Question 450961
The other tutor's solution is correct, but there is a faster way to find the area. Given the parallelogram (I have used the other tutor's as a reference),


{{{drawing(400,550/3,-2,22,-2,9,
locate(8.2,4.3,"65°"), locate(9,6,5),
line(0,0,13.87400479,0),
line(13.87400479,0,20.79342001,7.185658223),
line(6.91941522,7.185658223,20.79342001,7.185658223),
line(6.91941522,7.185658223,0,0),
line(0,0,20.79342001,7.185658223),
line(6.91941522,7.185658223,13.87400479,0),
locate(6,2.1,11), locate(11.2,2,5),
locate(13.5,4.5,11)
 )}}}


We see that the parallelogram is consisted of:

*Two triangles with sides 5, 11, and a 65 degree angle in between
*Two triangles with sides 5, 11, and a 115 degree angle in between


If you know the area of the triangle can be expressed as


*[tex \LARGE A = \frac{1}{2}ab \sin{\gamma}]


where *[tex \gamma] is the angle in between a and b, we can find the area of the parallelogram (denoted *[tex A_p] this way. Here, we have two of each triangle, so


*[tex \LARGE A_p = 2(\frac{1}{2}(5)(11)\sin{(65)}) + 2(\frac{1}{2} (5)(11)\sin{(115)})]


*[tex \LARGE A_p = 55\sin(65) + 55\sin(115)]


Noting that sin(x) = sin(180-x), this is equal to


*[tex \LARGE A_p = 55\sin(65) + 55\sin(65) = 110\sin(65) = 99.694] (sq in)