Let the equal sides of the isosceles Δ ABC be x.
Given that the perimeter of Δ ABC = 50m.
Therefore, 2x + AC = 50 --- (1)
It is also given that the perimeter of Δ ABD = 40m.
Therefore, x + BD + AD = 40
BD is the median of the Δ ABC. Therefore, D is the midpoint of AC.
So AD = CD.
Or, AD = [tex]\frac{1}{2}[/tex] AC
Therefore, [tex]x + BD + \frac{1}{2} AC = 40[/tex]
Multiply both sides by 2.
2x + 2BD + AC = 80
From (1), 2x + AC = 50.
Therefore, 2BD + 50 = 80
2BD = 80 - 50
2BD = 30
BD = 15m.
