Question

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

• A. 26.6
• B. 27
• C. 26
• D. 420/17

Answer 1

Answer: (D)

Given, sides of $\Delta ABC$ are $17,25$ and $28$.

Area of $\Delta ABC=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$
$s=\frac{17+25+28}{2}=35$
$\Delta =\sqrt{35\left(35-17\right)\left(35-25\right)\left(35-28\right)}$
$\Delta =\sqrt{35\times 18\times 10\times 7}$
$\Delta =210$ ...(i)
Also, area of $\Delta =\frac{1}{2}\times BC\times AD$
$=\frac{1}{2}\times 17\times AD$ ...(ii)
From Eqs. $(i)$ and $(ii)$,
$210=\frac{1}{2}\times 17\times AD$
$\therefore AD=\frac{420}{17}$
Hence, length of largest altitude is $\frac{420}{17}$.