ABCD is a quadrilateral with measurements of its sides shown in the above diagram. If AB is parallel to CD What is the area of the quadrilateral?
One way (and I don't claim that this is the easiest way):
Draw a line segment from A perpendicular to CD ending at point X on CD.
Draw a line segment from B perpendicular to CD ending at point Y on CD.
You now have triange(ADX), rectangle(AXYB), and triangle(BYC).
Combine triangle(ADX) and triangle(BYC) so that points A and B coincide and points X and Y coincide, creating a triangle
with sides of lengths 13, 15, and 14. [CY + DX = CD - 10 = 14]
Using Heron's formula to find the area of this new triangle: s = (13 + 14 + 15)/2 = 21
---> Area = sqrt[ 21(21-13)(21-14)(21-15) ] = 84.
To find the height of this triangle (which is the width of the rectangle):
---> Area = ½·base·height ---> 84 = ½·(14)·height ---> height = 12.
Therefore, the area of the rectangle is: 10·12 = 120
and the total area is 84 + 120 = 204