Probably -Area

Probability Level 3

A piece of wire of length 4L is bent at random,to form a rectangle. The probability that its area is at most $L^2/4$ is?

1/4

1 -\frac{\sqrt{3}}{2}

\frac{\sqrt{3} }{2}

4 - \sqrt{3}

1 solution

Deepak Jindal
Dec 19, 2014

Actually the question is slightly wrong.we need to find the probability of area less than or equal to (L^2)/4 To solve this question we can take the help of graphs. Let 'a'=length,'b'=breadth So, a+b=2L ab<=(L^2)/4 Let L=1 Now plot a+b=2 and ab=1/4 We will get two points of intersection. Now a+b=2 So (a+b)^ 2=4

a^2+b^2=4-1/2=7/2 (a-b)^ 2=7/2-1/2=3 a-b=3^(0.5) So a=1+(root3)/2, b=1-(root3)/2 Or a=1-(root3)/2, b=1+(root3)/2 Now with the help of graph Total length of hypotenuse=root(2^2+2^2)=2root2 Favourable length =2root2(1-(root3)/2) So probability= favorable length/total= 1-(root3)/2 Hope you might be able to understand my language.

Thanks, I have updated the problem accordingly.

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Calvin Lin Staff - 6 years, 5 months ago

I m not able to use equation editor

Rahul Kamble - 6 years, 5 months ago

To type Latex, all that you need to do is to place brackets $\backslash ( \quad \backslash )$ around your code. I've updated your question so that L^2 / 4 is displayed as $L^2 / 4$ . You can edit the question to take a look at how it is done.

Calvin Lin Staff - 6 years, 5 months ago

@Calvin Lin – Thank you sir

Rahul Kamble - 6 years, 5 months ago

Probably -Area

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