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Expert: Steve Holleran - 9/13/2008


Question

opto
We know area of BCD = 200 sin 2x.   Can we find an expression for BD^2  and can we use this to show that the area of triangle BAD is 200 - 200cos2x and hence show that area of kite is  A(x) = 200(1 - cos2x + sin2x)

Answer
Hello Jack,

Well , to get started, using the Law of Cosines, on triangle DBC, you have

      BD^2 = 20^2 + 20^2 - 2(20)(20)cos 2x

           = 800 - 800 cos 2x

In triangle ABD, since angle A is 90 and AB = AD, then angles
ABD and ADB are 45 degrees.

Since BD^2 = 800 - 800 cos 2x, BD = sqrt(800 - 800 cos 2x)

Also, triangle ABD is a 45-45-90 right triangle, so to get the sides
AB and AD, divide BD by sqrt(2):

  AB = AD = (sqrt(800 - 800 cos 2x)/sqrt(2))

          = (sqrt(400(1-cos 2x)) each.

The area of triangle ABD is 1/2 * AB * AD which would be

    Area ABD = 1/2 * (sqrt(400(1- cos 2x))(sqrt(400(1- cos 2x))

             = 1/2 * 400 * (1 - cos 2x)  = 200 * (1 - cos 2x)

The the area of the kite is the area of triangle ABD + triangle DBC

  A(x) = 200( 1 - cos 2x) + 200 sin 2x

       = 200( 1 - cos 2x + sin 2x)



Hope this is okay.
Steve
    Questioner's Rating
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    CommentExcellent response Thanks a million


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Steve Holleran

Expertise

I can help with all math questions from basic math to Calculus. Whether it`s consumer questions, or questions from high school or college students, I have probably dealt with it at some time in my career.

Experience

33 years teaching experience in NJ public schools

Education/Credentials
B.S. Mathematics : Wake Forest University 1972 M.S. Mathematics : Monmouth University 1981

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