Question

The perimeter of a triangle is $7p^2-5p+11$ and two of its sides are $p^2+2p-1$ and $3p^2-6p+3$. Find the third side of the triangle.

• A. $3p^2+p+9$
• B. $3p^2-p+9$
• C. $3p^2-p-9$
• D. $3p^2-p+5$

Answer 1

The correct option is

C

3p2−p+93{ p }^{ 2 }-p+9

Given,

Perimeter of △=7p2−5p+11\triangle =7{ p }^{ 2 }-5p+11△=$7p^2-5p+11$

And p2+2p−13p2−6p+3p^2+2p-1\\3p^2-6p+3p2+2p−13p2−6p+3

Let, the third side of the triangle is $y$.

We know that:

Perimeter of the triangle $=$ Sum of all its p2+2p−1,3p2−6p+3p^2+2p-

⇒p2+2p−1+3p2−6p+3+y=7p2−5p+11\Rightarrow \quad { p }^{ 2 }+2p-1+3{ p }^{ 2 }-6p+3+y=7{ p }^{ 2 }-5p+11⇒p2+2p−1+3p2−6p+3+y=⇒y=7p2−5p+11−4p2+4p−2⇒y=3p2−p+9\Rightarrow p^2+2p-1+3p^2-6p+3+y=\\\Rightarrow y=7p^2-5p+11-4p^2+4p-2\\\Rightarrow y=3p^2-p+9⇒p2+2p−1+3p2−6p+3+y=⇒y=7p2−5p+11−4p2+4p−2⇒y= 3p2−p+9