Difference between revisions of "2007 Cyprus MO/Lyceum/Problem 7"

m (+)
m
 
Line 7: Line 7:
 
Using <math>30-60-90</math> [[right triangle]] [[ratio]]s, the lengths of the sides of the rectangle are <math>\frac{d}{2}</math> and <math>\frac{d\sqrt{3}}{2}</math>.
 
Using <math>30-60-90</math> [[right triangle]] [[ratio]]s, the lengths of the sides of the rectangle are <math>\frac{d}{2}</math> and <math>\frac{d\sqrt{3}}{2}</math>.
  
The area of the rectangle is <math>\frac{d^2\sqrt{3}}{4}\Rightarrow\mathrm{ A}</math>.
+
The area of the rectangle is <math>\frac{d^2\sqrt{3}}{4}\Longrightarrow\mathrm{ A}</math>.
  
 
==See also==
 
==See also==

Latest revision as of 20:43, 6 May 2007

Problem

If a diagonal $d$ of a rectangle forms a $60^\circ$ angle with one of its sides, then the area of the rectangle is

$\mathrm{(A) \ } \frac{d^2 \sqrt{3}}{4}\qquad \mathrm{(B) \ } \frac{d^2}{2}\qquad \mathrm{(C) \ } 2d^2\qquad \mathrm{(D) \ } d^2 \sqrt{2}\qquad \mathrm{(E) \ } \mathrm{None\;of\;these}$

Solution

Using $30-60-90$ right triangle ratios, the lengths of the sides of the rectangle are $\frac{d}{2}$ and $\frac{d\sqrt{3}}{2}$.

The area of the rectangle is $\frac{d^2\sqrt{3}}{4}\Longrightarrow\mathrm{ A}$.

See also

2007 Cyprus MO, Lyceum (Problems)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30