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

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

___units²

Jan 7, 2018

#1
+7348
+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
+7348
+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

hectictar Jan 7, 2018
edited by hectictar  Jan 7, 2018
#2
+96367
+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

Jan 8, 2018