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The sides of a triangle are 35 cm, 54 cm and 61 cm respectively. Find the length of its longest altitude.

Answer» Let a = 35 cm, b = 54 cm and c = 61 cm. Then,
`s=(1)/(2)(a+b+c)=(1)/(2)(35+54+61)cm=75cm`.
` therefore" "(s-a)=(75-35)cm = 40 cm,(s-b)=(75-54)cm = 21cm and (s-c)=(75-61)cm = 14 cm`.
`therefore" ""area of the triangle"= sqrt(s(s-a)(s-b)(s-c))cm^(2)`
`= sqrt(75 xx 40 xx 21 xx 14)cm^(2)`
`= sqrt(15 xx 15 xx 14 xx 14 xx 4 xx 5)cm^(2)`
`=(15 xx 14 xx 2) sqrt(5)cm^(2) = 420 sqrt(5) cm^(2)`.
The longest altitude will be on smallest base.
Let the longest altitude be h cm. Then,
base = 35 cm and height = h cm.
`therefore" "(1)/(2) xx 35 xx h = 420 sqrt(5) rArr h = ((2 xx 420 xx sqrt(5))/(35))cm = 24 sqrt(5)` cm.
Hence, the longest altitude is `24 sqrt(5)` cm.


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