Question

ABC is an isosceles triangle with AB=AC=13cm and BC=10cm. Calculate the length of the perpendicular from A to BC.

Answer 1

since this is an isosceles triangle we could split and have a perpendicular bisector. since it is a bisector we know the bottom part of it would be 5 (half of 10) and the hypothenuse would be 13. this is a right triangle since the line we got was a perpendicular bisector. 

using the pythagorean theorem we get height perpendicular bisector is 12. 
13^2 - 5^2 = 144 = 12^2 so side is 12 and it would be the altitude to BC 

now we find the area which is 1/2 ( b ) ( h ) we got the height/altitude which is 12 when base is 10 

so area will be 

1/2 ( 12 ) ( 10 ) = 60 

using AB as the base the perpendicular which is the height would be solved by 

2 A / b = h 

120 / 13 would be the altitude to BC 

you could use heron's formula/ hero's formula if you want but it would be too tedious

Answer 2

   let a iso. triangle ABC , 
   in which BC = b ( base )
                 h = height ( altitude )
   S = $\sqrt{ \frac{a+a+b}{2} }$
      = $\frac{13+13+10}{2}$
      = 36 
     A  = $\sqrt{s(s-a)(s-a)(s-b)}$
         = [tex] \sqrt{18(18-13)(18-13)(18-10)} [/tex]
         = $\sqrt{18*5*5*8}$
        = 60 cm^2
   area = $\frac{1}{2} *b*h$
     60 = $\frac{1}{2} *10*h$
     $\frac{60*2}{10} = h$
     therefore height(altitude) = 12 cm