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In triangle $ABC$, $\angle ABC = 90^\circ$ and $AD$ is an angle bisector. If $AB = 90,$ $BC = x$, and $AC = 2x - 6,$ then find the area of $\triangle ADC$. Round your answer to the nearest integer.

 Nov 6, 2019
 #1
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By the Pythagorean Theorem

 

AC^2  = AB^2  + BC^2

 

(2x - 6)^2 =  90^2  + x^2

 

4x^2 - 24x + 36  =  8100 + x^2    rearrange as

 

3x^2  - 24x - 8064  = 0       divide through by 3

 

x^2  -  8x  - 2688  =  0       complete the square on x

 

x^2   - 8x  + 16  =  2688 + 16     

 

(x - 4)^2  =  2704        take the positive root

 

x - 4  =   52

 

x  =  56

 

So    BC  =  56     and  AC   =  2(56) - 6  =  106

 

So...since AD is an angle bisector  we have that

 

AB / BD  =  AC / DC

90 / BD  =  106 / DC

So

BD/ DC  =  90  / 106   

 

So...there are  90 + 106  =  196 equal parts  of BC....and   DC   is  106  of them

So    DC   =  56 ( 106 /196)  = 212/ 7

 

So  the area  of triangle  ADC  =  (1/2) (DC) ( AB)   =  (1/2) (212/7) * 90  ≈  1363 units^2  

 

Here's a pic :

 

 

cool cool cool

 Nov 7, 2019

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