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In quadrilateral ABCD, AB = 8, BC = 20, CD = 5*sqrt(2), angle ABC = 90, and angle BCD = 45. What is the area of ABCD?

 Jul 20, 2022
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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 

 

cool cool cool

 Jul 21, 2022

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