Question

find the area of the quadrilatral whose vertices are (1,1),(5,2),(4,4)and (2,3)

Answer 1

First we draw the quadrilateral:

(1,1), (5,2), (4,4) and (2,3)

So we must divide the quadrilateral into two triangles, for

there is a determinant formula for finding the area of a

triangle when the coordinates of the three vertices are

given:

The area of a triangle with the vertices

(x1,y1),(x2,y2), (x3,y3)

is equal to one half the absolute  

value of this 3×3 determinant

|x1 y1  1|          

|x2 y2  1|                

|x3 y3  1|                    

Now we draw in either diagonal we choose, either of

these will do, either this

or this  

If we use the left graph, the determinant for the

triangle with vertices (1,1), (2,3), and (4,4) is  

| 1 1 1 |

| 2 3 1 | = -3

| 4 4 1 |

I am assuming you already know how to find the

value of a determinant. If you don't. then post

again asking how.

So the area of that triangle is one half of the  

absolute value of -3.  The absolute value of -3  

is 3, and half of 3 is 1.5.

Again, using the left graph, the determinant for the

triangle with vertices (1,1), (4,4), and (5,2) is  

| 1 1 1 |

| 4 4 1 | = -9

| 5 2 1 |

So the area of that triangle is one half of the  

absolute value of -9.  The absolute value of -9  

is 9, and half of 9 is 4.5.

So the sum of the areas of those two triangles

is  

1.5 + 4.5 or 6 square units of area for the

quadrilateral.

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

Now let's check it by using the graph on the right

above.

If we use the graph on the right above, the  

determinant for the triangle with vertices

(2,3), (4,4), and (5,2) is  

| 2 3 1 |

| 4 4 1 | = -5

| 5 2 1 |

So the area of that triangle is one half of the  

absolute value of -5  The absolute value of -5

is 5, and half of 5 is 2.5.

Again, using the graph on the right, the  

determinant for the triangle with vertices  

(2,3), (5,2), and (1,1) is  

| 2 3 1 |

| 5 2 1 | = -7

| 1 1 1 |

So the area of that triangle is one half of the  

absolute value of -7. The absolute value of -7

is 7 and half of 7 is 3.5.

So the sum of the areas of those two triangles

is  

2.5 + 3.5 or, again, we get 6 square units of  

area for the quadrilateral.

Answer 2

2+7+8+5=22 and each vertical =3 as 21 division by three but not 4