Question

In DABC, AB = 3 cm, BC = 4 cm and AC = 5 cm, find area of DABC.​

Answer 1

$\textbf{Given:}$

$\text{In \triangle ABC, AB=3cm, BC=4cm, AC=5cm}$

$\textbf{To find:}$

$\text{Area of \triangle ABC}$

$\textbf{Solution:}$

$\boxed{\begin{minipage}{6cm} \\\text{Area of triangle having sides a,b,c is}\\\\\text{Area}\mathrm{=\sqrt{s(s-a)(s-b)(s-c)}}\\\\\text{where}\;\mathrm{s=\dfrac{a+b+c}{2}} \end{minipage}}$

$\text{Take the sides of the triangle as }$

$\text{a=3,\;b=4,\;c=5}$

$\mathrm{s=\dfrac{a+b+c}{2}=\dfrac{3+4+5}{2}=\dfrac{12}{2}=6}$

$\textbf{Area of triangle ABC}$

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

$\mathrm{=\sqrt{6(6-3)(6-4)(6-5)}}$

$\mathrm{=\sqrt{6{\times}3{\times}2{\times}1}}$

$\mathrm{=\sqrt{2{\times}3{\times}3{\times}2}}$

$\mathrm{=\sqrt{2^2{\times}3^2}}$

$\mathrm{=2{\times}3}$

$\mathrm{=6}\;\text{square units}$

$\textbf{Answer:}$

$\text{Area of given triangle is 6 square cm}$

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