Question

A triangle has integer sides. One side is three times a second side. The length of then third side is $15$. the greatest possible perimeter of the triangle is

• A. $43$
• B. $46$
• C. $48$
• D. $52$

Answer 1

The correct option is A $43$

Let second side be 'a'

So first side will be ${}^{\prime }3a^{\prime }$

Third side is given $15$

Now to form a triangle sum of any two sides must be greater than third side, so

$15+a>3a$

So maximum possible integer value is $7$

$=3a+a+15$

$=3\left(7\right)+\left(7\right)+15$

$=21+7+15$

$=43$.