To solve this we have to apply the Pythagorean theorem.
c^2= a^2+b^2
Where c is the hypotenuse of a right triangle (the longest side) and a and b are the other legs of the triangle.
Each of the triangles is mission their hypotenuse (c) so:
Smaller triangle (shortest wire)
c1^2 = 10^2+8^2
Solving for c1
c1^2 = 100+64
c1^2= 164
c1=√164
c1= 12.1
For the second triangle the sides are:
a2= 6+10= 16
Since both triangles are proportional
a1/b1= a2/b2
10/8 = 16/b2
Solving for b2
b2= 16/(10/8)
b2= 12.8
Back with the original formula for triangle 2:
c2^2= a2^2+b2^2
c2^2= 16^2+12.8^2
c2^2= 256+163.84
c2^2= 419.84
c2=√419.84
c2= 20.6
The shortest wire is 12.10 and the longest wire is 20.6
Option B.