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Triangle ABC is isosceles with AB = BC. If AB = 20 and [ABC] = \(160\), then find the perimeter of triangle ABC.

 Apr 15, 2022
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AB = BC =  20

 

Let the altitude  =  BM

And this altitude will bisect the base

Let (1/2) of the base =  AM

So  by the Pythagorean Theorem, the height  =    sqrt [ 20^2 - AM^2]

 

And

 

Area =  (1/2) base * height

 

160  = AM * sqrt  [ 20^2  - AM^2]           square both sides

 

25600 = AM^2  [ 400 - AM^2]

 

25600  = 400AM^2  - AM^4

 

AM^4  - 400AM^2 + 25600  =  0    factor  as

 

(AM^2 - 320) ( AM^2 - 80)  = 0

 

So  either

 

AM^2  -  320  =  0                    or               AM^2  - 80  =  0

AM^2  = 320                                              AM^2   = 80       

AM = sqrt (320)  =  8sqrt( 5 )                        AM =  sqrt (80) =  4sqrt (5)

 

We have two possible perimeters

 

20 + 20 + 2 (8sqrt 5)  =    40 + 16sqrt (5)   units

 

   or

 

20 + 20  + 2 ( 4 sqrt 5)  =   40 + 8 sqrt (5)    units

 

 

cool cool cool

 Apr 15, 2022

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