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Triangle ABC has side lengths AB = 6, BC = 10, CA = 8.

If D is the midpoint of  BC and angle ADE = 90 degrees, what is the area of triangle  ADE?

 

 Jan 21, 2021
 #1
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Since  DB  = CD   then  DB  = 5

 

If we let   A  = (0,0)  , then using similar triangles   we  have  that   the coordinates of  D   =  (3 , 4)

 

 Then AD  =   sqrt(3^2 + 4^2)  = sqrt (25)  = 5

 

And the area of  triangle ADE   = (1/2)  AD  ( ED)  = (1/2) (5) ED = (5/2)ED     ( 1)

Also  the altitude   of  ADE  = 3

So its area also  =   (1/2)  3(EA) =  (3/2) EA   (2)

 

Equating (1)  and (2)   we  have  that 

 

5ED = 3EA

 

ED  =  (3/5) EA

 

Using the  Pythagorean Theorem

 

EA^2   - ED^2   = AD^2

 

EA^2  - [( 3/5) EA ] ^2   = 5^2

 

EA^2   - (9/25)EA^2  =   25

 

(16/25) EA^2  =25

 

EA^2 =  (25* 25)  /16   =  625/16

 

EA =  sqrt (625/16)  =   25 / 4  

 

So....the area of  ADE = (1/2)(EA)(altitude)   =   (1/2) (25/4)(3)   = 75/8  units ^2 = 9.375  units^2

 

 

cool cool cool

 Jan 21, 2021

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