Question

Suppose p denotes the perimeter of a triangle with sides a, b and c and s is the semi perimeter. If A denotes the area of the triangle, then which of the given options is correct?

• A. $\frac{\sqrt{p(p+a)(p+b)(p+c)}}{A}$ = 4
• B. $\frac{\sqrt{p(p-2a)(p-2b)(p-2c)}}{A}$ = 4
• C. $\frac{\sqrt{p(p-2a)(p-2b)(p-2c)}}{A}$ = 9
  
• D. $\frac{\sqrt{p(p+a)(p+b)(p+c)}}{A}$ = 9
  

Answer 1

The correct option is B $\frac{\sqrt{p(p-2a)(p-2b)(p-2c)}}{A}$ = 4

We know that,
$\text{Area}=\sqrt{s(s-a)(s-b)(s-c)}$
$A=\sqrt{\left(\frac{a+b+c}{2}\right)(\frac{a+b+c}{2}-a)(\frac{a+b+c}{2}-b)(\frac{a+b+c}{2}-c)}$
$A=\sqrt{\left(\frac{a+b+c}{2}\right)\left(\frac{a+b-c}{2}\right)\left(\frac{a-b+c}{2}\right)\left(\frac{-a+b+c}{2}\right)}$
$A=\frac{1}{4}(\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}$)
$A=\frac{1}{4}(\sqrt{(a+b+c)\left(\right(a+b+c)-2c)\left(\right(a+b+c)-2b)\left(\right(a+b+c)-2a)}$)
$A=\frac{1}{4}(\sqrt{p(p-2a)(p-2b)(p-2c)}$)
$\therefore \frac{\sqrt{p(p-2a)(p-2b)(p-2c)}}{A}=4$