Mathematics, 10.07.2019 12:10 minecraftpro64p7t123

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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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

Answer from: Siebee23

The length of the height triangle ABC is BD i.e., $15\;units$ .

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

Let the length of the equal sides of the isosceles triangle be $x \;meters$ and the length of the base of the isosceles triangle is $2y\;meters$ .

$Perimeter_{ABC}=x+x+2y\\2x+2y=50\\x+y=25\\x=25-y----(i)$

Also,

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

So,

$x=25-8\\x=17\;units$

Now,

$BD=40-x-y\\BD=40-17-8\\BD=15\;units$

Hence, the length of the height triangle ABC is BD i.e., $15\;units$ .

Learn more about isosceles triangles here:

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In isosceles ∆abc the segment bd (with d ∈ ac ) is the median to the base ac . find bd, if the perim...