In quadrilateral ABCD, AB = 8, BC = 20, CD = 5*sqrt(2), angle ABC = 90, and angle BCD = 45. What is the area of ABCD?
+128474 Something like this (with the placement of D inexact )
A D
8 5sqrt (2)
B 20 C
Draw AC
AC = sqrt [ 20^2 + 8^2 ] = sqrt [464 ] = 4sqrt (29)
tan (ACB) = 8/20 = 2/5
arctan (2/5) = measure of angle ACB
Measure of angle ACD = measure of angle BCD - measure of angle ACB = 45 - arctan(2/5)
Area of figure = area of right triangle ABC + area of triangle ACD =
(1/2) (AB) (BC) + (1/2) (AC) ( DC) sin ( ACD) =
(1/2) (8) (20) + (1/2) (4sqrt (29) ) ( 5sqrt (2) ) sin [ 45 - arctan (2/5) ] =
80 + 10 sqrt ( 58) * sin [45 - arctan (2/5) ] = 110 units^2
