# Given the equilateral triangle ABC and the isosceles triangle DBC, what is AB+DB if the perimeter of DBC is 50 and BC=24?

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Consider the isosceles triangle DBC with BC = 24 and a perimeter of 50. Assuming 24 is the unique side of the triangle, you can write the following equation:

CD + DB + BC = 50

CD + DB + 24 = 50

CD + DB + 24 = 50

CD + DB = 26

These sides are equal because it is an isosceles triangle, so they each equal 13.

Therefore, DB = 13

Since triangle ABC is equilateral and BC = 24, then AB = 24.

AB + DB

24 + 13 = 37

**Solution: 37**

****Note****

This was assuming BC is the unique side of the isosceles triangle. If BC is one of the matching sides of the isosceles triangle, the answer changes.

CD + DB + BC = 50

24 + DB + 24 = 50

DB + 48 = 50

DB = 2

AB + DB

24 + 2 = 26

Another possible answer is 26.

Since the triangle ABC is equilateral, then the lengths of it's sides are equal. We know that BC=24, therefore AB=BC=AC=24.

We'll suppose that BC is the base of the isosceles triangle DBC, therefore the lengths of the sides DB and DC are equal.

The perimeter of DBC is:

P = 2DB + BC

50 = 2DB + 24

2DB = 50 - 24

2DB = 26

DB = 13

**The requested sum of the sides is: AB+DB = 24 + 13 = 37.**