Question

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

Answer 1

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 = $\frac{1}{2}$ AC

Therefore, $x + BD + \frac{1}{2} AC = 40$

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.

Answer 2

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.