Question

The perimeter of a triangle is 8y2-8y+4 and its two sides are 3y2-5y and 4y2 +12. Find its third side​

Answer 1

Step-by-step explanation:

$\huge\blue{Answer:-}$

Let the triangle =ABC

the perimeter of triangle given is =${8y}^{2}-8y+4$

the two sides namely AB=${3y}^{2}-4\:and\\AC={4y}^{2}+12$

Let the other side =BC

we know that

$\large\blue{\boxed{The\:perimeter\:of\:triangle=3a}}$

according to formula,

AB+BC+AC=${8y}^{2}-8y+4\\=>{3y}^{2}-4+{4y}^{2}+12+BC={8y}^{2}-8y+4\\=>{7y}^{2}+8+BC={8y}^{2}-8y+4\\=>BC={8y}^{2}-8y+4-{7y}^{2}-8\\={y}^{2}-8y-4\\=y(y-8)-4$

Hence your answer is y(y-8)-4