So with similar figures, the corresponding sides will all be proportional. But before I can use ratios, I have to use the distance formula, [tex] \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} [/tex] on BC and IJ.
B = (-3,0)
C = (-1,1)
[tex] \sqrt{(-1-(-3))^2+(1-0)^2} [/tex]
Firstly, solve the parentheses: [tex] \sqrt{(2)^2+(1)^2} [/tex]
Next, solve the exponents: [tex] \sqrt{4+1} [/tex]
Next, solve the addition, and your answer will be √5.
(The process with IJ will be similar, so I'll just go through it real quickly.)
I = (5,0)
J = (6,1)
[tex] \sqrt{(6-5)^2+(1-0)^2}\\ \sqrt{(1)^2+(1)^2}\\ \sqrt{1+1}\\ \sqrt{2} [/tex]
Now we can do the ratios:
[tex] \frac{BC}{IJ} =\frac{CD}{JK} =\frac{DE}{KL} =\frac{EF}{LM} =\frac{FG}{MN} =\frac{GA}{NH} =\frac{AB}{HI}\\ \frac{\sqrt{5}}{\sqrt{2}}=\frac{4}{2} =\frac{2}{1} =\frac{2}{1} =\frac{1}{1} =\frac{4}{2} =\frac{2}{1} [/tex]
From the ratios above, we can see that the sides are all not proportionally the same, therefore these figures are not similar.
