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

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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(6937) 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:
drawing%28400%2C400%2C-13%2C13%2C-5%2C21%2C%0D%0Alocate%287%2C10%2CX%29%2C+locate%28-8%2C10%2CX%29%2C%0D%0Aline%28-12%2C0%2C12%2C0%29%2C+line%2812%2C0%2C1%2C11sqrt%283%29%29%2C%0D%0A+%0D%0Aline%281%2C11sqrt%283%29%2C-1%2C11sqrt%283%29%29%2C+line%28-1%2C11sqrt%283%29%2C-12%2C0%29%2C%0D%0A+%0D%0Alocate%289.3%2C1.5%2C%2760%B0%27%29%2C+++%0D%0A+%0D%0Alocate%28-11%2C1.5%2C%2760%B0%27%29%2C+locate%28-.3%2C20.4%2C2%29+%0D%0A+%0D%0A%29
 

Draw in these two altitudes of the trapezoid, and
let their lengths be H:
 
drawing%28400%2C400%2C-13%2C13%2C-5%2C21%2Clocate%28-2%2C10%2CH%29%2C%0D%0Alocate%287%2C10%2CX%29%2C+locate%28-8%2C10%2CX%29%2C%0D%0Aline%28-12%2C0%2C12%2C0%29%2C+line%2812%2C0%2C1%2C11sqrt%283%29%29%2C%0D%0A+%0D%0Aline%281%2C11sqrt%283%29%2C-1%2C11sqrt%283%29%29%2C+line%28-1%2C11sqrt%283%29%2C-12%2C0%29%2C%0D%0A+%0D%0Alocate%289.3%2C1.5%2C%2760%B0%27%29%2C+++%0D%0Arectangle%28-1%2C0%2C1%2C11sqrt%283%29%29%2C%0D%0Alocate%28-11%2C1.5%2C%2760%B0%27%29%2C+locate%28-.3%2C20.4%2C2%29+%0D%0A+%0D%0A%29
 
Now we look at only the right triangle on the 
left, letting its bottom side's length be Z
 
drawing%28400%2C400%2C-13%2C13%2C-5%2C21%2C%0D%0A+locate%28-8%2C10%2CX%29%2C+locate%28-2%2C10%2CH%29%2C%0D%0Atriangle%28-12%2C0%2C-1%2C0%2C-1%2C11sqrt%283%29%29%2C%0D%0Alocate%28-6%2C-.5%2CZ%29%2C%0D%0Alocate%28-11%2C1.5%2C%2760%B0%27%29+%0D%0A+%0D%0A%29
 
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
  
drawing%28400%2C400%2C-13%2C13%2C-5%2C21%2C%0D%0Alocate%28-8%2C10%2CX%29%2C+locate%28-3%2C10%2CX%2Asqrt%283%29%2F2%29%2C%0D%0Atriangle%28-12%2C0%2C-1%2C0%2C-1%2C11sqrt%283%29%29%2C%0D%0Alocate%28-6%2C-.5%2CX%2F2%29%2C%0D%0Alocate%28-11%2C1.5%2C%2760%B0%27%29+%0D%0A+%0D%0A%29

Now we go back to the trapezoid:
 
drawing%28400%2C400%2C-13%2C13%2C-5%2C21%2C%0D%0Alocate%287%2C10%2CX%29%2C+locate%28-8%2C10%2CX%29%2C+locate%28-3%2C10%2CX%2Asqrt%283%29%2F2%29%2C%0D%0Aline%28-12%2C0%2C12%2C0%29%2C+line%2812%2C0%2C1%2C11sqrt%283%29%29%2C%0D%0Alocate%28-3%2C10%2CX%2Asqrt%283%29%2F2%29%2C++%0D%0Aline%281%2C11sqrt%283%29%2C-1%2C11sqrt%283%29%29%2C+line%28-1%2C11sqrt%283%29%2C-12%2C0%29%2C%0D%0A+%0D%0Alocate%289.3%2C1.5%2C%2760%B0%27%29%2C+locate%28-6%2C-.5%2CX%2F2%29%2C++%0D%0Arectangle%28-1%2C0%2C1%2C11sqrt%283%29%29%2C%0D%0Alocate%28-11%2C1.5%2C%2760%B0%27%29%2C+locate%28-.3%2C20.4%2C2%29+%0D%0A+%0D%0A%29

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.
 
drawing%28400%2C400%2C-13%2C13%2C-5%2C21%2C%0D%0Alocate%287%2C10%2CX%29%2C+locate%28-8%2C10%2CX%29%2C+locate%28-3%2C10%2CX%2Asqrt%283%29%2F2%29%2C%0D%0Aline%28-12%2C0%2C12%2C0%29%2C+line%2812%2C0%2C1%2C11sqrt%283%29%29%2C%0D%0Alocate%28-3%2C10%2CX%2Asqrt%283%29%2F2%29%2C++%0D%0Aline%281%2C11sqrt%283%29%2C-1%2C11sqrt%283%29%29%2C+line%28-1%2C11sqrt%283%29%2C-12%2C0%29%2C%0D%0A+%0D%0Alocate%289.3%2C1.5%2C%2760%B0%27%29%2C+locate%28-6%2C-.5%2CX%2F2%29%2C+locate%286%2C-.5%2CX%2F2%29%2C+%0D%0Arectangle%28-1%2C0%2C1%2C11sqrt%283%29%29%2Clocate%28-.3%2C-.5%2C2%29%2C%0D%0Alocate%28-11%2C1.5%2C%2760%B0%27%29%2C+locate%28-.3%2C20.4%2C2%29%0D%0A%29

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