Question

Find the area of triangle whose sides are $5$ cm, $12$ cm and $13$ cm. Also, find the length of its altitude corresponding to the longest side$.$

• A. $52c{m}^2$ and $4.61$ cm
• B. $30c{m}^2$ and $4.61$ cm
• C. $62c{m}^2$ and $4.61$ cm
• D. $70c{m}^2$ and $4.61$ cm

Answer 1

Answer: (B)

$30c{m}^2$ and $4.61$ cm

Given, sides of triangle $5$ cm, $12$ cm, $13$ cm.
Now semi perimeter, $s=$ $=\frac{\text{sum of the sides of triangle}}{2}$

$=\frac{5+12+13}{2}=15$ cm
Using heron's formula, Area of triangle$=\sqrt{s(s-a)(s-b)(s-c)}$

$=\sqrt{15(15-5)(15-12)(15-13)}$

$=\sqrt{15\times 10\times 3\times 2}=30{cm}^2$
Using altitude, area of triangle $=\frac{1}{2}\times$ base $\times$ altitude $=30{cm}^2$
$=\frac{1}{2}\times 13\times$ altitude $=30$

$=$ altitude $=\frac{30\times 2}{13}=4.61$ cm
So, altitude corresponding to largest side is $4.61$ cm.