Problem: The perimeter of an isosceles right triangle is 2p. Its area is:
Answer Choices:
A. (2+2)p
B. (2−2)p
C. (3−22)p2
D. (1−22)p2
E. (3+22)p2
Solution:
Let x be one of the equal sides of the triangle. 2p=2x+x2;
∴x=2+22p,A=21x2=21⋅6+424p2⋅6−426−42=p2(3−22)