Find the area of a triangle whose sides are 42 cm, 34 cm and 20 cm.

To find the area of the triangle, we will first find the semiperimeter of the triangle.

Thus, we have:

$s=\frac{1}{2}(a+b+c)=\frac{1}{2}(42+34+20)=\frac{1}{2} \times 96=48 \mathrm{~cm}$

Now,

Area of the triangle $=\sqrt{s(s-a)(s-b)(s-c)}$ $=\sqrt{48(48-42)(48-34)(48-20)}$

$=\sqrt{48 \times 6 \times 14 \times 28}$

$=\sqrt{112896}$

$=336 \mathrm{~cm}^{2}$