MATH SOLVE

2 months ago

Q:
# 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.

Accepted Solution

A:

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

(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