Question

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

perimeter of AABC?
10 units
10 + 2/10 units
20 units
20+ 2/10 units

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}`