SOLUTION: Given: AB∥DC,DE⊥AB,AD = BC m∠ADC = 134° AD = 40, DC = 32 Find: Area of ABCD https://www.bing.com/images/blob?bcid=T8FJaMRAB4YAoQ

Algebra ->  Trigonometry-basics -> SOLUTION: Given: AB∥DC,DE⊥AB,AD = BC m∠ADC = 134° AD = 40, DC = 32 Find: Area of ABCD https://www.bing.com/images/blob?bcid=T8FJaMRAB4YAoQ      Log On


   

Question 1147593: Given: AB∥DC,DE⊥AB,AD = BC
m∠ADC = 134°
AD = 40, DC = 32
Find: Area of ABCD
https://www.bing.com/images/blob?bcid=T8FJaMRAB4YAoQ
Answer by Edwin McCravy(20065) About Me  (Show Source):
You can put this solution on YOUR website!


AB∥DC tells us that ABCD is a trapezoid (trapezium in the UK).
AD = BC tells us it is an isosceles trapezoid. 
m∠ADC = 134°, ∠ADC and ∠A are supplementary, so ∠A = 180°-134°=46°

DE%2FAD=sin%28%22%22%3CA%29

DE%2FAD=sin%2846%5Eo%29

DE=AD%2Asin%2846%5Eo%29

DE=40%2A0.7193398003

DE=28.77359201

AE%2FAD=cos%28%22%22%3CA%29

AE%2FAD=cos%2846%5Eo%29

AE=AD%2Acos%2846%5Eo%29

AE=40%2A0.6946583705

AE=27.78633482



Area of triangle BCF is also 399.7563308

Area of rectangle DEFC = EF∙DE = 32∙28.77359201 = 920.7549443

Adding the two triangles' and the rectangle's areas:

399.7563308 + 399.7563308 + 920.7549443 = 1720.267606

Edwin