Question

In the diagram, AB = 10 and AC = 2V10. What is the

Answer 1

For this case we have that by definition, the distance between two points is given by:

$d = \sqrt {(y2-y1) ^ 2 + (x2-x1) ^ 2}$

We find the distance AB:

$A: (3,4)\\B: (- 5, -2)$

$AB = \sqrt {(x2-x1) ^ 2 + (y2-y1) ^ 2}\\AB = \sqrt {(- 5-3) ^ 2 + (- 2-4) ^ 2}\\AB = \sqrt {(- 8) ^ 2 + (- 6) ^ 2}\\AB = \sqrt {64 + 36}\\AB = \sqrt {100}\\AB = 10$

We have the BC distance:

$B: (- 5, -2)\\C: (5, -2)$

$BC = \sqrt {(5 - (- 5)) ^ 2 + (- 2 - (- 2)) ^ 2}\\BC = \sqrt {(5 + 5) ^ 2 + (- 2 + 2) ^ 2}\\BC = \sqrt {(10) ^ 2 + (0) ^ 2}\\BC = \sqrt {100}\\BC = 10$

We find the CA distance:

$C: (5, -2)\\A: (3,4)$

$CA = \sqrt {(3-5) ^ 2 + (4 - (- 2)) ^ 2}CA = \sqrt {(- 2) ^ 2 + (4 + 2) ^ 2}\\CA = \sqrt {4 + 36}\\CA = \sqrt {40}\\CA = \sqrt {4 * 10}\\CA = 2 \sqrt {10}$

Thus, the perimeter is given by:

$10 + 10 + 2 \sqrt {10} = 20 + 2 \sqrt {10}$

ANswer:

$20 + 2 \sqrt {10}$