ΔEFI is dilated by a scale factor of one third with the center of dilation at point F. Then, it is reflected over line a to create ΔHFG. Based on these transformations, which statement is true?

Line segments EG and HI intersect at point F, forming triangles EFI and HFG. Line a intersects with both triangles at point F.

segment FE = segment FH over 3, segment FI = segment FG over 3, and segment EI = segment HG over 3; ΔEFI ~ ΔHFG

segment FE = segment FG over 3, segment FI = segment FH over 3, and segment EI = segment HG over 3; ΔEFI ~ ΔGFH

segment FE over 3 = segment FG, segment FI over 3 = segment FH, and segment EI over 3 = segment HG; ΔEFI ~ ΔGFH

segment FE over 3 = segment FH, segment FI over 3 = segment FG, and segment EI over 3 = segment HG; ΔEFI ~ ΔHFG

Also by reflection over a line, the relative positions of the y-coordinates of the points remain the same while the x coordinates changes sign such that an upright mirror image of the dilated triangle is formed

Therefore, since we have;

[tex]\dfrac{sin\alpha}{sin\beta } \ and \ \dfrac{sin\gamma}{sin\alpha }[/tex] the same for both triangles, that is given the following relation;

FG/FH = (1/3×FI)/(1/3×FE) = FI/FE

Then, ΔEFI is similar to ΔHFG (triangle similarity requirement AAA)

Hence, the correct option is that the

Segment FE over 3 = Segment FH

Segment FI over 3 = Segment FG, and

Segment EI over 3 = Segment HG

ΔEFI ~ ΔHFG.

akposevictor akposevictor · Answer 2 · 2021-12-10T13:25:21+01:00

If ΔEFI is dilated by using a scale factor of 1/3 with the center of dilation at point F, which is reflected over a line to create ΔHFG, the statement that is true about this transformation is:

D. [tex]\frac{FE}{3} = FH[/tex], [tex]\frac{FI}{3} = FG[/tex], [tex]\frac{EI}{3} = HG[/tex]; ΔEFI ~ ΔHFG

Recall:

To get the new lengths of a figure after a dilation, multiply the scale factor by the original lengths of the figure.

Scale factor of dilation of ΔEFI, which is reflected over point F to get ΔHFG is given as: 1/3.

FE corresponds with FH
EI corresponds with HG
FI corresponds with FG

Therefore,

FH = 1/3(FE)

[tex]\frac{FE}{3} = FH[/tex]

HG = 1/3(EI)

[tex]\frac{EI}{3} = HG[/tex]

FG = 1/3(FI)

[tex]\frac{FI}{3} = FG[/tex]

Also, since the same shape is maintained, ΔEFI ~ ΔHFG

Therefore, if ΔEFI is dilated by using a scale factor of 1/3 with the center of dilation at point F, which is reflected over a line to create ΔHFG, the statement that is true about this transformation is:

D. [tex]\frac{FE}{3} = FH[/tex], [tex]\frac{FI}{3} = FG[/tex], [tex]\frac{EI}{3} = HG[/tex]; ΔEFI ~ ΔHFG

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