Difference between revisions of "1981 AHSME Problems/Problem 2"

(Solution)
(Solution)
 
Line 4: Line 4:
  
 
==Solution==
 
==Solution==
Note that <math>\triangle BCE</math> is a right triangle. Thus, we do Pythagorean theorem to find that side <math>BC=\sqrt{3}</math>. Since this is the side length of the square, the area of <math>ABCD</math> is <math>3, \boxed{\textbf{(C)}\ 3}</math>.  
+
Note that <math>\triangle BCE</math> is a right triangle. Thus, we do Pythagorean theorem to find that side <math>BC=\sqrt{3}</math>. Since this is the side length of the square, the area of <math>ABCD</math> is <math>\boxed{\textbf{(C)}\ 3}</math>.  
  
 
~superagh
 
~superagh

Latest revision as of 22:16, 7 August 2020

Point $E$ is on side $AB$ of square $ABCD$. If $EB$ has length one and $EC$ has length two, then the area of the square is

$\textbf{(A)}\ \sqrt{3}\qquad\textbf{(B)}\ \sqrt{5}\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 2\sqrt{3}\qquad\textbf{(E)}\ 5$

Solution

Note that $\triangle BCE$ is a right triangle. Thus, we do Pythagorean theorem to find that side $BC=\sqrt{3}$. Since this is the side length of the square, the area of $ABCD$ is $\boxed{\textbf{(C)}\ 3}$.

~superagh

Invalid username
Login to AoPS