Problem: In the adjoining figure ABCD is a square and CMN is an equilateral triangle. If the area of ABCD is one square inch, then the area of CMN in square inches is
Answer Choices:
A. 23ββ3
B. 2β/3
C. 1β3β/3
D. 4β23β
E. 3β/4
Solution:
Let DM=NB=x : then AM=AN=1βx. Denoting the length of each side of the equilateral triangle CMN by y and using the Pythagorean Theorem we see
x2+12=y2 and (1-x)^{2}+(1-x)^{2}=y^
Substituting the first equation into the second we get
2(1βx)2=x2+1 or x2β4x+1=0.
The roots of this equation are 2β3β and 2+3β. Since 2+3β>1 we must choose x=2-\sqrt
Now, area β³CMN= area ABCD - area β³ANM - area β³NBC - area β³CDM
=1β21β(1βx)2β2xββ2xβ=21β(1βx2).
Substituting x=2β3β we obtan area β³CMN=23ββ3.