Math, asked by ZzyetozWolFF, 4 months ago

In the figure, ∆ABC and ∆ABD are equilateral triangles, then find the coordinates of points C and D.

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Answered by ApprenticeIAS
80

 \rm \red{Given : }

 \sf{\Delta ABC  \: and  \: \Delta ABD \:  are \:  equilateral  \: triangles}

 \underline{ \underline{ \rm \red{Properties \:  of \:  Equilateral \:  triangle}}}  \red:

 \sf \bullet  \: All  \: sides \: are \: equal \: i.e. \: AB = BC = CA

 \sf \bullet  \: Each  \: and  \: every  \: angle \:  measures  \: 60°

 \sf \: the \: coordinates \: of \: equilateral \: triangle \: are :

 \sf \: A = ( - a,0) , \: B = (a,0)

 \rm \red{Let : }

 \sf \: C = (0,Y_1),  \: D = (0,Y_2) \\

 \underline{ \underline{ \rm \red{case \: i}}} \red{:} \sf{From  \: \Delta ABC}

 \rm{AB = BC = CA}

 \rm{The  \: distance \:  between  \: points \:  A(-a,0), \:  B(a,0)}

 \rm{AB= d =  \sqrt{(x_2 - x_1)^{2}  + (y_2 - y_1)^{2} } }

 \rm{AB =  \sqrt{ {(a + a)}^{2}  + (0  - 0) ^{2} } }

 \rm{AB \:  =  \sqrt{4 {a}^{2} } }

 \rm{AB = 2a}

 \rm{The  \: distance \:  between  \: points \:  B(a,0) \: ,C(0,y_1)}

 \rm{BC= d =  \sqrt{(x_2 - x_1)^{2}  + (y_2 - y_1)^{2} } }

 \rm{BC = \sqrt{( {0 - a)}^{2} +  {(y_1 - 0)}^{2}  }  }

 \rm{BC =  \sqrt{ {a}^{2}  +  y_1^{2} } }

 \sf \red{From \:  the  \: property \:  of  \: Equilateral \:  triangle : }

 \rm \: AB = BC

 \rm{2a =  \sqrt{ {a}^{2}  +  {y_1}^{2} } }

 \sf \red{Squaring \:  on  \: both  \: sides : }

 \rm \: 4 {a}^{2}  =   {a}^{2}  + y_1^{2}

 \rm \: {y_1 ^{2}  = 3 {a}^{2} }

 \rm \: y_1 =   \pm\sqrt{3} a

 \rm{As  \: C \:  is  \: in \:  postive \:  y-axis \:  we \:  get  \: y_1 = \sqrt{3}a}

 \boxed {\boxed {\rm \therefore \: cordinates \: of \: C = (0, \sqrt{3}a)}} \\

 \underline{ \underline{ \rm \red{case \: ii}}} \red{:} \sf{From  \: \Delta ABD}

 \rm{AB = B D= DA}

 \rm{The  \: distance \:  between  \: points \:  A(-a,0), \:  B(a,0)}

 \rm{AB= d =  \sqrt{(x_2 - x_1)^{2}  + (y_2 - y_1)^{2} } }

 \rm{AB =  \sqrt{ {(a + a)}^{2}  + (0  - 0) ^{2} } }

 \rm{AB \:  =  \sqrt{4 {a}^{2} } }

 \rm{AB = 2a}

 \rm{The  \: distance \:  between  \: points \:  B(a,0) \: ,D(0,y_2)}

 \rm{BD= d =  \sqrt{(x_2 - x_1)^{2}  + (y_2 - y_1)^{2} } }

 \rm{BD = \sqrt{( {0 - a)}^{2} +  {(y_2 - 0)}^{2}  }  }

 \rm{BD =  \sqrt{ {a}^{2}  +  y_2^{2} } }

 \sf \red{From \:  the  \: property \:  of  \: Equilateral \:  triangle : }

 \rm \: AB = BD

 \rm{2a =  \sqrt{ {a}^{2}  +  {y_2}^{2} } }

 \sf \red{Squaring \:  on  \: both  \: sides : }

 \rm \: 4 {a}^{2}  =   {a}^{2}  + y_2^{2}

 \rm \: {y_2 ^{2}  = 3 {a}^{2} }

 \rm \: y_2 =   \pm\sqrt{3} a

 \rm{As  \: D \:  is  \: in \:  negative \:  y-axis \:  we \:  get  \: y_2 =  - \sqrt{3}a}

 \boxed{ \boxed{ \rm \therefore \: cordinates \: of \: D= (0,  - \sqrt{3}a)}}

Answered by brainlychallenger19
15

\large\underline\orange{GIVEN ⤵}

Here, \:  \:  \:  AC = 2a \: and \: AO = a \\  {OC}^{2}  =  {AC}^{2}  -  {AO}^{2}  \\  =  {4a}^{2}  -  {a}^{2}  \\  =  {3a}^{2}  \\  ➡ \: OC \:  =  a\sqrt{3}  \\  Therefore \: ,coordinates \: of \: C \:  \\ are(0,a\sqrt{3}) \: and \: coordinates \: of \\ D \: are \: (0, - a\sqrt{3})

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