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.

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14-10-2017
Mathematics

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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.

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meerkat18 meerkat18 · Answer 1 · 2017-10-25T17:20:56+02:00

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

ernikhil ernikhil · Answer 2 · 2021-10-19T16:25:03+02:00

The length of the height triangle ABC is BD i.e., [tex]15\;units[/tex].

According to the question, In isosceles [tex]∆ABC[/tex] the segment BD is the median to the base AC . Also, the perimeter of ∆ABC is 50m, and the perimeter of the triangle [tex]ABD[/tex] is 40m.

Let the length of the equal sides of the isosceles triangle be [tex]x \;meters[/tex] and the length of the base of the isosceles triangle is [tex]2y\;meters[/tex].

[tex]Perimeter_{ABC}=x+x+2y\\2x+2y=50\\x+y=25\\x=25-y----(i)[/tex]

Also,

[tex]Perimeter_{ABD}=x+y+BD\\x+y+\sqrt{x^2-y^2}=40\\x+y+\sqrt{(25-y)^2-y^2}=40\\25+\sqrt{625-50y}=40\\\sqrt{625-50y}=15\\625-50y=225\\y=8\;units[/tex]

So,

[tex]x=25-8\\x=17\;units[/tex]

Now,

[tex]BD=40-x-y\\BD=40-17-8\\BD=15\;units[/tex]

Hence, the length of the height triangle ABC is BD i.e., [tex]15\;units[/tex].

Learn more about isosceles triangles here:

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