Find the area of triangle whose sides are 10 cm, 12 cm, and 18 cm.

$40\sqrt{2}cm^2$

Formula Used

Heron's formula = \(\sqrt {s(s\;-\;a)(s\;-\;b)(s\;-\;c)}\)

s = \(\frac{a\;+\;b\;+\;c}{2}\)

s = semi perimeter, a, b and c are the sides of triangle

Calculation

Putting value in formula

s = \(\frac{10\;+\;12\;+\;18}{2}\) = \(\frac{40}{2}\)

= Area of triangle = \(\sqrt {20(20\;-\;10)(20\;-\;12)(20\;-\;18)}\)

= \(\sqrt {20(10)(8)(2) }\) = \(\sqrt {3200 }\) = 40\(\sqrt {2}\)

Therefore, area of triangle is 40\(\sqrt {2}\)\( {cm }^{2 } \).