Problem:
In the figure, the length of side AB of square ABCD is 50,E is between B and H, and BE=1. What is the 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=(C)36.
OR
As in the first solution, BH=7. Now note that △ABH,△BCE,△CDF, and △DAG are congruent right triangles, so
Area(EFGH)=Area(ABCD)−4Area(△ABH)=50−4(21⋅1⋅7)=36
Answer: C.
The problems on this page are the property of the MAA's American Mathematics Competitions
