Chapter 8.4, Problem 48AYU

To determine

To show: $\frac{1}{h_1}+\frac{1}{h_2}+\frac{1}{h_3}=\frac{s}{K}$ where K is the area of the triangle and $s=\frac{1}{2}\left(a+b+c\right)$ .

Explanation of Solution

Given: K is the area of the triangle and $s=\frac{1}{2}\left(a+b+c\right)$ .

  

Area of triangle, $K=\frac{1}{2}\times \text{base}\times \text{Height}$

  $K=\frac{1}{2}a{h}_1\Rightarrow \frac{1}{h_1}=\frac{a}{2K}$

Similarly, $\frac{1}{h_2}=\frac{b}{2K}\text{ and }\frac{1}{h_3}=\frac{c}{2K}$

Taking left side and substitute the value of heights

  $\begin{array}{c}\frac{1}{h_1}+\frac{1}{h_2}+\frac{1}{h_3}=\frac{a}{2K}+\frac{b}{2K}+\frac{c}{2K}\\ =\frac{1}{2K}\left(a+b+c\right)\left[\because s=\frac{1}{2}\left(a+b+c\right)\right]\\ =\frac{s}{K}=RHS\end{array}$

Hence proved