Question 356942
First, we need to find the length of cd. So we need to find the distance between (-5,0) and (-2,4).



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(-5,0\right)]. So this means that {{{x[1]=-5}}} and {{{y[1]=0}}}.

Also, *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(-2,4\right)].  So this means that {{{x[2]=-2}}} and {{{y[2]=4}}}.



{{{d=sqrt((x[1]-x[2])^2+(y[1]-y[2])^2)}}} Start with the distance formula.



{{{d=sqrt((-5--2)^2+(0-4)^2)}}} Plug in {{{x[1]=-5}}},  {{{x[2]=-2}}}, {{{y[1]=0}}}, and {{{y[2]=4}}}.



{{{d=sqrt((-3)^2+(0-4)^2)}}} Subtract {{{-2}}} from {{{-5}}} to get {{{-3}}}.



{{{d=sqrt((-3)^2+(-4)^2)}}} Subtract {{{4}}} from {{{0}}} to get {{{-4}}}.



{{{d=sqrt(9+(-4)^2)}}} Square {{{-3}}} to get {{{9}}}.



{{{d=sqrt(9+16)}}} Square {{{-4}}} to get {{{16}}}.



{{{d=sqrt(25)}}} Add {{{9}}} to {{{16}}} to get {{{25}}}.



{{{d=5}}} Take the square root of {{{25}}} to get {{{5}}}.



So our answer is {{{d=5}}} 



So the distance between the two points is 5 units. 



So the length of cd is 5 units.



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Now we need to find the length of ef. So we need to find the distance between the points (2,2) and (5,-2)



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(2,2\right)]. So this means that {{{x[1]=2}}} and {{{y[1]=2}}}.

Also, *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(5,-2\right)].  So this means that {{{x[2]=5}}} and {{{y[2]=-2}}}.



{{{d=sqrt((x[1]-x[2])^2+(y[1]-y[2])^2)}}} Start with the distance formula.



{{{d=sqrt((2-5)^2+(2--2)^2)}}} Plug in {{{x[1]=2}}},  {{{x[2]=5}}}, {{{y[1]=2}}}, and {{{y[2]=-2}}}.



{{{d=sqrt((-3)^2+(2--2)^2)}}} Subtract {{{5}}} from {{{2}}} to get {{{-3}}}.



{{{d=sqrt((-3)^2+(4)^2)}}} Subtract {{{-2}}} from {{{2}}} to get {{{4}}}.



{{{d=sqrt(9+(4)^2)}}} Square {{{-3}}} to get {{{9}}}.



{{{d=sqrt(9+16)}}} Square {{{4}}} to get {{{16}}}.



{{{d=sqrt(25)}}} Add {{{9}}} to {{{16}}} to get {{{25}}}.



{{{d=5}}} Take the square root of {{{25}}} to get {{{5}}}.



So our answer is {{{d=5}}} 



So the distance between the two points is also 5 units. 



So the length of ef is also 5 units.



This means that cd and ef are congruent (since they are the same length).



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