Question

PYTHAGOREAN THEOREM
What is the length of the shortest side of a triangle that has vertices at (-3, -2), (1, 6), and (5, 3)?
A.SQUARE ROOT OF 7UNITS
B.9 UNITS
C.5 UNITS
D. 2 TIMES THE SQUARE ROOT OF 5 UNITS
NEED HELP A.S.A.P

Answer 1

Sides are:
\[\sqrt{(1--3)^2 + (6--2)^2} = \sqrt(16+64) = \sqrt80\ = 4\sqrt5\]
\[\sqrt{(5-1)^2+(3-6)^2} = \sqrt(16+9) = 5\]
\[\sqrt{(5--3)^2+(3--2)^2} = \sqrt{64+25} = \sqrt89\]
Therefore clearly the shortest side is 5 units.

Answer 2

THANKS