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The diagram above shows two congruent right triangles 12-16-20 overlapping to form a dark blue region.

Find the area of this dark blue region.

 

 Dec 20, 2019
 #1
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Let the bottom  let  vertex of the  blue triangle  =A

Let the bottom right vertex of the  blue triangle  = B

Let the remaining vertex of the blue triangle  = C

Let the bottom right vertex of the figure  =  D

Let  DE  be  the edge  that is perpendicular to AD

 

And traingles 

 

ABC  and ADE   are similar triangles

 

So

 

BC / AB   =  DE / AD

 

BC / 12  =  12 / 16

 

16* BC  = 12 * 12

 

16* BC  = 144

 

BC  = 144/16  =  9

 

So  the dark  area is the area of  triangle ABC  =

 

(1/2) (AB) ( BC)  = 

 

(1/2)(12) (9)  =  

 

(1/2) 108  =

 

54 units^2

 

 

 

cool cool cool

 Dec 20, 2019
edited by CPhill  Dec 20, 2019

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