I need to solve this problem and I'm having trouble. The textbook didn't give an image.
In trapezoid ABCD, AB = CD = 2 cm, BC = 6 cm, segment
AD is parallel to segment BC, and the angle BAD = 60˚. What
is the area of the trapezoid? Express your answer in
simplest radical form.
2006-11-26 06:53:12 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
.............6
........______
....../.|...........|\
...../..|...........|..\ 2
..../_|_____|__\
The triangle on the right is a 30-60-90, so the shorter side must be 2/1 = 1. So the length of the bottom part is 6+1+1 or 8. The height must be 1√3, so the area is 1/2(6+8)√3 = 7√3.
Gopal screwed up when he did 2*2cos60. That's 2*2*1/2 = 2, not 4.
2006-11-26 07:03:01 · answer #1 · answered by Jim Burnell 6 · 1⤊ 0⤋
AD=BC+2*2cos60*
=6+4=10
sothe sum of the parallel sides=16cm
height or distance bet the prallel sides=2sin60*
=2rt3/2=rt3cm
so the area=1/2(10+6)rt3
8rt3 squarecm.
2006-11-26 06:58:58 · answer #2 · answered by raj 7 · 0⤊ 1⤋
AB= height 2 cm. CD = minor base, BC = bigger base= 6 cm., so the area is:
A= (2 + 6)*2/2 now use your calculator and solve this.
2006-11-26 07:01:27 · answer #3 · answered by Anonymous · 0⤊ 1⤋