Trapezoid ABCD has area of 45 . Its diagonals intersect at M. Find the area of AMB if AB is parallel to CD and AB=2CD.
+128475 45 = (1/2) (height) (2CD + CD)
90 = (height) ( 3CD)
30 = (height) ( CD)
30 / CD = height of trapezoid
AMB and BMC are similar triangles
Since AB = 2CD
Then height of AMB = 2 (height of BMC)
So there are three equal parts to the height of the trapezoid and the height of AMB is two of them
So height AMB = (2/3) (30) /( CD) = 20 / CD
Area of AMB = (1/2) (AB) [height (AMB) ] = (1/2) (2CD) [ 20/CD] = 20
