The perpendicular bisector of side AB of ∆ABC intersects side BC at point D. Find AB if the perimeter of ∆ABC is with 12 cm larger than the perimeter of ∆ACD.

Respuesta :

Answer:

Hence, AB=12.

Step-by-step explanation:

We are given that the perpendicular bisector of side AB of ∆ABC intersects side BC at point D.

this means that side AE=BE.

Also we could clear;ly observe that

ΔBED≅ΔAED

( since AE=BE, side ED common, ∠BED=∠AED

so by SAS congruency the two triangles are congruent)

Now we are given that:

the perimeter of ∆ABC is 12 cm larger than the perimeter of ∆ACD.

i.e. AB+AC+BC=AC+AD+CD+12

AB+BC=AD+CD+12

as AD=BD

this means that AD+CD=BD+CD=BC

AB+BC=BC+12

AB=12

Hence AB=12



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Answer:

The required length of [tex]AB[/tex] is [tex]12\rm\;{cm}[/tex].

Step-by-step explanation:

Given: The perpendicular bisector of side [tex]AB[/tex] of [tex]\bigtriangleup{ABC}[/tex] intersects side [tex]BC[/tex] at point [tex]D[/tex] and the perimeter of  [tex]\bigtriangleup{ACD}[/tex].

From the figure,

[tex]AE=BE[/tex]         .......(1)              (as [tex]DE[/tex] is perpendicular bisector of side [tex]AB[/tex])

Now, In [tex]\bigtriangleup{BED}[/tex] and [tex]\bigtriangleup{AED}[/tex]

     [tex]AE=BE[/tex]                                     ( from equation 1 )

[tex]\angle {BED} =\angle {AED}[/tex]                               ( Both [tex]90^\circ[/tex] )

    [tex]ED=ED[/tex]                                     ( Common side)

[tex]\bigtriangleup{BED}\cong\bigtriangleup{AED}[/tex]                              ( by SAS congruence rule)

      [tex]BD=AD[/tex]    .........(2)                   (by CPCT)

As per question,

The perimeter of ∆ABC is with 12 cm larger than the perimeter of ∆ACD.

[tex]AB+BC+AC=AC+CD+AD+12[/tex]

         [tex]AB+BC=AD+CD+12\\AD+CD=BD+CD\\AB+BC=BC+12\\[/tex]

                   [tex]AB=12\rm\;{cm}[/tex]

Hence, the length of [tex]AB[/tex] is [tex]12\rm\;{cm}[/tex].

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