SOLUTION: The shorter base of a trapezoidal lot measures 12m. The non-parallel sides make 45 degrees and 60 degrees interior angles with the longer base. Find the area of the lot if the s

Algebra ->  Trigonometry-basics -> SOLUTION: The shorter base of a trapezoidal lot measures 12m. The non-parallel sides make 45 degrees and 60 degrees interior angles with the longer base. Find the area of the lot if the s      Log On


   



Question 1184911: The shorter base of a trapezoidal lot measures 12m. The non-parallel sides
make 45 degrees and 60 degrees interior angles with the longer base.
Find the area of the lot if the side inclined at an angle of 45 degrees with
the longer base measures 9m.

Found 2 solutions by greenestamps, Edwin McCravy:
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Ugly numbers; you can do the details to get the answer.

The 9m side is the hypotenuse of a 45-45-90 right triangle, so the length of each leg is (9/sqrt(2)) meters. One leg is part of the longer base; the other is the height of the trapezoid.

The height of the trapezoid is the long leg of a 30-60-90 right triangle; the length of the short leg is (9/sqrt(2)) divided by sqrt(3). That short leg is the remaining part of the longer base.

Now you have expressions for the height of the trapezoid, the length of the shorter base, and the lengths of the three pieces of the longer base; the area of the trapezoid is the height times the average of the two bases.


Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
We draw perpendiculars to the longer base from the ends of the shorter
base (in green), and let their lengths be h.  We have right triangles on
each side of the trapezoid.  We let x = the length of the bottom side of the
right triangle on the right side of the trapezoid.



In the right triangle on the right side, for the 60° angle,

sin%2860%5Eo%29=OPPOSITE%2FHYPOTENUSE=h%2F9
sqrt%283%29%2F2=h%2F9
Cross-multiply:
2h=9sqrt%283%29
h=9sqrt%283%29%2F2   <--the height of the trapezoid

cos%2860%5Eo%29=ADJACENT%2FHYPOTENUSE=x%2F9
1%2F2=x%2F9
Cross-multiply:
2x=9
x=9%2F2    <--the bottom side of the right triangle on the right.

Now we can put in the values for h and x:



We don't need to solve the right triangle on the left because we know that a
45-45-90 right triangle is isosceles and its two legs are the same length,
so we have this:



%28matrix%284%2C1%2CAREA%2COF%2CTHE%2CTRAPEZOID%29%29%22%22=%22%22%28matrix%288%2C1%2CAREA%2COF%2CTHE%2CRIGHT%2CTRIANGLE%2CON%2CTHE%2CLEFT%29%29%22%22%2B%22%22%28matrix%287%2C1%2CAREA%2COF%2CTHE%2CRECTANGLE%2CIN%2CTHE%2CMIDDLE%29%29%22%22%2B%22%22%28matrix%288%2C1%2CAREA%2COF%2CTHE%2CRIGHT%2CTRIANGLE%2CON%2CTHE%2CRIGHT%29%29%22%22=%22%22

%281%2F2%29%289sqrt%283%29%2F2%29%289sqrt%283%29%2F2%29%22%22%2B%22%2212%289sqrt%283%29%2F2%29%22%22%2B%22%22%281%2F2%29%289%2F2%29%289sqrt%283%29%2F2%29%22%22=%22%22

81%2A3%2F8%22%22%2B%22%2254sqrt%283%29%22%22%2B%22%2281sqrt%283%29%2F8%22%22=%22%22

243%2F8%22%22%2B%22%22%28432sqrt%283%29%29%2F8%22%22%2B%22%2281sqrt%283%29%2F8%22%22=%22%22

243%2F8%22%22%2B%22%22%28513sqrt%283%29%29%2F8%22%22=%22%22

%28243%2B513sqrt%283%29%29%2F8%22%22=%22%22141.442758 m²

Edwin