D. 20 + 2âš10 units To solve this, you simply need to calculate the length of each side of the triangle with the vertexes of A(3,4), B(-5,-2), and C(5,-2). The length of each side is simply calculated using the pythagoras theorem. Note that it doesn't matter what order you do the subtraction. The absolute value will be the same and if it happens to be negative, not a problem since it will become positive once you square the values. So the length of side AB is sqrt((3-(-5))^2 + (4-(-2))^2) = sqrt(8^2 + 6^2) = sqrt(64 + 36) = sqrt(100) = 10. The length of side BC is sqrt((-5 - 5)^2 + (-2 - (-2))^2) = sqrt(-10^2 + 0^2) = sqrt(100+0) = sqrt(100) = 10. And finally, the length of side AC is sqrt((3-5)^2 + (4-(-2))^2) = sqrt(-2^2 + 6^2) = sqrt(4+36) = sqrt(40) = 2 * sqrt(10) Finally, add all the lengths together. 10 + 10 + 2âš10 = 20 + 2âš10