Here is a polygon. The side lengths are: AB= 18 units, BC= 12 units, CD= 6 units, and DA= 24 units. 1. If polygon ABCD is dilated using center A and scale factor LaTeX: \frac{1}{2}1 2, what will be the side lengths of dilated polygon EFGH?
EF= , FG= , GH= , HE=

2. If polygon ABCD is dilated using center D and scale factor LaTeX: \frac{1}{3}1 3, what will be the side lengths of dilated polygon IJKL?

IJ= , JK= , KL= , LI=

3. How do you know that polygons EFGH and IJKL are similar?

Given : Polygon side lengths are: AB= 18 units, BC= 12 units, CD= 6 units, and DA= 24 units.  

polygon ABCD is dilated using center A and scale factor  1/2

polygon ABCD is dilated using center D and scale factor  1/3

To Find :

side lengths of dilated polygon EFGH

EF= , FG= , GH= , HE=

side lengths of dilated polygon IJKL?

IJ= , JK= , KL= , LI=

Solution:

AB= 18 units, BC= 12 units, CD= 6 units, and DA= 24 units.  

EF=  AB/2  = 18/2 = 9

FG = BC/2 = 12/2 = 6

GH = CD/2 = 6/2 = 3

HI = DA/2 = 24/2 = 12

IJ= AB/3 = 18/3 = 6

JK = BC/3 = 12/3 = 4

KL= CD/3 = 6/3 = 2

LI =  DA/3 = 24/3 = 8

polygon  EFGH is Dilated from polygon ABCD

Hence polygon  EFGH ≈ polygon ABCD

polygon  IJKL is Dilated from polygon ABCD

Hence polygon   IJKL ≈ polygon ABCD

As polygon ABCD   ≈  polygon   IJKL  & polygon  EFGH

=>   polygon  EFGH ≈ polygon   IJKL

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Step-by-step explanation:

infinitely many solutions

step-by-step explanation:

we have been given the equation;

2(2x + 3) = -4 + 2(2x + 5)

we need to determine the value of x.

2(2x + 3), we expand this expression to obtain 4x + 6

similarly 2(2x + 5) will become 4x + 10

now we have;

4x + 6 = -4 + 4x + 10

4x + 6 = 4x + 10 - 4

4x + 6 = 4x + 6

the expression on the left hand side is identical to the one on the right hand side. therefore, we are looking at just a single straight line whose equation is;

y = 4x + 6

consequently, we shall have infinitely many solutions