Question

Два рівносторонніх трикутники АВС і АВК мають спільну сторону АВ, довжина якої 20 см. Площини цих трикутників перпендикулярні. Знайдіть відстань між вершинами С і К.

Answer 1

Пусть M - середина AB

CM, KM - медианы и высоты в равносторонних треугольниках

CM=KM =AB √3/2 =10√3

CM и KM - перпендикуляры к линии пересечения плоскостей, следовательно угол CMK - угол между плоскостями. Дано, что плоскости перпендикулярны, то есть угол между ними - прямой, CMK=90.

CK =CM √2 =10√6 (из треугольника CMK)

Answer 2

Problem Analysis

We are given two equilateral triangles, ABC and ABK, which share a common side AB with a length of 20 cm. The areas of these triangles are perpendicular to each other. We need to find the distance between the vertices C and K.

Solution

To find the distance between vertices C and K, we can use the Pythagorean theorem. Let's denote the distance between C and K as x.

We can divide triangle ABC into two right-angled triangles by drawing a perpendicular from vertex C to side AB. Similarly, we can divide triangle ABK into two right-angled triangles by drawing a perpendicular from vertex K to side AB.

Let's consider triangle ABC. The perpendicular from vertex C to side AB divides triangle ABC into two right-angled triangles, ACB and BCA. The hypotenuse of triangle ACB is side AB, which has a length of 20 cm. The base of triangle ACB is x cm (the distance between vertices C and K), and the height is the altitude of triangle ABC. Since triangle ABC is equilateral, the altitude is given by the formula (sqrt(3)/2) * side length. Therefore, the height of triangle ABC is (sqrt(3)/2) * 20 cm.

Using the Pythagorean theorem, we can write the equation for triangle ACB as:

x^2 + ((sqrt(3)/2) * 20)^2 = 20^2

Similarly, considering triangle ABK, we can write the equation as:

x^2 + ((sqrt(3)/2) * 20)^2 = 20^2

Simplifying the equations, we get:

x^2 + (10 * sqrt(3))^2 = 400

x^2 + 300 = 400

x^2 = 400 - 300

x^2 = 100

Taking the square root of both sides, we get:

x = sqrt(100)

x = 10 cm

Therefore, the distance between vertices C and K is 10 cm.

Conclusion

The distance between the vertices C and K of the equilateral triangles ABC and ABK, which share a common side AB with a length of 20 cm, is 10 cm.