Question

Suppose p denotes the perimeter of a triangle with sides a, b and c and s is the semi perimeter and r is the inradius. If A denotes the area of the triangle, select the statements that are true.

• A. $\frac{\sqrt{p(p-2a)(p-2b)(p-2c)}}{A}$ = 4
• B. $r=\frac{A}{s}$
• C. $\frac{\sqrt{p(p-2a)(p-2b)(p-2c)}}{A}$ = 9
• D. $r=\frac{5A}{s}$

Answer 1

Answer: (A), (B)

The correct options are
A

$\frac{\sqrt{p(p-2a)(p-2b)(p-2c)}}{A}$ = 4

B

$r=\frac{A}{s}$

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

(ii) The radius of the incircle of a triangle is equal to its area divided by half the perimeter.
$\Rightarrow r=\frac{A}{s}$