Question

Two vertices of a square are (12,0) and (12,12). What is the smallest possible area the square can have?

A. 24√2
B. 36
C. 48
D. 72

Answer 1

Explanation:

Make the points diagonal corners of the square...the distance,d between = 12 units .... then the side length using Pyhtagorean theorem is found via :

d^2 = s^2 *+ s^2

12^2 = 2s^2

144 = 2 s^2

s^2 = 72

s = sqrt (72)

and then, the area = s x s = 72 units^2