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
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
Similar to Fox 191.
ReplyDeleteA=5
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?
ReplyDeleteAjit
MIN{ 3/sin(2θ) } gives θ=45, gives Area=5.
ReplyDeleteYou may be making a minor error.
At θ=45, sin(2θ)=sin(90)=1 and hence A= 3/sin(2θ)= 3. Is that not so?
ReplyDeleteThis must be wrong: 3/sin(2θ)
ReplyDeleteIf θ=45 => area in the picture is 5.
Forget about the equation, please look at the picture.
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.
ReplyDeleteThanks are due to Anonymous for being patient with me.
Ajit