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 :
We have, sides of triangle 11 cm, 15 cm and 16 cm. \[s=\frac{11+15+16}{2}=21\] \[\therefore \]Area of triangle \[=\sqrt{21(21-11)(21-15)(21-16)}\] \[=30\sqrt{7}\,c{{m}^{2}}\] Let altitude to the largest side be h cm \[\therefore \] \[\frac{1}{2}\times 16\times h=30\sqrt{7}\Rightarrow 8h=30\sqrt{7}\] \[\Rightarrow \] \[h=\frac{15\sqrt{7}}{4}\,cm\]
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