A parallelogram has sides A, B, C, and D. Sides A and B have a length of #2 # and sides C and D have a length of # 9 #. If the angle between sides A and C is #(3 pi)/8 #, what is the area of the parallelogram?

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Answer 1 · 2017-03-09T14:27:41

#" The Reqd Area="9(sqrt(2+sqrt2)).#

From Trigonometry, we know that,

The Area #a# of #Delta# formed by the sides #A and C# of the

#||^(grm)# is #=(1/2)(A)(C)sin/_(A,C)=(1/2)(2)(9)sin(3pi/8).#

#:.# Area of the #||^(grm)# = #2a=18sin(3pi/8)#

Here, #sin(3pi/8)=+sqrt{1/2(1-cos(2*3pi/8))#

#=sqrt{1/2(1-cos(3pi/4))}#

Since, #cos(3pi/4)=cos(pi-pi/4)=-cos(pi/4)=-1/sqrt2,#

#:. sin(3pi/8)=sqrt{1/2(1+1/sqrt2)}=sqrt{(sqrt2+1)/(2sqrt2)}#

#=sqrt[{(sqrt2+1)/(2sqrt2)}(sqrt2/sqrt2)]=sqrt(2+sqrt2)/2.#

#:." The Reqd Area="9(sqrt(2+sqrt2)).#

Enjoy Maths.!