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given,
ABCD is a square and
EB=6-2b and AD=6
so AE=2b

area of triangle =(1/2)*2b*2b=\(2b^2\)
area of square =\(6^2\)=36

probability of finding a point in triangle region=\(2b^2/36\)=\(1/3\) (given)

by solving b(quant A) =√6=2.44>2(quant B)

option A is correct
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Since ABCD is a square, it is clear that AB=6.
Therefore, if EB= 6-2b then AE= 2b.
AE is the base of the shaded triangle.
If the probability of a randomly selected point lying in the shaded triangle is 1/3. This will equal the (Area of the shaded triangle/ area of the square).
The area of the square ABDE is 6^2= 36.
Determined by 1/3= Area of triangle/ 36. This will give the area of the triangle to be 12.

The area of the triangle = (2b*2b)/2, which is 2b^2=12.
b^2=6 and therefore the value of b= √6 > 2.
Giving us the the answer A

Thanks and best of luck of everyone

We are given that side of the square is 6, and that the shaded region is composed of an isos. triangle.

Area of the square is therefore 36. If there is a 1/3 chance of randomly going to the shaded region, it implies that the area of the isos. triangle must be 1/3 of the squares area, which is 12.

If the area of the isos triangle is 12, then that implies that the two legs of the triangle are:
\(x^2/2=24\)
\(x=2\sqrt{6}\)

Since the side of the angle must add up to 6, we know that:
\(2\sqrt{6}+6-2b=6\)
\(2\sqrt{6}=2b\)
\(\sqrt{6}=b\)
Since 2 is equal to root 4, that implies that b is greater than 2, so QA is larger than QB.

A.