What is the area of the triangle whose vertices are D(−7, 3) , E(−7, 9) , and F(−11, 7) ?
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___units²
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
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
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