Question

Triangle ABC is an isosceles triangle, with AB=AC, altitude AM=11cm. Let a point D be on AM so that AD=10cm. If angleBDC=3× angleBAC. The perimeter of the triangle is?​​

Answer 1

Question  : -

                       Triangle ABC is an isosceles triangle, with AB=AC, altitude AM=11cm. Let a point D be on AM so that AD=10cm. If angleBDC=3× angleBAC. The perimeter of the triangle is?​​

Answer:-

                    given ⬇

                               AB = AC

                               AM = 11

                               AD = 10

                             < BDC= 3* < BAC

               To find :- ⬇

                            perimeter of triangle

                                = AB + BC + AC=?

SO,

      here is your solution , ⬇

   In triangle ABM ,

                              <AMB = 90

                                AM = 11cm

so by SAS rule BM= MC

         by using pythagorus theroem

                   AB^2 = AM ^2 + BM^2

                   BM^2 = 121- AB ^2    (1)

AS

     $\sqrt{3} /2=AM/ AB \\AB=22/\sqrt{3} \\AC=22/\sqrt{3}$

                         

again ,

           $BC=22/\sqrt{3}$

        SO AB + BC + CA =

            = $22/\sqrt{3} + 22/\sqrt{3} + 22/\sqrt{3} \\= 66/\sqrt{3} \\=22\sqrt{3}$             

HOPE IT WILL HELP YOU PLESE MARK IOT AS BRIANLIST

thank you

Answer 2

Step-by-step explanation:

74h

$\sqrt[9]{?}$

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