SICP exercise 1.3 - sum of squares of two largest of three numbers
First of all, the name sum-of-squares is misleading.  Longer names-separated-with-hyphens are quite acceptable in Scheme, so why not tell the whole truth?  I suggest sum-of-squares-except-min or sum-of-squares-of-greatest-two.

The first solution is a bit of a brute-force approach.  It works, but it is not obvious at a glance how it works unless you trace all paths of the decision tree.

The second solution, if the indentation and naming were improved, would be a bit clearer.

(define (sum-of-squares-except-min a b c)
    (let ((max-of-bc (if (&gt; b c) b c))
          (max-of-ac (if (&gt; a c) a c)))
         (if (&gt; a b)
             (+ (square a) (square max-of-bc))
             (+ (square b) (square max-of-ac)))))


Personally, I would adapt the second approach and go "all the way" with it.  Then, all the possibilities would be laid out logically, one per line.

(define (sum-of-squares-except-min a b c)
    (let ((least (min a b c)))
         (cond ((= least a) (+ (square b) (square c)))
               ((= least b) (+ (square a) (square c)))
               (else        (+ (square a) (square b))))))


min should be built in, but you can implement it as a helper function yourself if you need to.

Alternatively, sum all the squares, then subtract the square of the least.

Which approach is fastest?  It's hard to say, without benchmarking, whether it's better to define fewer intermediate variables, do less branching, or do less arithmetic.  I wouldn't worry about performance for such a simple problem.