Question

The sides of a triangle are $50cm,78cm$ and $112cm.$ The smallest altitude is

Answer 1

Given: Sides of a triangle are $50cm,78cm$ and $112cm$

$s=\frac{50+78+112}{2}=\frac{240}{2}=120cm$

$\therefore$ Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$

$=\sqrt{120(120-50)(120-78)}$
$\sqrt{(120-112)}$

$=\sqrt{120\times 70\times 42\times 8}$

$=\sqrt{10\times 3\times 2\times 2\times 10\times }$
$\sqrt{7\times 7\times 3\times 2\times 2\times 2\times 2}$

$=10\times 7\times 3\times 2\times 2\times 2$

$=1680c{m}^2$

The smallest altitude will be the one corresponding to the longest side i.e., $112cm$

$\therefore$ Shortest altitude $=\frac{Area\times 2}{longestside}$

$=\frac{1680\times 2}{112}=30cm$

Thus, the shortest altitude is $30cm.$

Hence, $\left(b\right)$ is the correct answer.