Question 148655
Let's name the parallel sides as {{{a[1] =9inches}}} and {{{b[1]=12inches}}}
And the non parallel sides as "c=?" and "d=?".
Before I proceed farther, I suggest you draw it on your own while we discuss it by words so you can follow better (just a suggestion).
.
Now from that starting point and also the endpoint of that 9 inches which is {{{a[1]}}}, draw a line all the way down to {{{b[1]}}} which is perpendicular right? --> mark these lines as "h" (for height),likewise the 2 lines are parallel, and that line inside these parallel lines on {{{b[1]}}} also measures 9 inches right (of course) ---> mark it {{{a[2]}}}
.
Now you have remaining length of 3 inches,{{{b[1]-a[2]=12-9=3}}}
And these 3 inches will be cut into half, 1-1/2" on one side, likewise 1-1/2" on the other. Why? Because you have opposite similar 65 deg angle right? (If not the same angle, the length won't be the same)
The 1-1/2" length mark one side as {{{b[2]}}} and the other as {{{b[3]}}}
.
Can you follow still?
Now, side "c", "h" and {{{b[2]}}} forms a right triangle (if you mark them properly) with angle 65 deg. To get "c",
{{{cos65=b[2]/c}}}
{{{c=b[2]/cos65}}}= 1.5/cos65
c=3.55 inches = d -------------> length of the non parallel sides
.
For area in the trapezoid,
A=1/2(h){{{(a[1]+b[1])}}}
To get "h", {{{sin65=h/c}}}
h=(sin65)(3.55)=3.22 inches
Therefore,
A=(1/2)(3.22){9+12}
A=33.81 sq inches
<font><font size=4><b>Thank you,
Jojo</font>