The sides of a triangle are $11\ cm, 15\ cm$ and $16\ cm$. The altitude to the largest side is:
A$30\sqrt{7}\text{cm}$
B$\frac{15\sqrt{7}}{2}\text{cm}$
C$\frac{15\sqrt{7}}{2}\text{cm}$
D$60\text{cm}$
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D$60\text{cm}$
$\text{s}=\frac{11+60+61}{2}=66\text{m}$
Area of $\triangle=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=\sqrt{21\times10\times6\times5}=30\sqrt{7}\text{cm}^2$
Also if we choose largest side and its Altitude, the area would be
$\text{A}=\frac12\times\text{largest side}\times\text{h}$
$330=\frac12\times11\times\text{Height}$
$\Rightarrow\text{Height}=\frac{2\times330}{11}=60\text{m}$
Hence, correct option is $(d)$.
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