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# Immpossible stuff

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Find  Apr 13, 2022

#1
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This might not be the easiest approach....but.....

By Leg - Leg ,  the two right triangles   are congruent.....so DC = DB   =11

Since angle ABC is bisected, then AB  =(25/11)BC

And, since angles ADB and CDB are supplementary  the  cosine of  ADC  = - cosCDB

By the Law of Cosines

(  BC )^2  = 11^2  + 11^2  - 2 (11)(11) (cos CDB)

(BC)^2  = 242   - 242 ( cos CBD)

[(BC)^2 - 242 ] / [ -242]   = ( cos  CBD )         (1)

And

[ (25/11)BC]^2  = 11^2 + 25^2  - 2 (11)(25) (-cos CBD)

[ (625 /121)BC^2 ] = 746 + 550 (cos CBD)

[ (625/121)BC^2  - 746 ] /   =  (cos CBD)   (2)

Equate  ( 1) and (2)

[ (BC)^2 - 242 ] / [ -242]  =  [ (625 /121) BC^2 - 746 ] / 550         cross-multiply

550 [ (BC)^2  - 242 ] =  (-242)[ (625/ 121) (BC^2) - 746]        simplify

550BC^2  - 133100 =  -1250BC^2 + 180532

1800BC^2  = 313632

BC^2  = 313632 / 1800

BC  =  sqrt [ 313632 / 1800 ]  =  13.2

So   AB  = (25/11) (13.2)   =  30

The semi-perimeter of   ABD  =  (25 + 11 + 30) / 2  =  33

So  ....using Heron's Formula, [ ABD ]  is

sqrt  [ 33 * ( 33-25) * (33 - 11) * ( 33-30) ] =

sqrt [ 33 * 8 * 22 * 3]  =

sqrt [ 17424 ]   =

132  units ^2   Apr 13, 2022