Big Ideas Math Integrated I, 2016
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Big Ideas Math Integrated I, 2016 View details
4. Perimeter and Area in the Coordinate Plane
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Exercise 22 Page 410

The quadrilateral can be cut up into two triangles and a quadrilateral.

26 square units.

Practice makes perfect

Quadrilateral ABCD can be broken up into two triangles, △ ABF and △ CDE, and one quadrilateral, BCEF. By calculating the areas of these three individual figures, we can determine the total area.

Area of the Triangles

To find the area of a triangle, we need its base and height. The two triangles look like they might be right triangles, which means their legs would make up the base and height. Let's make sure by calculating the slope of AD and comparing it to the slopes of BF and CE. If we can prove that AD ⊥ BF and AD ⊥ CE, then we will know that △ ABF and △ CDE are right triangles. The slope can be calculated using the Slope Formula.
Segment Points y_2-y_1/x_2-x_1 m
AD ( - 5,4) & ( 4,- 5) 4-( - 5)/- 5- 4 - 1
BF ( 0,3) & ( - 2,1) 3- 1/0-( - 2) 1
CE ( 4,- 1) & ( 2,- 3) - 1-( - 3)/4- 2 1

Since 1 and - 1 are opposite reciprocals, we know that AD is perpendicular to both BF and CE. Therefore, △ ABC and △ DCE are right triangles which means we can calculate their areas using the lengths of their legs. To find these lengths, we can use the Distance Formula.

d_(AF) = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)
d_(AF) = sqrt(( - 2 - ( - 5))^2 + ( 1 - 4)^2)
Evaluate right-hand side
d_(AF) = sqrt((- 2 + 5)^2 + (1 - 4)^2)
d_(AF) = sqrt(3^2 + (- 3)^2)
d_(AF) = sqrt(9 + 9)
d_(AF) = sqrt(18)
We can calculate the rest of the segment lengths in a similar manner and will summarize the results in the following table.
Segment Points sqrt((x_2-x_1)^2+(y_2-y_1)^2) d
BF ( 0,3) & ( - 2,1) sqrt(( 0-( - 2))^2+( 3- 1)^2) sqrt(8)
DE ( 4,- 5) & ( 2,- 3) sqrt(( 4- 2)^2+( - 5-( - 3))^2) sqrt(8)
CE ( 4,- 1) & ( 2,- 3) sqrt(( 4- 2)^2+( - 1-( - 3))^2) sqrt(8)

The area of a triangle is calculated by multiplying its base and height and then dividing by 2. Area △ ABF: sqrt(18)*sqrt(8)/2&=6 Area △ CDF: sqrt(8)*sqrt(8)/2&=4

Area of the Rectangle

We can see that the quadrilateral resembles a rectangle. If it is, then opposing sides should be parallel and adjacent sides should be perpendicular. We already know that BF and CE are perpendicular to AD. Since FE is a piece of AD, we only need to prove that BC ∥ EF. Let's calculate the slope of BC. m_(BC)=- 1-3/4-0=- 1 Since the slope of BC is the same as that of FE, these sides are in fact parallel. Therefore, we can conclude that BCEF is a rectangle. To find its area, we multiply its length by its width. From calculating the triangle's area, we already know the width. Using the Distance Formula, we can calculate the length.
d_(BC) = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)
d_(BC) = sqrt(( 4 - 0)^2 + ( - 1 - 3)^2)
Evaluate right-hand side
d_(BC) = sqrt(4^2 + (- 4)^2)
d_(BC) = sqrt(16 + 16)
d_(BC) = sqrt(32)
Now we can calculate the area of the rectangle. Area BCEF: sqrt(32)* sqrt(8)=16

Total Area

Now that we have the area of the individual pieces, the total area of the quadrilateral can be found by adding them together. Area ABCD: 6+4+16=26