draw the trap ABCD where AD=10,DC = 7, BC=17 and BA = 6
Draw AE parallel to DC, then AE=7 and BE = 7
We now have a parallelogram and a triangle.
Let's find angle AEB which is also angle C , call it Ø
by cosine law in triangle ABE
6^2 = 7^2 + 7^2 - 2(7)(7)cosØ
cosØ = .632653
Ø = 50.75386°
area = triangle + parallelogram
= (1/2)(7)(7)sin 50.75386 + (10)(7)sin50.75386
= 73.184
How do I find the area of a trapazoid with side lenghts 6 & 7, bottom 17 and top 10.
I know the perimeter is 40.
Now I need the area.
1 answer
1 answer