Problem:
Right triangles T1 and T2 have areas 1 and 2, respectively. A side of T1 is congruent to a side of T2, and a different side of T1 is congruent to a different side of T2. What is the square of the product of the lengths of the other (third) sides of T1 and T2?
Answer Choices:
A. 328
B. 10
C. 221
D. 332
E. 12
Solution:
Let a and b, with a<b, be the shared side lengths. Then T1 has hypotenuse b and legs a and b2−a2, and T2 has hypotenuse a2+b2 and legs a and b. Thus 21ab2−a2=1 and 21ab=2. Multiplying the first equation by 2 and then squaring gives a2b2−a4=4. From the second equation, a2b2=16, so 16−a4=4, which means a4=12. Then
b4=(a4)4=a444=12256=364
Therefore the square of the product of the other sides is