Friday, November 27, 2009

Fox 195

6 comments:

  1. Similar to Fox 191.

    A=5

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  2. If we assume the inclination of the shortest side to the horizontal as θ then A=(1/2)(1/sin(θ))(2/cos(θ)+1/cos(θ))= 3/sin(2θ) and the minimum value of this would be 3. So my answer is Option E. Is there some error in my calculation?
    Ajit

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  3. MIN{ 3/sin(2θ) } gives θ=45, gives Area=5.

    You may be making a minor error.

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  4. At θ=45, sin(2θ)=sin(90)=1 and hence A= 3/sin(2θ)= 3. Is that not so?

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  5. This must be wrong: 3/sin(2θ)

    If θ=45 => area in the picture is 5.
    Forget about the equation, please look at the picture.

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  6. Yes, of course. I stand corrected. In fact, A =(1/2)*(1/sin(θ))[3/cos(θ)+2/cos(θ))= 5/sin(2θ) and therefore A(min)= 5 leading to Option A.
    Thanks are due to Anonymous for being patient with me.
    Ajit

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