Question

find the area of a triangle whose side are 10 centimeter ,12cm,14cm​

Answer 1

$\bf{\underline{\underline{Answer:}}}$

$\bf{\therefore Area \:of \:Triangle = 24 \sqrt{6}\:cm^2 \:or \:58.78\: cm^2}$

$\bf{\underline {Given:}}$

• ABC is a triangle.
• Side a = 10 cm
• Side b = 12 cm
• Side c = 14 cm

$\bf{\underline {To\:Find:}}$

• The semi perimeter (s) of triangle

$\bf{\underline{\underline{Step\: by\: step \:explanation:}}}$

Semi perimeter of triangle,

$s = \dfrac{(a+b+c)}{2}\: unit\\ \\</p><p>=\dfrac{(10+12+14)}{2} cm\\ \\</p><p>=\dfrac{36}{2}cm \\ \\= 18 cm$

Finding the area of triangle by using the heron's formula :-

$\sf= \sqrt{s(s - a)(s - b)(s - c)}$

Area of triangle

$\begin{lgathered}= \sqrt{18(18 - 10)(18 - 12)(18 - 14)} \\ \\ = \sqrt{18 \times 8 \times 6 \times 4} \\ \\ = \sqrt{2 \times 3 \times 3 \times 2 \times 2 \times 2 \times 2 \times 3 \times 2 \times 2} \\ \\ = 2 \times 2 \times 2 \times 3 \times \sqrt{2 \times 3 } \\ \\ = 24 \sqrt{6}\end{lgathered}$

$\bf{\therefore Area \:of \:Triangle = 24 \sqrt{6}\:cm^2 \:or \:58.78\: cm^2}$