A) \[\frac{3\sqrt{15}}{4}c{{m}^{2}}\]
B) \[\frac{15\sqrt{3}}{4}c{{m}^{2}}\]
C) \[\frac{15}{4}c{{m}^{2}}\]
D) \[\frac{3\sqrt{3}}{4}c{{m}^{2}}\]
Correct Answer: B
Solution :
Given that \[2b=a+c\]and \[c=7cm\] \[\cos C=\frac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}\Rightarrow -\frac{1}{2}=\frac{{{a}^{2}}+\frac{{{a}^{2}}+{{c}^{2}}+2ac}{4}-{{c}^{2}}}{2a\frac{(a+c)}{2}}\] On simplification and putting the value of c, we get \[a=3\]and\[b=5\]. Hence the area is\[\frac{15\sqrt{3}}{4}c{{m}^{2}}\].
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