Problem:
Square EFGH is inside square ABCD so that each side of EFGH can be extended to pass through a vertex of ABCD. Square ABCD has side length 50,E is between B and H, and BE=1. What is the area of the inner square EFGH?
Answer Choices:
A. 25
B. 32
C. 36
D. 40
E. 42 Solution:
The symmetry of the figure implies that △ABH,△BCE,△CDF, and △DAG are congruent right triangles. So
BH=CE=BC2−BE2=50−1=7
and EH=BH−BE=7−1=6. Hence the square EFGH has area 62=36.
OR
As in the first solution, BH=7. Now note that △ABH,△BCE,△CDF, and △DAG are congruent right triangles, so
