Question

The sides of a triangle are ( -12x^2– 5) , (36x^2 + 10) and (25x^2 + 15). Find its perimeter.

Answer 1

Perimeter of triangle = a + b + c

$\sf \\ \sf let \\ \sf a = ( - 12 {x}^{2} - 5) \\ \sf \: b = (36 {x}^{2} + 10) \\ \sf \: c = (25 {x}^{2} + 15)$

Add them to get perimeter,

$( - 12 {x}^{2} - 5) + (36 {x}^{2} + 10) + (25 {x}^{2} + 15)$

$- 12 {x}^{2} - 5 + 36 {x}^{2} + 10 + (25 {x}^{2} + 15)$

$- 12 {x}^{2} - 5 + 36 {x}^{2} + 10 + 25 {x}^{2} + 15$

Collect like terms,

$49 {x}^{2} - 5 + 10 + 15$

Add first then subtract.

$49 {x}^{2} + 20$

Hope it's correct.