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Inside a square with side length $10$, two congruent equilateral triangles are drawn such that they share one side and each has one vertex on a vertex of the square.  What is the total area of the two equilateral triangles?

 

 Aug 7, 2023
 #1
avatar+128826 
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The height of  one  of these triangles = (1/2) diagonal of the square =  (1/2) 10 *sqrt (2) =  5sqrt 2

 

Half the base length =  5 sqrt (2) / sqrt (3) ) ...so the whole base length = 10sqrt (2) / sqrt (3)

 

Total area of both triangles =  2 (1/2) ( 5 sqrt (2)) (10 sqrt (2) / sqrt (3) = 

 

50 * 2 / sqrt (3)  = 

 

100 /sqrt (3) ≈ 57.735

 

cool cool cool

 Aug 7, 2023

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