# Find the area of a polygon with the given vertices? A(1, 4), B(-2, -2) C(-7, -2), D(-4, 4) Please show work.

Jan 1, 2016

$S = 30$

#### Explanation:

Consider that the polygon ABCD is composed of the triangle ABC and ACD.

To find the area of a triangle whose vertices coordinates are given we can use the Cramer's Rule, described in:
Finding the area of a triangle using the determinant of a matrix

Evaluating the determinant of the Cramer's Rule we get:
${S}_{\triangle} = \left(\frac{1}{2}\right) | {x}_{1} {y}_{2} + {x}_{2} {y}_{3} + {x}_{3} {y}_{1} - {x}_{1} {y}_{3} - {x}_{2} {y}_{1} - {x}_{3} {y}_{2} |$
${S}_{\triangle} = \left(\frac{1}{2}\right) | {x}_{1} \cdot \left({y}_{2} - {y}_{3}\right) + {x}_{2} \cdot \left({y}_{3} - {y}_{1}\right) + {x}_{3} \cdot \left({y}_{1} - {y}_{2}\right) |$

For $\triangle$ABC
A(1, 4)
B(-2, -2)
C(-7, -2)
${S}_{\triangle A B C} = \left(\frac{1}{2}\right) | 1 \cdot \left(- 2 + 2\right) + \left(- 2\right) \left(- 2 - 4\right) + \left(- 7\right) \left(4 + 2\right) |$
${S}_{\triangle A B C} = \left(\frac{1}{2}\right) | 0 + 12 - 42 | = \left(\frac{1}{2}\right) \cdot 30 = 15$

For $\triangle$ACD
A(1, 4)
C(-7, -2)
D(-4, 4)
${S}_{\triangle A C D} = \left(\frac{1}{2}\right) | 1 \cdot \left(- 2 - 4\right) + \left(- 7\right) \left(4 - 4\right) + \left(- 4\right) \left(4 + 2\right) |$
${S}_{\triangle A C D} = \left(\frac{1}{2}\right) | - 6 + 0 - 24 | = \left(\frac{1}{2}\right) \cdot 30 = 15$

${S}_{A B C D} = {S}_{\triangle A B C} + {S}_{\triangle A C D} = 15 + 15 = 30$

Jan 2, 2016

S=30

#### Explanation:

Repeating the points
A(1,4)
B(-2,-2)
C(-7,-2)
D(-4,4)

If we plot those points we'll see that A and D are in the same line ($y = 4$) parallel to the x-axis and that B and C also are in the same line ($y = - 2$) also parallel to the x-axis.

Beyond that, since A and D are in the same line and also B and C are in the same line
$D A = | {x}_{A} - {x}_{D} | = | 1 + 4 | = 5$
$B C = | {x}_{B} - {x}_{B} | = | - 7 + 2 | = 5$
=> $D A = B C$

Two segments of line of the same size in lines parallel to each other, yet the segments are not aligned: it means that the polygon is a parallelogram, whose equation of area is $b a s e \cdot h e i g h t$.

The separation or distance between the two lines ($y = 4$ and $y = - 2$) give us the height. The separation is $4 - \left(- 2\right) = 6$ linear units.

So the area of the polygon ABCD, a parallelogram, is
${S}_{A B C D} = b a s e \cdot h e i g h t = 5 \cdot 6 = 30$