3. Two similar triangles, their perimeters are 36 and 12, respectively. The largest side of the triangle with the larger perimeter is 15, and the smallest side of the triangle with the smaller perimeter is 3. Then the area of the triangle with the larger perimeter is ( ).
(A) 52
(B) 54
(C) 56
(D) 58
3. B.

Let the smallest side of the larger perimeter triangle be $x$. Then $\frac{x}{3}=$ $\frac{36}{12}$. Solving for $x$ gives $x=9$.
The other side should be $36-15-9=12$.
Since $15^{2}=12^{2}+9^{2}$, both triangles are right triangles, and the area of the larger perimeter triangle is $\frac{1}{2} \times 9 \times 12=54$.