Question 186035
Given an isosceles trapezoid with a smaller base 0f 2, a perimeter of 70, and acute base angles of 60 degrees, find the trapezoid's area. 
<pre><font size = 4 color = "indigo"><b> 
Let the slanted sides of the trapezoid be X:
{{{drawing(400,400,-13,13,-5,21,
locate(7,10,X), locate(-8,10,X),
line(-12,0,12,0), line(12,0,1,11sqrt(3)),
 
line(1,11sqrt(3),-1,11sqrt(3)), line(-1,11sqrt(3),-12,0),
 
locate(9.3,1.5,"60°"),   
 
locate(-11,1.5,"60°"), locate(-.3,20.4,2) 
 
)}}}
 

Draw in these two altitudes of the trapezoid, and
let their lengths be H:
 
{{{drawing(400,400,-13,13,-5,21,locate(-2,10,H),
locate(7,10,X), locate(-8,10,X),
line(-12,0,12,0), line(12,0,1,11sqrt(3)),
 
line(1,11sqrt(3),-1,11sqrt(3)), line(-1,11sqrt(3),-12,0),
 
locate(9.3,1.5,"60°"),   
rectangle(-1,0,1,11sqrt(3)),
locate(-11,1.5,"60°"), locate(-.3,20.4,2) 
 
)}}}
 
Now we look at only the right triangle on the 
left, letting its bottom side's length be Z
 
{{{drawing(400,400,-13,13,-5,21,
 locate(-8,10,X), locate(-2,10,H),
triangle(-12,0,-1,0,-1,11sqrt(3)),
locate(-6,-.5,Z),
locate(-11,1.5,"60°") 
 
)}}}
 
Now we know that 
 
{{{H/X=sin(60)}}} and {{{Z/X=cos(60)}}}
 
or
 
{{{H=X*sin(60)}}} and {{{Z=X*cos(60)}}} 
 
We also know that {{{cos(60)=1/2}}} and {{{sin(60)=sqrt(3)/2}}}
 
so
 
{{{Z = X*(1/2)}}}  and {{{H=X*(sqrt(3)/2)}}}
 
or
 
{{{Z = X/2}}}  and {{{H=(X*sqrt(3)/2)}}}
 
so we replace {{{Z}}} by {{{X/2}}} and H by {{{(X*sqrt(3))/2}}}
  
{{{drawing(400,400,-13,13,-5,21,
locate(-8,10,X), locate(-3,10,X*sqrt(3)/2),
triangle(-12,0,-1,0,-1,11sqrt(3)),
locate(-6,-.5,X/2),
locate(-11,1.5,"60°") 
 
)}}}

Now we go back to the trapezoid:
 
{{{drawing(400,400,-13,13,-5,21,
locate(7,10,X), locate(-8,10,X), locate(-3,10,X*sqrt(3)/2),
line(-12,0,12,0), line(12,0,1,11sqrt(3)),
locate(-3,10,X*sqrt(3)/2),  
line(1,11sqrt(3),-1,11sqrt(3)), line(-1,11sqrt(3),-12,0),
 
locate(9.3,1.5,"60°"), locate(-6,-.5,X/2),  
rectangle(-1,0,1,11sqrt(3)),
locate(-11,1.5,"60°"), locate(-.3,20.4,2) 
 
)}}}

We know the small segment on the bottom is 2 because
the top side of the trapezoid is 2.  And we know that
the right segment on the bottom is also {{{X/2}}} because
the right triangle on the left and the one on the right 
are congruent.
 
{{{drawing(400,400,-13,13,-5,21,
locate(7,10,X), locate(-8,10,X), locate(-3,10,X*sqrt(3)/2),
line(-12,0,12,0), line(12,0,1,11sqrt(3)),
locate(-3,10,X*sqrt(3)/2),  
line(1,11sqrt(3),-1,11sqrt(3)), line(-1,11sqrt(3),-12,0),
 
locate(9.3,1.5,"60°"), locate(-6,-.5,X/2), locate(6,-.5,X/2), 
rectangle(-1,0,1,11sqrt(3)),locate(-.3,-.5,2),
locate(-11,1.5,"60°"), locate(-.3,20.4,2)
)}}}

Now we are told that the perimeter is 70.

So to form the perimeter, we add up all four sides of the 
trapezoid.

The left side of the trapezoid is {{{X}}}
The bottom side is this sum {{{X/2+2+X/2}}}
The right side is {{{X}}}
The top side is {{{2}}}.

So the sum of all those must equal 70:

{{{X+(X/2+2+X/2)+X+2=70}}}

We solve that for X and get 22

So now we know that the bottom side is

{{{X/2+2+X/2 = X+2 = 22+2 = 24}}} 

and that is the larger base of the trapezoid.

And its height is {{{(X*sqrt(3))/2=(24sqrt(3))/2=12sqrt(3)}}}.

Now we can find the area using {{{B[1]=2}}}, {{{B[2]=24}}}, and
{{{H=12sqrt(3)}}}, in the formula

{{{Area=(B[1]+B[2])(H/2)}}}

{{{Area=(2+24)(12sqrt(3)/2)=26(6sqrt(3))= 156sqrt(3)=270.199926}}}.

Edwin</pre>