The correct option is b, 10.
Given to us,
Sides of the triangle x, (x+4) and, 20.
According to the Triangle Inequality Theorem, the sum of any 2 sides of a triangle must be greater than the third side.
Thus,
![x+(x+4)>20\\2x+4>20\\2x> 16\\x>8](https://tex.z-dn.net/?f=x%2B%28x%2B4%29%3E20%5C%5C2x%2B4%3E20%5C%5C2x%3E%2016%5C%5Cx%3E8)
Now, let us assume it as a right-angle triangle with the longest side as 20.
Thus,
![x^2+(x+4)^2=20^2\\](https://tex.z-dn.net/?f=x%5E2%2B%28x%2B4%29%5E2%3D20%5E2%5C%5C)
using the expression, (a+b)² = a² + b² + 2ab;
![x^2+x^2+16+8x=400](https://tex.z-dn.net/?f=x%5E2%2Bx%5E2%2B16%2B8x%3D400)
Diving both sides by 2,
![x^2+8+4x=200\\](https://tex.z-dn.net/?f=x%5E2%2B8%2B4x%3D200%5C%5C)
subtracting 200 on both sides,
![x^2+8+4x-200=200-200\\2x^2+4x-192=0](https://tex.z-dn.net/?f=x%5E2%2B8%2B4x-200%3D200-200%5C%5C2x%5E2%2B4x-192%3D0)
Solving further, using ;
x = 12, -16
As negative value can be neglected.
for an Acute angle triangle,
X should be 12 >x >8. Therefore the only feasible option is 10.
Hence, the correct option is b, 10.
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