Instructions: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.
Accepted Solution
A:
to compare the
triangles, first we will determine the distances of each side
Distance = ((x2-x1)^2+(y2-y1)^2)^0.5 Solving
∆ABC A(11, 6),
B(5, 6), and C(5, 17)
AB = 6 units BC = 11 units AC = 12.53 units ∆XYZ X(-10, 5), Y(-12, -2), and Z(-4, 15) XY = 7.14 units YZ = 18.79 units XZ = 11.66 units
∆MNO M(-9, -4), N(-3, -4), and O(-3, -15).
MN = 6 units NO = 11 units MO = 12.53 units ∆JKL J(17, -2), K(12, -2), and L(12, 7). JK = 5 units KL = 9 units JL = 10.30 units ∆PQR P(12, 3), Q(12, -2), and R(3, -2) PQ = 5 units QR = 9 units PR = 10.30 units Therefore we have the ∆ABC and the ∆MNO with all three sides equal ---------> are congruent we have the ∆JKL and the ∆PQR with all three sides equal ---------> are congruent
let's check
Two plane figures are congruent if and only if
one can be obtained from the other by a sequence of rigid motions (that is, by
a sequence of reflections, translations, and/or rotations).
1) If ∆MNO ----
by a sequence of reflections and translation --- It can be obtained ------->∆ABC
then ∆MNO ≅ ∆ABC
a) Reflexion (x axis)
The coordinate notation for the Reflexion is
(x,y)---- >(x,-y)
∆MNO
M(-9, -4), N(-3, -4), and O(-3, -15).
M(-9, -4)-----------------> M1(-9,4)
N(-3, -4)------------------ > N1(-3,4)
O(-3,-15)----------------- > O1(-3,15)
b) Reflexion (y axis)
The coordinate notation for the Reflexion is
(x,y)---- >(-x,y)
∆M1N1O1
M1(-9, 4), N1(-3, 4), and O1(-3, 15).
M1(-9, -4)-----------------> M2(9,4)
N1(-3, -4)------------------ > N2(3,4)
O1(-3,-15)----------------- > O2(3,15)
c)
Translation
The coordinate notation for the Translation is
(x,y)---- >(x+2,y+2)
∆M2N2O2
M2(9,4), N2(3,4), and O2(3, 15).
M2(9, 4)-----------------> M3(11,6)=A
N2(3,4)------------------ > N3(5,6)=B
O2(3,15)----------------- > O3(5,17)=C
∆ABC A(11, 6),
B(5, 6), and C(5, 17)
∆MNO reflection------- > ∆M1N1O1 reflection---- >
∆M2N2O2
translation -- --> ∆M3N3O3
The ∆M3N3O3=∆ABC
Therefore ∆MNO ≅ ∆ABC - > check list
2) If ∆JKL -- by a sequence of rotation and translation--- It
can be obtained ----->∆PQR
then ∆JKL ≅
∆PQR
d) Rotation 90 degree anticlockwise
The coordinate notation for the Rotation is
(x,y)---- >(-y, x)
∆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)
e) translation
The coordinate notation for the translation is
(x,y)---- >(x+10,y-14)
∆J1K1L1 J1(2, 17), K1(2, 12), and L1(-7, 12).
J1(2, 17)-----------------> J2(12,3)=P
K1(2, 12)------------------
> K2(12,-2)=Q
L1(-7, 12)-----------------
> L2(3,-2)=R
∆PQR P(12, 3), Q(12, -2), and R(3, -2) ∆JKL rotation------- > ∆J1K1L1 translation -- --> ∆J2K2L2=∆PQR
Therefore ∆JKL ≅ ∆PQR - > check list