The sides of a triangle are $50\ cm, 78\ cm$ and $112\ cm$. The smallest altitude is:
A$20\ cm$
B$30\ cm$
C$40\ cm$
D$50\ cm$
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B$30\ cm$
The smallest altitude is $\perp$ drawn to the largest side of a $\triangle$ From opposite point. i. e. BD Area of $\triangle=\frac{1}{2} \times \mathrm{AC} \times \mathrm{BD}=\frac{1}{2} \times 112 \times \mathrm{BD}=56 \times \mathrm{BD}$
$\mathrm{s}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2}=\frac{50+78+115}{2}=120 \mathrm{~cm}$
$\mathrm{~s}-\mathrm{AB}=70 \mathrm{~cm}, \mathrm{~s}-\mathrm{BC}=42 \mathrm{~cm}, \mathrm{~s}-\mathrm{AC}=8 \mathrm{~cm}$
$\text { Area }=\sqrt{\mathrm{s}(\mathrm{~s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}$
$=\sqrt{120 \times 70 \times 42 \times 8}=1680 \mathrm{~cm}^2$
Now, $56 \times \mathrm{BD}=1680 \mathrm{~cm}^2$
$\Rightarrow \mathrm{BD}=\frac{1680}{56}=30 \mathrm{~cm}$
Hence, correct option is $(b)$.
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