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In triangle $ABC,$ $AB = 15,$ $BC = 10,$ and $AC = 12.$ Find the length of the shortest altitude in this triangle.

 Mar 12, 2024
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The shortest altitude is  drawn to the longest side

Let tha altitude we are looking for be CM

Let  AM  = x  and  BM = 15 - x

 

We can solve this  (equating squares of the altitude CM )

 

12^2  - x^2  =  10^2 - (15  - x)^2

 

144  - x^2  = 100 - (225 - 30x + x^2)

 

30x = 369

 

x = 269 / 30

 

The altitude is    sqrt [ 12^2 - (269/30)^2 ]  ≈  7.97

 

 

cool cool cool

 Mar 12, 2024

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