Albert has a unique letter obviously. Those are bolded below.
CAT
DOG
HAS
MAX
DIM
TAG

TAG has no unique letters so it can't be that.

Next, Bernard. He had to think, so he didn't know it from having a unique letter himself. But he knows that Albert knows, so it can't be TAG.

Bernard has G. Knowing that TAG is eliminated, it must be DOG.

Cheryl knows that they both know, looks down at her D, and gets the right answer as well.

Albert: o
Bernard: g
Crystal: d

dog

You're right but if you want to be complete then you've also got to consider something else. Imagine that Albert had C. He'd know immediately that the word was Cat. Bernard having the T would know from Albert's reaction that the word is cat. Cheryl has an A and from her point of view the word can be CAT or HAS. Because Bernard paused for a second it has to be CAT rather than HAS.

What about that?

DOG

For Albert, it has to be one of the unique letters on the board {C, O, H, S, X, I}, so Tag is out, since none of those are unique.

Any word that has A is out because if Albert gets them, then he wouldn't know the word of course. If Bernard gets them, they he also can't know the word since there's still 4 options (3 if we ignore TAG).

It can't be DIM since Albert has to get I first, and in either case of D or M Bernard will not know the word since it could have gone ID/OD or IM/XM to get to him.

It is DOG since Albert would have gotten O, then Bernard wold have gotten G (but knowing that it can't be TAG) he would know that it's DOG, which means that Cheryl gets D (slut), but she knows that it didn't go IMD (for MID) because when Bernard got M he wouldn't have known if it was XM or IM so he wouldn't have know the word.

DOG. tag doesn't work for first guy. dim and max don't work. Cat and has have the same problem for the third person.

First person gets o. Knows its dog.

Second person gets g. Knows it can't be tag, because first person wouldn't be able to know if they got t-a-g.

Third person gets d. It can't be dim, because if second person got i, first wouldn't have known with m (max or dim), and if second person got m, first could've had x for max or i for dim, so 2nd person wouldn't know if it was dim or max.

cat dog has max dim tag

Albert knows the word, so as everybody says, he has one of c,o,h,s,x,i. And the word is one of

cat dog has max dim

Bernard cannot have the c, otherwise the word would be cat, but Albert wouldn't know that. Similarly, Bernard doesn't have any of o, x, i. Therefore, he has one of t,g,s,h. So the word is one of

cat dog has

Cheryl can't have c,t,o,g,h,s for the same reasons Bernard's letters were restricted. If she has the a, she can't differentiate between cat and has so she has d for dog, with Albert's o and Bernard's g.

Let's call C the center of the larger circle of which quarter of is showing. Draw a line from C to B. Lets call x = AB, y = AC and z = BC. Notice that y is the diameter of the smaller circle and that z is the radius of the larger circle. Then by Pythagorean theorem we have that sqrt(x^2+y^2)=z. Now since z is the radius of the larger circle, then its area is pi*z^2 and thus the area of the quarter circle is (1/4)*pi*z^2. Area of the smaller circle is (1/4)*pi*y^2 (this time we're using the formula with the diameter).

Hence the area of the shaded shape is (1/4)*pi*z^2-(1/4)*pi*y^2 = (1/4)*pi*(sqrt(x^2+y^2)^2-y^2) = (1/4)*pi*(x^2+y^2-y^2) = (1/4)*pi*x^2 = ((12^2)/4)*pi = 36*pi.

What I think is interesting that if you take another quarter circle with the radius AB, it has the same area as the shaded shape. And like nich said the AB can be anything, which means that you can draw the smaller circle anywhere inside the quarter circle and draw the line parallel to the line here to get the same result.