Coordinate plane with quadrilaterals ABCD and EFGH with A at 0 comma 0, B at 2 comma 0, C at 3 comma negative 2, D at negative 1
comma negative 2, E at 2 comma 4, F at 6 comma 4, G at 7 comma 2, and H at 1 comma 2
Are quadrilaterals ABCD and EFGH similar?
Yes, quadrilaterals ABCD and EFGH are similar because a translation of (x + 2, y + 4) and a dilation by the scale factor of 2 from point D′ map quadrilateral ABCD onto EFGH.
Yes, quadrilaterals ABCD and EFGH are similar because a translation of (x + 3, y + 4) and a dilation by the scale factor of 2 from point A′ map quadrilateral ABCD onto EFGH.
No, quadrilaterals ABCD and EFGH are not similar because their corresponding angles are not congruent.
No, quadrilaterals ABCD and EFGH are not similar because their corresponding segments are not proportional.
Are quadrilaterals ABCD and EFGH similar?
Yes, quadrilaterals ABCD and EFGH are similar because a translation of (x + 2, y + 4) and a dilation by the scale factor of 2 from point D′ map quadrilateral ABCD onto EFGH.
Yes, quadrilaterals ABCD and EFGH are similar because a translation of (x + 3, y + 4) and a dilation by the scale factor of 2 from point A′ map quadrilateral ABCD onto EFGH.
No, quadrilaterals ABCD and EFGH are not similar because their corresponding angles are not congruent.
No, quadrilaterals ABCD and EFGH are not similar because their corresponding segments are not proportional.
2 answers:
Answer:
No, quadrilaterals ABCD and EFGH are not similar because their corresponding segments are not proportional
Step-by-step explanation:
Corresponding sides have the same slope, so corresponding angles are congruent.
Side AB is 1/2 the length of side EF; side BC is equal to the length of FG, so corresponding sides are not proportional.
The two quadrilaterals are not similar because their corresponding segments are not proportional (last choice).
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Given that line PN is a perpendcular bisector of line AB, then angles APN and BPN are equal to 90 degrees each.
Also, given that line PN is an angle bisector to angle CPD and angle CPD measures x, then the measure of angle DPN is
.
Thus, the measure of angle DPB = angle BPN - angle DPN
Therefore,
Also, given that line PN is an angle bisector to angle CPD and angle CPD measures x, then the measure of angle DPN is
Thus, the measure of angle DPB = angle BPN - angle DPN
Therefore,