Suppose a square has side length s. How could you use the formula for the area of a trapezoid to find the area of the square?

A. For a square, h=b1=b2=s.

Substitute into the trapezoid area formula. You get A=12s(s+s)=12s(2s)=s^2.

This result is consistent with the formula for the area of a square.

B. For a square, s=b1=b2. Substitute into the trapezoid area formula. You get A=12s^2=s^2/2. This result is consistent with the formula for the area of a trapezoid.

Atrain traveled for 1.5 hours to the first station, stopped for 30 minutes, then traveled for 4 hours to the final station where it stopped for 1 hour. the total distance traveled is a function of time. which graph most accurately represents this scenario? a graph is shown with the x-axis labeled time (in hours) and the y-axis labeled total distance (in miles). the line begins at the origin and moves upward for 1.5 hours. the line then continues upward at a slow rate until 2 hours. from 2 to 6 hours, the line continues quickly upward. from 6 to 7 hours, it moves downward until it touches the x-axis a graph is shown with the axis labeled time (in hours) and the y axis labeled total distance (in miles). a line is shown beginning at the origin. the line moves upward until 1.5 hours, then is a horizontal line until 2 hours. the line moves quickly upward again until 6 hours, and then is horizontal until 7 hours a graph is shown with the axis labeled time (in hours) and the y-axis labeled total distance (in miles). the line begins at the y-axis where y equals 125. it is horizontal until 1.5 hours, then moves downward until 2 hours where it touches the x-axis. the line moves upward until 6 hours and then moves downward until 7 hours where it touches the x-axis a graph is shown with the axis labeled time (in hours) and the y-axis labeled total distance (in miles). the line begins at y equals 125 and is horizontal for 1.5 hours. the line moves downward until 2 hours, then back up until 5.5 hours. the line is horizontal from 5.5 to 7 hours