Question

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

• A. $27$
• B. $26$
• C. $420/17$
• D. $26.6$

Answer 1

The correct option is C $420/17$

$F\text{ormula used}$
$\text{Area of triangle using Heron}{}^{\prime }\text{s formula}$
$\sqrt{s(s-a)(s-b)(s-c)}wheres=\frac{a+b+c}{2}$
$Givensidesoftriangleas17,25,28$
$s=\frac{17+25+28}{2}=\frac{70}{2}=35$
$Area=\sqrt{35(35-17)(35-25)(35-28)}$
$=\sqrt{35\left(18\right)\left(10\right)\left(7\right)}$
$=\sqrt{44100}$
$=210$
$WeknowAreaoftriangle=\frac{1}{2}\times base\times height$
$\frac{1}{2}\times 17\times height=210$
$\Rightarrow height=\frac{210\left(2\right)}{17}$
$=\frac{420}{17}$