Find the area of the quadrilateral with vertices (0,3), (3,0), (0,9), and (9,0).
Let A = (3,0) B = (0,3) C = (9,0) D = (0,9) O = (0,0)
This is a trapezoid with AB and CD being the parallel sides (the bases).
The formula for the area of a trapezoid: A = ½·h·(b1 + b2)
b1 = side AB b2 = side CD
Since triangle(OAB) is 45o-45o-90o right triangle with OA = 3 and OB = 3,, AB = 3·sqrt(2).
Since triangle(OCD) is 45o-45o-90o right triangle with OC = 9 and OD = 9,, CD = 9·sqrt(2).
Now to find the height -- draw line segment OXY perpendicular to both AB and CD, with
X the point of intersection with AB and Y the point of intersection with CD.
Since triangle(OAX) is 45o-45o-90o right triangle with OA = 3, OX = 3/sqrt(2).
Since triangle(OCY) is 45o-45o-90o right triangle with OC = 9, OY = 9/sqrt(2).
XY = OY - OX = 9/sqrt(2) - 3/sqrt(2) = 6/sqrt(2) or 3·sqrt(2).
This is the height of the trapezoid.
A = ½·h·(b1 + b2) = ½·[ 3·sqrt(2) ]·[ 3·sqrt(2) + 9·sqrt(2) ]
= ½·[ 3·sqrt(2) ]·[ 12·sqrt(2) ]
= 36