Question

The perimeter of a triangle is 7p^2-8p+9 and two of its sides are 2p^2-p+1 and 11p^2-3p+5. Find the third side of the triangle.

Answer 1

$\huge\mathbb{SOLUTION:-}$

$\sf\bullet Perimeter\:of\: triangle=Sum\:of\:all\:sides$

let the third side be x

$\sf\implies 7p{}^{2}-8p+9= 2p{}^{2}-p+1+11p{}^{2}-3p+5 +x\\ \\ \\ \sf\implies 7p{}^{2}-8p+9= 13p{}^{2}-4p+6 +x\\ \\ \sf\implies 7p{}^{2}-8p+9-(13p{}^{2}-4p+6)=x\\ \\ \sf\implies 7p{}^{2}-8p+9 -13p{}^{2}+4p-6=x\\ \\ \sf\implies -6p{}^{2}-4p+3=x$

So the third side of triangle is

$\boxed{\sf{\red{-6p{}^{2}-4p+3}}}$

Answer 2

Step-by-step explanation:

Let the sides of triangle be AB, BC, CA

now a/c to ques,

AB + BC + CA = 7P²-8P + 9

Given that two side... i.e

AB = 2P²-P+1

BC = 11P²-3P+5

SO third side i.e CA = Perimeter - (AB + BC)

= 7p²- 8p +9 - (13p²-4p +6)

=7p²-8p+9 - 13p² + 4p - 6

= - 6p² - 4p +3