SOLUTION: find the area of the triangle with vertices A(-1,3)B(2,3)C(2,6)

Question 461705: find the area of the triangle with vertices A(-1,3)B(2,3)C(2,6)

Answer by MathLover1(20850) (Show Source): You can put this solution on YOUR website!
the area of the triangle with vertices
A(-1,3)
B(2,3)
C(2,6)
first find a distance between vertices which is equal to the length of sides:
a distance between vertices
A(-1,3)
B(2,3)
is:

Solved by pluggable solver: Distance Formula to determine length on coordinate plane

The distance (d) between two points is given by the following formula:

Thus in our case, the required distance is

For more on this concept, refer to Distance formula.

so,
a distance between vertices
A(-1,3)
C(2,6)
is:

Solved by pluggable solver: Distance Formula to determine length on coordinate plane

The distance (d) between two points is given by the following formula:

Thus in our case, the required distance is

For more on this concept, refer to Distance formula.

so,
a distance between vertices
B(2,3)
C(2,6)
is:

Solved by pluggable solver: Distance Formula to determine length on coordinate plane

The distance (d) between two points is given by the following formula:

Thus in our case, the required distance is

For more on this concept, refer to Distance formula.

so,

Solved by pluggable solver: Hero's (or Heron's) Formula (Used to Find the Area of a Triangle Given its Three Sides)

In order to find the area of a triangle 'A' with side lengths of 'a', 'b', and 'c', we can use Hero's Formula:

where S is the semiperimeter and it is defined by

Note: "semi" means half. So the semiperimeter is half the perimeter.

So let's first calculate the semiperimeter S:

Start with the semiperimeter formula.

Plug in , , and .

Add.

Divide.

----------------------------------------

Now move onto Hero's Formula.

Plug in , , , and .

Subtract.

Multiply.

Take the square root of to get .

So the area of the triangle with side lengths of , , and is roughly square units.

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