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.

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

The length of the height triangle ABC is BD i.e., .

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

Let the length of the equal sides of the isosceles triangle be and the length of the base of the isosceles triangle is .

Also,

So,

Now,

Hence, the length of the height triangle ABC is BD i.e., .

Learn more about isosceles triangles here:

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