A) \[30\sqrt{7}\,cm\]
B) \[\frac{15\sqrt{7}}{2}cm\]
C) \[\frac{15\sqrt{7}}{4}cm\]
D) \[30\,cm\]
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\[=\frac{30}{8}\sqrt{7}=\frac{15\sqrt{7}}{4}cm.\]
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