a right triangle has a perimeter

[figs]

If a right triangle has a perimeter of 19 and a hypotenuse that is greater than 9, its area can be how many positive integers?

A. 1
B. 2
C. 3
D. 4
E. 5

can anybody explain the solution
OA:D

[dumb.doofus]

Well, if the hypotenuse is going to be greater than 9 then the sum of other two sides has to be greater than 9. If we are talking about only positive integers.. then the sum of other two sides can be just one value i.e. 10

so if a and b are the two sides then, a + b = 10 ---- (1)
and a^2 + b^2 = 81 ---(2)

squaring (1) and then subtracting (2) from it gives

2ab = 19 or ab = 19/2

or area of triangle = ab/2 = 19/4 = 4.75

Now since hypotenuse is greater than 9, it has to be less than 9.5 though, so in that case the value of a + b will also be lower than 10 but greater than 9.5

When that happens, you would see that ab/2 will be more close to 4 than to 5 and hence the answer is 4