#### Answer

yes

#### Work Step by Step

(13,14,15)
Using Heron's formula if three sides of a triangle have length a,b,c then area A of a triangle is
A = $\sqrt s(s-a)(s-b)(s-c)$
where s = $\frac{a+b+c}{2}$
Lets take a=13 b=14 and c=15
s = $\frac{13+14+15}{2}$
= 21
A = $\sqrt 21(21-13)(21-14)(21-15)$
=$\sqrt 21*8*7*6$
= 84 $unit^{2}$
If a,b,c be the integer length of the sides of a triangle . If the area of the triangle is also a integer then (a,b,c) is known as a heron triplet.
Since A = 84 is a integer
(a,b,c) = (13,14,15) integer
Therefore the triplet is a heron triplet.