The altitudes of a triangle are 10,12,15 cm each.Find the semiperimeter of the triangle.

Question

The altitudes of a triangle are 10,12,15 cm each.Find the semiperimeter of the triangle.

I think this can be easily done by a herons formula equation but i want other easy methods to do this sum.

Answer 1

Let $Δ$ be the area of triangle. Let $h_a=10, h_b=12, h_c=15$ be lengths of perpendiculars dropped on sides $a, b, c$ respectively. Hence we have $$Δ=\frac {1}{2}\cdot 10 \cdot a=\frac {1}{2}\cdot 12 \cdot b=\frac {1}{2}\cdot 15 \cdot c $$ Hence let $$10a=12b=15c=k$$ For some constant k. Hence we have $$a=\frac {k}{10} ; b=\frac {k}{12}; c=\frac {k}{15}; s=\frac {k}{8}$$

By cosine rule we get $\cos A= \frac {1}{8}$ We also know that $$Δ=\frac {1}{2} \cdot bc \sin A = \frac {1}{2} \cdot 12\cdot b$$ $$\Rightarrow \sin A =\frac {180}{k}$$

Using these information we get the equation as $$\frac {\sqrt {63}}{8}= \frac {180}{k}$$

$$\Rightarrow \frac {k}{8}= \frac {60}{\sqrt 7}$$ $$\Rightarrow s=\frac {k}{8}= \frac {60}{\sqrt {7}}$$

Hope it helped.

Answer 2

I do not get how the Cauchy-Schwarz inequality can be used to derive a length in explicit terms, but the lenghts of the altitudes give the area of the triangle through Heron's formula:

$$ \Delta = \frac{1}{4}\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)} $$ $$ \frac{1}{\Delta} = \sqrt{\left(\tfrac{1}{h_a}+\tfrac{1}{h_b}+\tfrac{1}{h_b}\right)\left(-\tfrac{1}{h_a}+\tfrac{1}{h_b}+\tfrac{1}{h_b}\right)\left(\tfrac{1}{h_a}-\tfrac{1}{h_b}+\tfrac{1}{h_b}\right)\left(\tfrac{1}{h_a}+\tfrac{1}{h_b}-\tfrac{1}{h_c}\right)}$$ hence $\Delta=\frac{240}{\sqrt{7}}$ and $a,b,c\in\left\{\frac{48}{\sqrt{7}},\frac{40}{\sqrt{7}},\frac{32}{\sqrt{7}}\right\}$, from which $$ s=\frac{a+b+c}{2}=\color{red}{\frac{60}{\sqrt{7}}}.$$