Question

find the area of an triangle who's sides are 3 cm,4cm,5cm. hence find the length of the altitude coreponding to the longest side.​

Answer 1

$\large \bf \underline{Applying \: Heron's \: Formula;}\\\\\rm{Semi-Perimeter = \dfrac{Sum \: of \: all \: sides}{2}}\\\\\\\longrightarrow \rm{s = \dfrac{a+b+c}{2}}\\\\\\\implies \rm{s = \dfrac{3+4+5}{2}}\\\\\\\implies \rm{s = \dfrac{12}{2}}\\\\\\\implies \boxed{\rm{s = 6 \: cm}}$

$\rm{Area = \sqrt{s(s-a)(s-b)(s-c)} }\\\\\\\longrightarrow \rm{Area = \sqrt{6(6-3)(6-4)(6-5)} }\\\\\\\dashrightarrow \: \rm{Area=\sqrt{6(3)(2)(1)} }\\\\\\\implies \rm{Area = \sqrt{6 \times 3 \times 2} }\\\\\\\implies \rm{Area = \sqrt{36} }\\\\\\\implies \rm{Area = \pm 6}\\\\\\\tt{Area \: cannot \: be \: negative}\\\\\\\therefore \boxed{\bf{Area = 6 cm^{2} }}$

$\tt{Longest \: Side=5 cm}\\\\\\\sf{Area =\dfrac{1}{2} \times Base \times Height}\\\\\\\implies \sf{36 = \dfrac{1}{2} \times 5 \times Height}\\\\\\\implies \sf{36= 2.5 \times Height}\\\\\\\implies \large \bf \underline{Height = 14.4cm}$