Question

The perimeter of triangle with vertices $A(0,0),B(5,7)$ and $C(9,5)$

• A. $\sqrt{74}+\sqrt{20}$
• B. $\sqrt{74}+\sqrt{106}$
• C. $\sqrt{74}+\sqrt{20}+\sqrt{106}$
• D. None of the above

Answer 1

Answer: (C)

$\sqrt{74}+\sqrt{20}+\sqrt{106}$

P $=AB+BC+CA$

$AB=\sqrt{5^2+7^2}$ and $BC=\sqrt{(5-9{)}^2+(7-5{)}^2}$
$=\sqrt{25+49}$ and $BC=\sqrt{16+4}=\sqrt{20}$
$=\sqrt{74}$

CA $=\sqrt{9^2+(5{)}^2}=\sqrt{81+25}$

CA $=\sqrt{106}$

$P=\sqrt{74}+\sqrt{20}+\sqrt{106}$

So, option C is correct.