Question 1125238


Ways to prove that a quadrilateral is a rhombus:

Show that it has four congruent sides. ( by definition of rhombus)

 Given: 
E({{{-2}}},{{{3}}}) 
F({{{-2}}},{{{-2}}})
G({{{3}}},{{{-2}}})
H({{{3}}},{{{3}}})
Prove: EFGH is a rhombus

find distance between  two points:{{{EF}}}, {{{FG}}},{{{GH}}}, and {{{EH}}}

E({{{-2}}},{{{3}}}) 
F({{{-2}}},{{{-2}}})

{{{EF=sqrt((x-x[1])^2+(y-y[1])^2)}}}
{{{EF=sqrt((-2)-(-2))^2+(3-(-2))^2)}}}
{{{EF=sqrt((-2+2)^2+(3+2)^2)}}}
{{{EF=sqrt(0^2+5^2)}}}
{{{highlight(EF=5)}}}

F({{{-2}}},{{{-2}}}) 
G({{{3}}},{{{-2}}})

{{{FG=sqrt((x-x[1])^2+(y-y[1])^2)}}}
{{{FG=sqrt((3-(-2))^2+(-2-(-2))^2)}}}
{{{FG=sqrt((3+2)^2+(-2+2)^2)}}}
{{{FG=sqrt(5^2+0^2)}}}
{{{FG=sqrt(5^2)}}}
{{{highlight(FG=5)}}}

G({{{3}}},{{{-2}}})
H({{{3}}},{{{3}}})

{{{GH=sqrt((x-x[1])^2+(y-y[1])^2)}}}
{{{GH=sqrt((3-3)^2+(3-(-2))^2)}}}
{{{GH=sqrt(0^2+(3+2)^2)}}}
{{{GH=sqrt(5^2)}}}
{{{highlight(GH=5)}}}


E({{{-2}}},{{{3}}}) 
H({{{3}}},{{{3}}})

{{{EH=sqrt((x-x[1])^2+(y-y[1])^2)}}}
{{{EH=sqrt((3-(-2))^2+(3-3)^2)}}}
{{{EH=sqrt((3+2)^2+0^2)}}}
{{{EH=sqrt(5^2)}}}
{{{highlight(EH=5)}}}

as you can see, all sides are same length which proves that a quadrilateral is a {{{rhombus}}}