SOLUTION: Question: Given an isosceles triangle with a smaller base 0f 2, a perimeter of 70, and acute base angles of 60 degrees, find the trapezoid's area. Work done so far: I know the

Algebra ->  Customizable Word Problem Solvers  -> Geometry -> SOLUTION: Question: Given an isosceles triangle with a smaller base 0f 2, a perimeter of 70, and acute base angles of 60 degrees, find the trapezoid's area. Work done so far: I know the      Log On

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Question 186035: Question: Given an isosceles triangle with a smaller base 0f 2, a perimeter of 70, and acute base angles of 60 degrees, find the trapezoid's area.

Work done so far: I know the formula for Area of trapezoid is %28B%5B1%5D%2BB%5B2%5D%29%28H%2F2%29.
b1 is 2 but I don't know if I draw altitudes for the trapezoid, the altitudes will equal 2 because the altitudes will make a square. I'm not sure on that part.

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
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.
 
Let the slanted sides of the trapezoid be X:

 

Draw in these two altitudes of the trapezoid, and
let their lengths be H:
 

 
Now we look at only the right triangle on the 
left, letting its bottom side's length be Z
 

 
Now we know that 
 
H%2FX=sin%2860%29 and Z%2FX=cos%2860%29
 
or
 
H=X%2Asin%2860%29 and Z=X%2Acos%2860%29 
 
We also know that cos%2860%29=1%2F2 and sin%2860%29=sqrt%283%29%2F2
 
so
 
Z+=+X%2A%281%2F2%29  and H=X%2A%28sqrt%283%29%2F2%29
 
or
 
Z+=+X%2F2  and H=%28X%2Asqrt%283%29%2F2%29
 
so we replace Z by X%2F2 and H by %28X%2Asqrt%283%29%29%2F2
  


Now we go back to the trapezoid:
 


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%2F2 because
the right triangle on the left and the one on the right 
are congruent.
 


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%2F2%2B2%2BX%2F2
The right side is X
The top side is 2.

So the sum of all those must equal 70:

X%2B%28X%2F2%2B2%2BX%2F2%29%2BX%2B2=70

We solve that for X and get 22

So now we know that the bottom side is

X%2F2%2B2%2BX%2F2+=+X%2B2+=+22%2B2+=+24 

and that is the larger base of the trapezoid.

And its height is %28X%2Asqrt%283%29%29%2F2=%2824sqrt%283%29%29%2F2=12sqrt%283%29.

Now we can find the area using B%5B1%5D=2, B%5B2%5D=24, and
H=12sqrt%283%29, in the formula

Area=%28B%5B1%5D%2BB%5B2%5D%29%28H%2F2%29

Area=%282%2B24%29%2812sqrt%283%29%2F2%29=26%286sqrt%283%29%29=+156sqrt%283%29=270.199926.

Edwin