Question

Дана равнобедренная трапеция ABCD с основанием 4 и 5 см.Угол BAD равен 45 градусов.Из вершнины трапеции С восстановлен перпендикуляр к плоскости трапеции СК длиной 3 см. Найти расстояние от вершины перпендикуляра К до точки А.

Answer 1

Problem Analysis

We are given a trapezoid ABCD with bases of length 4 cm and 5 cm. Angle BAD is 45 degrees. From vertex C of the trapezoid, a perpendicular CK of length 3 cm is drawn to the plane of the trapezoid. We need to find the distance from the vertex of the perpendicular K to point A.

Solution

To find the distance from point K to point A, we can use the properties of similar triangles. Let's denote the distance from point K to point A as x cm.

Since triangle ABC is an isosceles triangle (AB = BC), we can conclude that triangle ABC is also a right triangle because angle BAD is 45 degrees. Therefore, angle ABC is also 45 degrees.

Now, let's consider triangle ABC and triangle KCA. These two triangles are similar because they share angle ABC and angle KCA is a right angle. Therefore, we can set up the following proportion:

AB / BC = KA / KC

Substituting the given values, we have:

4 / 5 = x / 3

Cross-multiplying, we get:

4 * 3 = 5 * x
12 = 5x
x = 12 / 5

Therefore, the distance from point K to point A is 12/5 cm.

Answer

The distance from the vertex of the perpendicular K to point A is 12/5 cm.