Find-the-area-S-of-a-triangle-ABC-as-a-function-of-the-heights-h-a-h-b-and-h-c-

Question Number 76272 by Maclaurin Stickker last updated on 25/Dec/19

F i n d t h e a r e a S o f a t r i a n g l e A B C

a s a f u n c t i o n o f t h e h e i g h t s

h_{a}, h_{b} a n d h_{c} .

Commented by mr W last updated on 25/Dec/19

i g o t t h i s r e s u l t :

S = \frac{1}{\sqrt{(\frac{1}{h_{a}} + \frac{1}{h_{b}} + \frac{1}{h_{c}}) (- \frac{1}{h_{a}} + \frac{1}{h_{b}} + \frac{1}{h_{c}}) (\frac{1}{h_{a}} - \frac{1}{h_{b}} + \frac{1}{h_{c}}) (\frac{1}{h_{a}} + \frac{1}{h_{b}} - \frac{1}{h_{c}})}}

Commented by Maclaurin Stickker last updated on 25/Dec/19

T h a n k y o u . I g o t t h e s a m e r e s u l t

Answered by Maclaurin Stickker last updated on 25/Dec/19

Commented by Maclaurin Stickker last updated on 25/Dec/19

l e t A B = c, B C = a, A C = b, B E = h_{b},

A D = h_{a} a n d C F = h_{c} .

B y {H e r o n}^{'} f o r m u l a, w e h a v e :

a = \frac{2 S}{h_{a}}, b = \frac{2 S}{h_{b}} a n d c = \frac{2 S}{h_{c}}

a + b + c = 2 p

p = (\frac{S}{h_{a}} + \frac{S}{h_{b}} + \frac{S}{h_{c}})

F o r a r e a :

S = \sqrt{(\frac{S}{h_{a}} + \frac{S}{h_{b}} + \frac{S}{h_{c}}) (\frac{S}{h_{b}} + \frac{S}{h_{c}} - \frac{S}{h_{a}}) (\frac{S}{h_{a}} - \frac{S}{h_{b}} + \frac{S}{h_{c}}) (\frac{S}{h_{a}} + \frac{S}{h_{b}} - \frac{S}{h_{c}})}

\frac{1}{S^{2}} = (\frac{1}{h_{a}} + \frac{1}{h_{b}} + \frac{1}{h_{c}}) (\frac{1}{h_{b}} + \frac{1}{h_{c}} - \frac{1}{h_{a}}) (\frac{1}{h_{a}} - \frac{1}{h_{b}} + \frac{1}{h_{c}}) (\frac{1}{h_{a}} + \frac{1}{h_{b}} - \frac{1}{h_{c}})

S = \frac{1}{\sqrt{(\frac{1}{h_{a}} + \frac{1}{h_{b}} + \frac{1}{h_{c}}) (\frac{1}{h_{b}} + \frac{1}{h_{c}} - \frac{1}{h_{a}}) (\frac{1}{h_{a}} - \frac{1}{h_{b}} + \frac{1}{h_{c}}) (\frac{1}{h_{a}} + \frac{1}{h_{b}} - \frac{1}{h_{c}})}}

S = \frac{1}{\sqrt{2 (\frac{1}{h_{a}^{2} h_{b}^{2}} + \frac{1}{h_{a}^{2} h_{c}^{2}} + \frac{1}{h_{b}^{2} h_{c}^{2}}) - (\frac{1}{h_{a}^{4}} + \frac{1}{h_{b}^{4}} + \frac{1}{h_{c}^{4}})}}

Commented by benjo last updated on 26/Dec/19

sir, what the application for

drawing this triangle in a cellphone ?

Commented by Maclaurin Stickker last updated on 26/Dec/19

M e d i B a n g