Exercise 83 Page 115

Practice makes perfect

a We will begin by plotting ABCDE.

To enlarge the polygon by a factor of 2, we have to double the distance between each vertex and the origin in the same direction. We can do this if we multiply each point's coordinates by 2.

Original point	(x,y)	(2x,2y)
A	(- 3,- 2)	(- 6,- 4)
B	(5,- 2)	(10,- 4)
C	(5,3)	(10,6)
D	(1,6)	(2,12)
E	(- 3,3)	(- 6,6)

With this information we can plot the dilated polygon.

b In Part A we multiplied the coordinates of each of the vertices of ABCDE to get the coordinates of the corresponding vertex of A'B'C'D'E'. Therefore, the coordinates of each of the vertices of A'B'C'D'E' are twice as large as the coordinates of the corresponding vertex of ABCDE.

c The perimeter adds up the length of a figure's sides. Any horizontal and vertical parts are the easiest to measure. The length of a horizontal segment is the absolute value of the difference of the x-coordinates. Similarly, the length of a vertical segment is the absolute value of the difference of the y-coordinates.

To measure the distance of the remaining sides, we have to use the Distance Formula.

Points	sqrt((x_2-x_1)^2+(y_2-y_1)^2)	d
D( 1,6), E( - 3,3)	sqrt(( 1-( - 3))^2+( 6- 3)^2)	5
C( 5,3), E( 1,6)	sqrt(( 5- 1)^2+( 3- 6)^2)	5
D'( 2,12), E'( - 6,6)	sqrt(( 2-( - 6))^2+( 12- 6)^2)	10
C'( 10,6), E'( 2,12)	sqrt(( 10- 2)^2+( 6- 12)^2)	10

Let's mark the last side's length in the diagram.

Now we have enough information to calculate the perimeter of the polygons. ABCDE:& 8+5+5+5+5=28 A'B'C'D'E':& 16+10+10+10+10=56 To find the area of the polygons, we will divide them into a rectangle and a triangle. Note that the area of a triangle is half its base times its height. Therefore, we will also mark the triangle heights.