Question

Боковая сторона равнобедренного треугольника делится точкой касания вписанной окружности в отношении 2:7считая от вершины угла при основании треугольника найдите стороны треугольник если периметр равен 110

Answer 1

Примем коэффициент пропорциональности отрезков, на которые боковая сторона равнобедренного треугольника делится точкой касания вписанной окружности, за х, а основание - за у.
Тогда периметр треугольника равен 2*(2х+7х) + у = 110.
По свойству точки касания 2х = у/2 или у = 4х (так как треугольник равнобедренный).
Подставим эту зависимость в первое уравнение.
2*9х + 4х = 110,
22х = 110,
х = 110/22 = 5.
Отсюда находим стороны треугольника:
- боковые стороны равны 2*5+7*5 = 10 + 35 = 45,
- основание равно 110 - 2*45 = 110 - 90 = 20.

Answer 2

Решение в приложении:

Answer 3

Given Information:

- The lateral side of an isosceles triangle is divided by the point of tangency of the inscribed circle in a ratio of 2:7, counting from the vertex of the angle at the base of the triangle. - The perimeter of the triangle is 110.

Solution:

Let's assume that the base of the isosceles triangle is divided into two segments by the point of tangency of the inscribed circle. The ratio of these two segments is 2:7.

Let the lengths of these two segments be 2x and 7x, respectively.

Since the triangle is isosceles, the lengths of the two equal sides are also 2x and 7x.

The perimeter of the triangle is the sum of the lengths of all three sides. According to the given information, the perimeter is 110.

Therefore, we can write the equation:

2x + 7x + 7x = 110

Simplifying the equation:

16x = 110

Dividing both sides by 16:

x = 110/16 = 6.875

Now, we can find the lengths of the sides of the triangle:

- Length of the base = 2x = 2 * 6.875 = 13.75 - Length of the equal sides = 7x = 7 * 6.875 = 48.125

Therefore, the lengths of the sides of the triangle are approximately: - Base: 13.75 - Equal sides: 48.125

Please note that these values are approximate due to rounding.

I hope this helps! Let me know if you have any further questions.