Select the correct answer from each drop-down menu. ∆ABC has vertices at A(11, 6), B(5, 6), and C(5, 17). ∆XYZ has vertices at X(-10, 5), Y(-12, -2), and Z(-4, 15). ∆MNO has vertices at M(-9, -4), N(-3, -4), and O(-3, -15). ∆JKL has vertices at J(17, -2), K(12, -2), and L(12, 7). ∆PQR has vertices at P(12, 3), Q(12, -2), and R(3, -2). can be shown to be congruent by a sequence of reflections and translations. can be shown to be congruent by a single rotation.

Respuesta :

The option  that is shown to be congruent by a sequence of reflections and translations is Option B: Triangle ABC and Triangle MNO

  • Option  C:  Triangle JKL and Triangle PQR can be shown to be congruent by a single rotation.

How are they congruent?

Note that:

If ∆MNO  can be gotten  from ∆ABC  then ∆MNO ≅ ∆ABC  

To solve for Reflection (x axis), it will be:

The coordinate note for the reflection is (x, y) will be: (x,-y)

Since ∆MNO  M(-9, -4), N(-3, -4), and O(-3, -15).

Then the reflections will be:

M(-9, -4)- M1(-9,4)

N(-3, -4)- N1(-3,4)

O(-3,-15)-O1(-3,15)

To solve for Reflection (y axis), it will be:    

The coordinate note for the Reflection  is (x,y) will be (-x,y)

Since ∆M1N1O1  = M1(-9, 4), N1(-3, 4), and O1(-3, 15).

Then M1(-9, -4) - M2(9,4)

N1(-3, -4) - N2(3,4)

O1(-3,-15)- O2(3,15)

To solve for Translation

The coordinate note for the Translation is (x,y) will be (x+2,y+2)

Since ∆M2N2O2  = M2(9,4), N2(3,4), and O2(3, 15).

Then M2(9, 4) - M3(11,6)=A

N2(3,4) -  N3(5,6)=B

O2(3,15) - O3(5,17)=C

Hence  ∆MNO ≅ ∆ABC  

To know the congruent by a single rotation.

 If ∆JKL  can be gotten from ∆PQR

Then ∆JKL ≅ ∆PQR  

For Rotation of 90 degree in an anticlockwise position, then:

The coordinate note for the Rotation is (x,y) will be (-y, x)

So, ∆JKL  = J(17, -2), K(12, -2), and L(12, 7).

J(17, -2)-  J1(2,17)

K(12, -2) - K1(2,12)

L(12,7) - L1(-7,12)

To solve for translation

The coordinate note for the translation is (x,y) will be (x+10,y-14)

Then ∆J1K1L1 =  J1(2, 17), K1(2, 12), and L1(-7, 12).

Hence J1(2, 17) - J2(12,3) = P

K1(2, 12)- K2(12,-2) = Q

L1(-7, 12)- L2(3,-2) = R

So, ∆JKL ≅ ∆PQR.

So therefore, The option  that is shown to be congruent by a sequence of reflections and translations is Option B: Triangle ABC and Triangle MNO

  • Option  C:  Triangle JKL and Triangle PQR can be shown to be congruent by a single rotation.

See options below

___ can be shown to be congruent by a sequence of reflections and translations.

1)Pick one to go with the sentence.

a Triangle ABC and Triangle XYZ

b Triangle ABC and Triangle MNO

c Triangle JKL and Triangle ABC

d Triangle PQR and Triangle XYZ

___ can be shown to be congruent by a single rotation.

Pick one to go with the sentence.

a Triangle ABC and Triangle XYZ

b Triangle PQR and Triangle XYZ

c Triangle JKL and Triangle PQR

d Triangle JKL and Triangle XYZ

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