+817 In the diagram, $AB\perp BC$ and $BC \perp CD$. $AB = 6'', BC = 4''$, and $CD = 10''$. $E$ is the midpoint of $AD$. How many square inches are in the area of $ABCD$?
+129771 Draw AC
Triangle BAC is right with angle ABC = 90
AC = sqrt [AB^2 + BC^2] = sqrt [ 6^2 + 4^2 ] = sqrt [ 52 ]
sin [angle BCA ] = 6/sqrt [52]
Angle BCD = 90
Angle BCD = angle BCA + angle ACD
sin [ angle BCA ] = cos angle [ACD ]
cos [angle ACD ] = 6/sqrt 52
sin [angle ACD ] = sqrt [ 1 - (6/sqrt 52)^2 ] = sqrt [ 1 - 36/52 ] = sqrt [ 16/52] = 4 / sqrt 52
[ ABCD] =
[ABC] + [ACD] =
(1/2)(BA)(BC) + (1/2) (AC) (CD) ( 4 /sqrt 52) =
(1/2)(6) (4) + (1/2) (sqrt 52) ( 10) (4 /sqrt 52) =
12 + 20 =
32
