Q:

In isosceles ∆ABC the segment BD (with D ∈ AC ) is the median to the base AC . Find BD, if the perimeter of ∆ABC is 50m, and the perimeter of ∆ABD is 40m.

Accepted Solution

A:
Given:

(Isosceles Trianlge) ----∆ABC Perimeter = 50 meters 
(Right Triangle) ---------∆ABD Perimeter = 40 meters
Segment BD bisects segment AC into two equal lenghts

Required: BD

Solution: 

We know that isosceles triangle has two equal legs and a base. Therefore,
For ∆ABC,
AB & BC are the two equal legs.
AC is the base.

For right triangle ∆ABD,
AB is the hypotenuse
AD is the base leg
BD is the other leg


Let us represent segments AB & BC as a
                           segment  AC as b
                           segment BD as h        
                           segment AD as b/2 

In equation we have,
 ∆ABC Perimeter = a + a + b
                       50 = 2a + b                   eq. (1)

∆ABD Perimeter = a + b/2 + h
                       40 = a + b/2 + h            eq. (2)

Dividing eq. (1) by 2, we get
                       25 = a + b/2                  eq. (3)

Substitute eq. (3) in eq. (2),
                        40 = (25) + h
                         h = 15 meters

 ANSWER: BD = 15 meters