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The three heights of a triangle are 1/3, 1/4, and 1/6. What is its area?

 Jul 19, 2022
 #1
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Let  the altitude of 1/3 be drawn to side A  

Let the altitude of 1/4 be drawn to side B   

Let the altitude of 1/6 be drawn to side C   

 

Equating areas

(1/2) (A) (1/3)  = (1/2)(B)(1/4)  →   B = (4/3)A

(1/2)(A) (1/3)  = (1/2)(C)(1/6)  →   C =  (2)A

 

The semi-perimeter  =  (A + (4/3)A + (2)A )  / 2  =     (13/6) A  

 

Using Heron's Formula

 

Area =  sqrt  [   (13/6)A *  [ (13/6)A - A ] * [ (13/6)A - (4/3)A ] * [ (13/6)A - 2A ]  ]  =

 

sqrt  [  A^4  (13/6)  (7/6) (5/6) ( 1/6)  ]   =

 

A^2 sqrt  [ 455 ]  /  36 

 

Equating areas

 

(1/2)A (1/3)  =  A^2  sqrt [ 455]  / 36

 

(1/6)A  = A^2 sqrt [ 455] / 36

 

(1/6)  = A sqrt [ 455 ]  / 36

 

6  /sqrt [ 455]  =  A

 

The area is   (1/2) (6 / sqrt [455])  ( 1/3)  =     1/ sqrt 455 units^2  ≈   .0468 units^2

 

 

cool cool cool

 Jul 20, 2022

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