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. 23​−3
B. 2​/3
C. 1−3​/3
D. 4−23​
E. 3​/4
Solution:
Let DM=NB=x : then AM=AN=1−x. 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(1−x)2=x2+1 or x2−4x+1=0.
The roots of this equation are 2−3​ and 2+3​. Since 2+3​>1 we must choose x=2-\sqrt
Now, area â–³CMN= area ABCD - area â–³ANM - area â–³NBC - area â–³CDM
=1−21​(1−x)2−2x​−2x​=21​(1−x2).
Substituting x=2−3​ we obtan area △CMN=23​−3.
