+0  
 
+1
44
5
avatar+1146 

What is the area, in square units, of triangle ABC?

 

 

Thank you!

 #1
avatar+86889 
+1

We need to  find the distance between

 

(0,6) and (-4,3)  = √[ ( -4)^2 + (6-3)^2 ] = √ [4^2 + 3^2 ]  =√[25]   = 5

 

Now, ACG....we could find the lengths of the other two sides and use something known as "Heron's Formula" to calculate the area, but it's messy.....here's an easier way....let the first distance we calculated be  the  base  of the triangle

 

We need to find the equation of the line between the first two points....

The slope is   [ 6 -3] / [ 0 - -4]  =  3/4

The equation is

y  = (3/4)(x - 0) + 6

y = (3/4)x  + 6      ....... now.... we need this form :  Ax + BY + C  = 0....so ....multiply through by  4.....

4y = 3x + 24

3x  - 4y + 24  = 0

 

Now the distance  between (2, -2)  and  this line will be the altitude of the triangle and  it is given by

 

l Ax + By + C l  /  √[ A^2 + B^2 ]       where  (x,y)  = (2, -2)   ....so we have

 

 l 3(2)  - 4(-2) + 24  l   / √ [ (3)^2 + (-4)^2 ]  =

 

l 6  + 8  + 24  l  / √25   =

 

38 / 5     units

 

So....the area  is

 

(1/2) base * altitude  =

 

(1/2) (5) * 38/ 5  =

 

19 units^2 

 

 


cool cool cool

CPhill  Jun 6, 2018
 #2
avatar+12561 
+1

As Chris mentioned:

 

I calculated the distances between the points (side lengths) as  

5    7.81   and    8.25

thus p = 10.53

 

sqrt(10.53(10.53-5)(10.53-7.81)(10.53-8.25)) = 19.0 units^2    

                                                               (just as Chris found!)

ElectricPavlov  Jun 6, 2018
 #3
avatar+86889 
0

Thanks for that answer, EP........

 

 

cool cool cool

CPhill  Jun 6, 2018
 #4
avatar+1146 
0

Thank you both!

 #5
avatar+9439 
+1

What is the area, in square units, of triangle ABC?

laugh

Omi67  Jun 6, 2018
edited by Omi67  Jun 6, 2018

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