Question

Which polygons can be mapped onto each other by similarity transformations?

Answer 1

Answer with explanation:

Polygon 1

AB=4.5 - 0.5=4

By distance formula

$BC=\sqrt{(2.5-0.5)^2+(6.5-4.5)^2}\\\\BC=\sqrt{4+4}\\\\BC=2 \sqrt{2}\\\\CD=\sqrt{(2.5-3.5)^2+(6.5-4.5)^2}\\\\CD=\sqrt{1+4}\\\\CD=\sqrt{5}\\\\DE=4,\\\\AE=3.5-0.5=3$

Polygon 2

G H=F J=G F=2 units,

$HI=\sqrt{(6-7)^2+(2.5-1)^2}\\\\HI=\sqrt{1+2.25}\\\\NO=\sqrt{3.25}\\\\JI=\sqrt{(7-6)^2+(1-0.5)^2}\\\\JI=\sqrt{1.25}$

Polygon 3

LM=3 units,

$MN=\sqrt{(8-7)^2+(4.5-5)^2}\\\\MN=\sqrt{1+(-0.5)^2}\\\\MN=\sqrt{1.25}\\\\NO=\sqrt{(8-7)^2+(4.5-3)^2}\\\\NO=\sqrt{1+2.25}\\\\NO=\sqrt{3.25}\\\\OK=3\\\\KL=2$

Polygon 4

PQ=9-7.5=1.5

PT=QR=3.5-1.5=2

$TS=\sqrt{(8-7.5)^2+(1.5-0.5)^2}\\\\TS=\sqrt{0.25+1}=\sqrt{1.25}\\\\SR=\sqrt{(8-9)^2+(1.5-0.5)^2}\\\\RS=\sqrt{1+1}=\sqrt{2}$

Now we will check which polygon have corresponding proportional sides.

→→→Corresponding sides of polygon 1 and Polygon 4

  $\frac{DE}{TP}=\frac{AE}{PQ}=\frac{AB}{QR}=\frac{BC}{RS}=\frac{CD}{ST}\\\\=\frac{4}{2}=\frac{3}{1.5}=\frac{4}{2}=\frac{2\sqrt{2}}{\sqrt{2}}\neq \frac{\sqrt{5}}{1.25}$

Not Similar.

→→Corresponding sides of polygon 1 and Polygon 2

$\frac{DE}{FJ}\neq \frac{AE}{GF}\\\\ \frac{4}{2}\neq\frac{3}{2}$

Not Similar.

→→Corresponding sides of polygon 1 and Polygon 3

$\frac{DE}{KO}\neq \frac{AE}{KL}\\\\\frac{4}{3}\neq\frac{3}{2}$

Not Similar.

→→Corresponding sides of polygon 2 and Polygon 4

$\frac{FJ}{TP}\neq\frac{GF}{PQ}\\\\\frac{2}{2}\neq\frac{2}{1.5}$

Not Similar.

→→⇒None of the four given Option are true.

Answer 2

Answer by Mimiwhatsup: D. Polygons 1 and 2

Why: They have the exact same shape all you need to do is move the points a little and it would be the same thing almost.