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 21​ab2−a2​=1 and 21​ab=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