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In right triangle $ABC$, $M$ and $N$ are midpoints of legs $\overline{AB}$ and $\overline{BC}$, respectively. Leg $\overline{AB}$ is 6 units long, and leg $\overline{BC}$ is 8 units long. How many square units are in the area of $\triangle APC$?

Guest Aug 16, 2018
 #1
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What  is  " P ??? "

 

 

cool cool cool

CPhill  Aug 16, 2018
 #2
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picture 

newuseeer  Aug 16, 2018
 #3
avatar+91019 
+1

Let  B  =(0,0)   A  = (0,6)   and C  = (8,0)   M  =(0,3)   N  = (4,0)

 

Let  the slope of MC  = [ 3 -0 ]  / [ 0 - 8]  =  -3/8

And the equation of MC is   y  = -(3/8) x + 3     (1)

 

Let the slpoe  of AN  = [ 6 - 0 ] / [ 0 - 4 ] = -6/4  = -3/2

And the  equation of An  is   y  = (-3/2)x + 6    (2)

 

Set (1)  = (2)  to find the x coordinate of P

 

(-3/8)x + 3  = (-3/2)x + 6

(-3/8 + 3/2)x  = 3

(9/8)x  = 3

24/9 = x  = 8/3

So...the y  coordinate of P  is  (-3/2)(8/3)  + 6 =  -24/6 + 6=  -4 + 6  = 2  = altitude of triangle PNC

 

The area of triangle APC  = area of triangle ANC  - area of triangle  PNC

Area of ANC  =(1/2) (AB) (NC)  = (1/2)(6)(4)   = 12  units^2

Area  of PNC  = (1/2) (2)(NC) = (1/2) (2) (4)  =4  units^2

 

So...area of triangle APC =  12 -  4  = 8 units^2

 

 

cool cool cool

CPhill  Aug 16, 2018

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