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 40mIn 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
2 answers:
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 = AC
Therefore,
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.
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 = AC
Therefore,
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.
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