A) \[30\sqrt{7}\,cm\]
B) \[30\,cm\]
C) \[\frac{15\sqrt{7}}{4}\,cm\]
D) \[\frac{15\sqrt{7}}{2}\,cm\]
Correct Answer: C
Solution :
By Heron's formula, we have area of triangle \[=\sqrt{s\,(s-a)\,(s-b)(s-c)}\] Here, \[s=\frac{11+15+16}{2}=\frac{42}{2}=21\,cm\] \[\therefore \] Area \[=\sqrt{21(10)\,(6)\,(5)}\] \[=30\sqrt{7}sq.\,cm\] \[\therefore \] Height \[=\frac{2\,\,area}{base}=\frac{2\times 30\sqrt{7}}{16}\] \[=\frac{15\sqrt{7}}{4}\,cm\]
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