Question

An isosceles trapezoid has vertices at (-2, 1), (2, 1), (5, -1), and
(-5, -1). Find the measure of each diagonal.

Answer 1

What is the distance from (-2,1) to (5,-1) ?

Answer 2

\[d = \sqrt{(\Delta y)^2 +(\Delta x)^2 }\]

Answer 3

ok, gimme a sec

Answer 4

I dont know how to apply your formula in the equation?

is Distance equals squareroot of triangle y^2 or some thing else?\[d=\sqrt{\triangle y}^2\]

Answer 5

Take two points (x1, y1) and (x2, y2)
The distance is sqrt( (y2-y1)^2 + (x2-x1)^2)

Answer 6

\[d=\sqrt{-2^2+7^2}\]
d=squareroot of 53

Answer 7

and then thats the answer right?