Question

sides of the triangle are 13cm, 14cm and 15 cm, find the length of the
smallest altitude of the triangle.

Answer 1

$\textbf{Given:}$

$\text{Sides of the triangle are 13cm, 14cm and 15 cm}$

$\textbf{To find:}$

$\text{Smallest altitude of the triangle}$

$\textbf{Solution:}$

$\text{Let the sides be}$

$a=13\,cm,\;b=14\,cm,\;c=15\,cm$

$s=\dfrac{a+b+c}{2}$

$s=\dfrac{13+14+15}{2}$

$s=\dfrac{42}{2}=21$

$\textbf{Area of the triangle}$

$=\sqrt{s(s-a)(s-b)(s-c)}$

$=\sqrt{21(21-13)(21-14)(21-15)}$

$=\sqrt{21{\times}8{\times}7{\times}6}$

$=\sqrt{3{\times}7{\times}4{\times}2{\times}7{\times}2{\times}3}$

$=2{\times}3{\times}2{\times}7$

$=84\,cm^2$

$\textbf{Case(i):}$

$\text{Take, base}\;b=13\,cm$

$\text{But, Area=}84\,cm^2$

$\implies\dfrac{1}{2}{\times}b{\times}h=84$

$\implies\dfrac{1}{2}{\times}13{\times}h=84$

$\implies\,h=\dfrac{168}{13}$

$\implies\boxed{\,h=12.9\,cm}$

$\textbf{Case(ii):}$

$\text{Take, base}\;b=14\,cm$

$\text{But, Area=}84\,cm^2$

$\implies\dfrac{1}{2}{\times}b{\times}h=84$

$\implies\dfrac{1}{2}{\times}14{\times}h=84$

$\implies\,h=\dfrac{168}{14}$

$\implies\boxed{\,h=12\,cm}$

$\textbf{Case(iii):}$

$\text{Take, base}\;b=15\,cm$

$\text{But, Area=}84\,cm^2$

$\implies\dfrac{1}{2}{\times}b{\times}h=84$

$\implies\dfrac{1}{2}{\times}15{\times}h=84$

$\implies\,h=\dfrac{168}{15}$

$\implies\boxed{\,h=11.2\,cm}$

$\text{Comparing the 3 cases,}$

$\textbf{The smallest altitude of the triangle is 11.2 cm}$

Answer 2

Answer:

11.2 is the answer hope it helps u