mathrm{(B)}\ 20\ \text{m}^2\qquad\mathrm{(C)}\ 16\ \text{m}^2\qquad\mathrm{(D)}\ 32\ \text{m}^2\qquad\mathrm{(E)}\ 10\sqrt{5}\ \text{m}^2$

Solution

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See also

2006 Cyprus MO, Lyceum (Problems)
Preceded by
Problem 13
Followed by
Problem 15
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