Problem:
Figure ABCD is a trapezoid with ABβ₯DC,AB=5,BC=32β, β BCD=45β and β CDA=60β. The length of DC is
Answer Choices:
A. 7+32β3β
B. 8
C. 921β
D. 8+3β
E. 8+33β
Solution:
Drop perpendiculars from A and B to DC, intersecting DC at F and E, respectively. β³BEC is an isosceles right triangle, so BE=EC=3. Since ABEF is a rectangle, FE=5 and AF=3. β³AFD is a 30β60β90 triangle, so DF=3β1βAF=3β. So DC=8+3β.