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What is the area of the triangle whose vertices are  D(−7, 3) ,  E(−7, 9) , and  F(−11, 7) ?

 

Enter your answer in the box.

 

 

___units²

 Jan 7, 2018

Best Answer 

 #1
avatar+9466 
+1

There are a few ways to do this... here's one way..

 

 

Draw a rectangle around the triangle, so...

 

area of the triangle in question  =  area of the rectangle  -  area of the triangles around it

 

area of triangle in question   =   (4 * 6)  -  ( 1/2 * 4 * 4 )  -  ( 1/2 * 4 * 2 )

 

area of triangle in question   =   24  -  8  -  4   =   12   sq. units

 Jan 7, 2018
edited by hectictar  Jan 7, 2018
 #1
avatar+9466 
+1
Best Answer

There are a few ways to do this... here's one way..

 

 

Draw a rectangle around the triangle, so...

 

area of the triangle in question  =  area of the rectangle  -  area of the triangles around it

 

area of triangle in question   =   (4 * 6)  -  ( 1/2 * 4 * 4 )  -  ( 1/2 * 4 * 2 )

 

area of triangle in question   =   24  -  8  -  4   =   12   sq. units

hectictar Jan 7, 2018
edited by hectictar  Jan 7, 2018
 #2
avatar+128460 
+2

Here's one more method of determining the area using something known as Pick's Theorem

 

We can use this whenever the vertices of the triangle are lattice points  [ the vertices have integer coordinates}

 

Area  =  B/2  +  I   -  1

 

Where B  is the number of  lattice points on the triangle's edge  = 12

And  I   is the number of lattice points in the interior of the triangle  = 7

 

So....we have

 

Area  =  12/2  + 7  -  1    =    6 +  7  - 1  =   12 units^2

 

 

cool cool cool

 Jan 8, 2018

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