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In the figure, what is the area of triangle ABD? Express your answer as a common fraction.

 

 Nov 20, 2020
 #1
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Let   A   =(0,5)    F  ( 3, 0)

Slope of AB =  -5/3

Equation of line through  this  segment =  y = -(5/3)x  + 5

 

Let E = ( 0,3)      B = ( 7,0)

Slope of BE =  -3/7

Equation of line through BE

y = (-3/7)x + 3

 

Find D  = the intersection of these two lines

 

(-5/3)x  + 5 = (-3/7)x + 3

 

5 -3 = ( 5/3 - 3/7)x

2  = (26/21)x

x = 21/13

y= (-3/7)(21/13) + 3

y = 30/13

 

Slope of AB =  -5/7

Equation of line through AB

y= (-5/7)x + 5   Put into standard form

 

7y= -5x + 35

5x + 7y  - 35  =  0

 

Distance  from D to AB  is  given by

 

l 5 (21/13)  + 7 (30/13)  -35  l               140/13          140 / [ 13sqrt (74) ]  =  altitude of ADB

_________________________   =    _________  =  

      sqrt ( 5^2 + 7^2)                             sqrt ( 74)

 

AB length =  sqrt  [ 5^2 + 7^2] = sqrt (74)

 

Area of  ADB  = (1/2) AB * altitude  =  

 

(1/2) sqrt (74)  *  140 / ( 13 sqrt (74) )  =    140/26  =  70/13

 

cool cool cool

 Nov 20, 2020

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