1). Two of the heights are less than 1/3

In this case the sides that act as hypotenuse to these heights may also very samll ie less that 1/3, so the perimeter might be or might not be greater that 1

INSUFF
2). One of the heights is greater than ½

Now if we visualise this height in a triangle which is not right angle then

this height will divide the triangle into 2 triangles each of whose hypotenuse is the side of the original triangle hence we can say that two sides itself are greater than 1

SUFF

C

I think it is B.
I've tried to explain it with reference to (w.r.t) attached doc for different possibilities of triangles.

1). Two of the heights are less than 1/3

wr.t attached doc figire 1, two of the heights are less than 1/3, in that case side a can be 1/3 but there is no restirction on the size of b (it can be more than 1 also). So perimiter can be > 1.

wr.t attached doc figure 2, two of the heights are less than 1/3, in that case side a, b & c can be 1/3 . So perimiter can be < 1.

Hence INSUFF.

2). One of the heights is greater than ½

wr.t attached doc figure 2, side b will be greater than 1/2. Minimum length of a can be 1/2 when it is height itself. Hence the perimiter will be greater than 1.

SUFF.

stmt 1 - two sides are less than 1/3...let's say .32...now third side can be .32 itself making the premieter < 1 or can be 0.5 making perimeter > 1....so insuff

stmt 2 - one side > 0.5..let's say 0.51...now the sum of the other two sides has to be atleast 0.52...so the perimeter will defenitely be > 1...Suff

So ans is B...

Sorry for the confusion the answer is B only, I solved it to show B is sufficient but by mistake wrote C instead of B