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BH = 2^2 * 3^2 * 5* 7

The power of 5 and 7 is just one in the product, this means only one of the dimensions (either B or H) would contain that power

Since the power of 2 and 3 is 2 in the product, it can be a part of B and H both individually and hence 6 would be able to divide B and H both.
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If the lengths of perpendicular sides of a park that is in the shape of a right-angled triangle park are integers and the area of the triangular park is 630 square units, which of the following can be the greatest number that divides the lengths of both the perpendicular sides?

I. 6
II. 10
III. 21


A. I only
B. II only
C. III only
D. I and II only
E. I, II and III

If the two perpendicular sides of the right triangle are a and b, we have:

½ x a x b = 630

a x b = 1260

Now let’s prime factorize 1260:

1260 = 126 x 10 = 6 x 21 x 2 x 5 = 2 x 3 x 3 x 7 x 2 x 5 = 2^2 x 3^2 x 5 x 7.

We see that 6 can be the greatest number that divides the lengths of both sides if, for example, a = 2 x 3 x 5 and b = 2 x 3 x 7. This is possible because 1260 has two factors of 2 and two factors of 3. Therefore, we see that neither 10 nor 21 can be the greatest number that divides the lengths of both the perpendicular sides since there is only one factor of 5 and one factor of 7 in 1260.

Answer: A
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