What is the perimeter of a triangle with corners at (1 ,4 ), (3 , 2 ), and (9 ,7 )?

Aug 30, 2016

The perimeter of $\Delta = 2 \sqrt{2} + \sqrt{61} + \sqrt{73} \approx 2.828 + 7.810 + 8.544 = 19.182$.

Explanation:

Let the given corners be $A \left(1 , 4\right) , B \left(3 , 2\right) , \mathmr{and} , C \left(9 , 7\right)$.

Then, using The Distance Formula,

$A B = \sqrt{{\left(1 - 3\right)}^{2} + {\left(4 - 2\right)}^{2}} = \sqrt{8} = 2 \sqrt{2}$.

$B C = \sqrt{{\left(3 - 9\right)}^{2} + {\left(2 - 7\right)}^{2}} = \sqrt{61}$.

$C A = \sqrt{{\left(9 - 1\right)}^{2} + {\left(7 - 4\right)}^{2}} = \sqrt{73}$.

Hence, the perimeter of $\Delta A B C$

$= A B + B C + C A = 2 \sqrt{2} + \sqrt{61} + \sqrt{73}$.

$\approx 2.828 + 7.810 + 8.544 = 19.182$.