A) \[\frac{1}{2}\sqrt{{{b}^{2}}{{c}^{2}}+{{c}^{2}}{{a}^{2}}+{{a}^{2}}{{b}^{2}}}\]
B) \[\frac{1}{2}(bc+ca+ab)\]
C) \[\frac{1}{2}abc\]
D) \[\frac{1}{2}\sqrt{{{(b-c)}^{2}}+{{(c-a)}^{2}}+{{(a-b)}^{2}}}\]
Correct Answer: A
Solution :
Length of sides are \[\sqrt{{{a}^{2}}+{{b}^{2}}},\,\sqrt{{{b}^{2}}+{{c}^{2}},}\sqrt{{{c}^{2}}+{{a}^{2}}}\] respectively. Now use \[\Delta =\frac{1}{2}\sqrt{s\,(s-a)\,(s-b)\,(s-c)}\]. Trick : Put \[a=2,\,\,b=2,\,\,c=2\], then sides will be \[2\sqrt{2},\,\,2\sqrt{2}\] and \[2\sqrt{2}\] i.e., equilateral triangle. So area of this triangle will be\[\Delta =\frac{\sqrt{3}}{4}\times {{(2\sqrt{2})}^{2}}=2\sqrt{3}\,\,sq.\,\,units\] Now option \[\,\Rightarrow \,\,\Delta =\frac{1}{2}\sqrt{16+16+16}=\]\[\frac{1}{2}\times 4\sqrt{3}\] \[=2\sqrt{3}\] . Hence the result.
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