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Q. In a triangle ABC, the sides are of length 17, 25 and 28 units. Then, the length of the largest altitude is

UPSEEUPSEE 2018

Solution:

Given, sides of $ΔABC$ are $17,\, 25$ and $28$.
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Area of $\Delta ABC=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$
$s=\frac{17+25+28}{2}=35$
$\Delta= \sqrt{35\left(35-17\right)\left(35-25\right)\left(35-28\right)}$
$\Delta=\sqrt{35\times18\times10\times7}$
$\Delta=210$ ...(i)
Also, area of $\Delta=\frac{1}{2} \times BC\times AD$
$=\frac{1}{2}\times17\times AD$ ...(ii)
From Eqs. $(i)$ and $(ii)$,
$210=\frac{1}{2}\times17\times AD$
$\therefore AD=\frac{420}{17}$
Hence, length of largest altitude is $\frac{420}{17}$.