Question

5. Найдите среднюю линию равнобедренного треугольника, параллельной боковой стороне, основание равно 10 см, а его периметр равен 34 см.

Answer 1

Ответ:

6 cм

Объяснение:

Дано: ΔАВС, АВ=ВС, АС=10 см, Р=34 см. КМ - средняя линия. Найти КМ.

Средняя линия треугольника равна половине стороны, которой она параллельна. КМ=1/2 АВ.

АВ=ВС=(34-10):2=12 см.

КМ=12:2=6 см.

Answer 2

To answer this question, we need to use some properties of the isosceles triangle and the midline. Let ABC be the isosceles triangle with AB = AC and base BC = 10 cm. Let M be the midpoint of BC and N be the midpoint of AC. Then MN is the midline parallel to the side AB. We need to find the length of MN.

First, we can use the fact that the perimeter of the triangle is 34 cm to find the length of the side AB. We have:

AB + BC + AC = 34

AB + 10 + AB = 34

2AB = 24

AB = 12

So, the side AB is 12 cm long.

Next, we can use the property that the midline of a triangle is half the length of the parallel side. We have:

MN = (1/2)AB

MN = (1/2)12

MN = 6

So, the midline MN is 6 cm long.

Therefore, the answer is: the midline of the isosceles triangle parallel to the side is 6 cm long. You can see a diagram of the triangle and the midline [here](https://reshak.ru/otvet/reshebniki.php?otvet=6-52&predmet=pogorelov7-9).