Question

The perimeter of a triangle having sides (p^2+2p-1) ; (3p^2-6p+3)and (3p^2-p+q)

Answer 1

Answer:

perimeter of triangle = sum of all sides

= (p^2+2p-1) + (3p^2-6p+3) + (3p^2-p+q)

= 7p^2 - 5p + (4+q)