+0  
 
+2
830
5
avatar+168 

help would be much appreciated

 Nov 11, 2019
 #1
avatar+29 
+3

why do you have to enter a more detailed answer if what you post is too short

 Nov 11, 2019
 #2
avatar+168 
+2

I would guess that it prevents spam

HoldMyBeer  Nov 11, 2019
 #3
avatar+29 
+2

oh k

 Nov 11, 2019
 #4
avatar+128406 
+2

Given  that the  equation through AB  has  the equation 5y + 2x  = 10

Then A  has the coordinates  (0, 2)

And B has the coordinates  (5, 0)

And the slope  of the line through  AB =  [ 2 -0 ] / [ 0 - 5 ]  = -2/5

So....the slope of the line through AD  is   5/2

 

So....the equation of the line through  AD is

 

y = (5/2) x +  2

2y - 5x  = 4

Since D lies on the x axis, then we have that

2(0) - 5x  = 4

x = -4/5

 

So....the distance  AD  =  sqrt [ (0 - - 4/5)^2  + (2 - 0)^2 ] = sqrt [16/25 + 4]  = sqrt [116] / 5

And the distance AB  = sqrt [ 5^2 + 2^2 ]  =  sqrt [ 29]

 

So.....the area of the rectangle  = AD * AB  =

 

sqrt (116 ) sqrt (29)   / 5  =  sqrt (3364)  / 5  =   58/5  units^2  =  11.6 units^2

 

 

cool cool cool

 Nov 11, 2019
 #5
avatar+168 
+2

I understand now, thank youuu coolcoolcool

HoldMyBeer  Nov 11, 2019

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