Question.
Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.
Solution:
Let the third side of the triangle be x.
Perimeter of the given triangle = 42 cm
$18 \mathrm{~cm}+10 \mathrm{~cm}+x=42$
$x=14 \mathrm{~cm}$
$s=\frac{\text { Perimeter }}{2}=\frac{42 \mathrm{~cm}}{2}=21 \mathrm{~cm}$
By Heron's formula,
Area of a triangle $=\sqrt{s(s-a)(s-b)(s-c)}$
Area of the given triangle $=(\sqrt{21(21-18)(21-10)(21-14)}) \mathrm{cm}^{2}$
$=(\sqrt{21(3)(11)(7)}) \mathrm{cm}^{2}$
$=21 \sqrt{11} \mathrm{~cm}^{2}$
Let the third side of the triangle be x.
Perimeter of the given triangle = 42 cm
$18 \mathrm{~cm}+10 \mathrm{~cm}+x=42$
$x=14 \mathrm{~cm}$
$s=\frac{\text { Perimeter }}{2}=\frac{42 \mathrm{~cm}}{2}=21 \mathrm{~cm}$
By Heron's formula,
Area of a triangle $=\sqrt{s(s-a)(s-b)(s-c)}$
Area of the given triangle $=(\sqrt{21(21-18)(21-10)(21-14)}) \mathrm{cm}^{2}$
$=(\sqrt{21(3)(11)(7)}) \mathrm{cm}^{2}$
$=21 \sqrt{11} \mathrm{~cm}^{2}$