Finding the area of the trapezoid

[isu23pink]

[attachment=0:3p8kmt99]Photo-015j.jpg[/attachment:3p8kmt99]

okay, so since that had rads i think that's where i screwed up.
so i'm not 100% sure my work is right. this is what i did:
(3rad5)^2 = 3^2 + x^2
15 = 9 + x^2
subtract 9 from both sides
6 = x^2
radr = x

?

 

[soroban]

Hello, isu23pink!

okay, so since that had rads i think that's where i screwed up.
so i'm not 100% sure my work is right. this is what i did:

. . \(\displaystyle x^2 + 3^2 \:=\3\sqrt{5})^2\)
. . \(\displaystyle x^2 + 9 \:=\:15\)
. . . . . . . . . \(\displaystyle \uparrow\)
. . . . . . . .Here!

\(\displaystyle (3\sqrt{5})^2 \;=\;3^2\cdot(\sqrt{5})^2 \;=\;9\cdot5 \;=\;45\)