The sides of a triangle are 56 cm,

(c) Since, the three sides of a triangle are $a=56 \mathrm{~cm}, b=60 \mathrm{~cm}$ and $c=52 \mathrm{~cm}$.

Then, semi-perimeter of a triangle,

$s=\frac{a+b+c}{2}=\frac{56+60+52}{2}=\frac{168}{2}=84 \mathrm{~cm}$

Area of a triangle $=\sqrt{s(s-a)(s-b)(s-c)}$ [by Heron's formula]

$=\sqrt{84(84-56)(84-60)(84-52)}$

$=\sqrt{84 \times 28 \times 24 \times 32}$

$=\sqrt{4 \times 7 \times 3 \times 4 \times 7 \times 4 \times 2 \times 3 \times 4 \times 4 \times 2}$

$=\sqrt{(4)^{6} \times(7)^{2} \times(3)^{2}}$

$=(4)^{3} \times 7 \times 3=1344 \mathrm{~cm}^{2}$

Hence, the area of triangle is $1344 \mathrm{~cm}^{2}$.