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The sides of a triangle are 35 cm, 54 cm and 61 cm. What will be the length of its longest altitude?
Given :
The sides of a triangle are 35 cm, 54 cm and 61 cm.
To find :
We have to find the length of the longest altitude.
Solution :
Let a = 35 cm, b = 54 cm, c = 61 cm and h the longest altitude of the triangle.
We know that,
Semi-perimeter of a triangle of sides of lengths a, b and c = $\frac{a+b+c}{2}$
Area of the triangle = $\sqrt{s( s-a)( s-b)( s-c)}$
Semi-perimeter of the given triangle = $\frac{35+54+61}{2}$ cm = $\frac{150}{2}$ cm = 75 cm.
Area of the given triangle = $\sqrt{75( 75-35)( 75-54)( 75-61)} cm^{2}$
$=\sqrt{75\times 40\times 21\times 14} \ cm^{2}$
$=\ \sqrt{75\times 40\times 21\times 14} \ cm^{2}$
$=\sqrt{5\times 3\times 5\times 5\times 8\times 7\times 3\times 7\times 2} \ cm^{2}$
$=\sqrt{5\times 3\times 3\times 5\times 5\times 7\times 7\times 16} \ \ cm^{2}$
$=3\times 4\times 5\times 7\sqrt{5} \ \ cm^{2}$
$=420\sqrt{5} \ \ cm^{2}$.
We also know that,
Area of a triangle = $\frac{1}{2}\times b\times h$
The longest altitude is on the smallest side in length.
Therefore,
$\frac{1}{2}\times b\times h$ = $420\sqrt{5} cm^{2}$
$ \begin{array}{l} \frac{1}{2} \times 35\times h=420\sqrt{5} cm^{2}\ \ h=\frac{420\sqrt{5} \times 2}{35}\ h=12\sqrt{5} \times 2\ h=24\sqrt{5} \ cm \end{array}$
The length of the longest altitude of the triangle is $24\sqrt{5}$ cm.
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