Pearson Geometry Common Core, 2011
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Pearson Geometry Common Core, 2011 View details
7. Polygons in the Coordinate Plane
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Exercise 38 Page 405

Practice makes perfect
a Let's examine the given diagram.

There can be more than one way to classify ABCD. One way is to classify by the slope of each side. To do so, we will use the Slope Formula.

Side Points y_2-y_1/x_2-x_1 Slope
AB A( -4, 4) and B( -1, -2) -2- 4/-1-( -4) -2
BC B( -1, -2) and C( -3, -3) -3-( -2)/-3-( -1) 1/2
CD C( -3, -3) and D( -6, 3) 3-( -3)/-6-( -3) -2
AD A( -4, 4) and D( -6, 3) 3- 4/-6-( -4) 1/2

Notice that the opposite sides are parallel and that the consecutive sides are perpendicular. This means that ABCD is a parallelogram with four right angles, so it is a rectangle.

b To classify EFGH, we will follow the same steps as we did in Part A.
Side Points y_2-y_1/x_2-x_1 Slope
EF E( 1, -3) and F( 4, 3) 3-( -3)/4- 1 2
FG F( 4, 3) and G( 6, 2) 2-( 3)/6-( 4) -1/2
GH G( 6, 2) and H( 3, -4) -4- 2/3- 6 2
EH E( 1, -3) and H( 3, -4) -4-( -3)/3- 1 -1/2

Because of the same reasons that we stated in Part A, EFGH is also a rectangle.

c To determine whether ABCD and EFGH are congruent, we will find their side lengths by the Distance Formula.
ABCD EFGH
Side Points sqrt((x_2-x_1)^2+(y_2-y_1)^2) Length Side Points sqrt((x_2-x_1)^2+(y_2-y_1)^2) Length
AB A( -4, 4) and B( -1, -2) sqrt(( -1-( -4))^2+( -2- 4)^2) s sqrt(45) EF E( 1, -3) and F( 4, 3) sqrt(( 4- 1)^2+( 3- 1)^2) sqrt(45)
BC B( -1, -2) and C( -3, -3) sqrt(( -3-( -1))^2+( -3-( -2))^2) s sqrt(5) FG F( 4, 3) and G( 6, 2) sqrt(( 6- 4)^2+( 2- 3)^2) sqrt(5)
CD C( -3, -3) and D( -6, 3) sqrt(( -6-( -3))^2+( 3-( -3))^2) sqrt(45) GH G( 6, 2) and H( 3, -4) sqrt(( 3- 6)^2+( -4- 2)^2) sqrt(45)
AD A( -4, 4) and D( -6, 3) sqrt(( -6-( -4))^2+( 3- 4)^2) sqrt(5) EH E( 1, -3) and H( 3, -4) sqrt(( 3- 1)^2+( -4-( -3))^2) sqrt(5)

Looking at the table above, we see that the lengths and widths of the rectangles are congruent. Therefore, ABCD and EFGH are congruent.