Question

Пожалуйста, срочно нужно с решением! Из вершины B параллелограмма ABCD проведены его высоты BM и BN. Известно, что угол A=60°, точеи M и N лежат на сторонах параллелограмма DM=4 DN=2. Найдите площадь треугольника BMN(Sbmn).

Answer 1

Решение во вложении........

Answer 2

Problem Analysis

We are given a parallelogram ABCD with a vertex B. From vertex B, the altitudes BM and BN are drawn. It is known that angle A is 60° and points M and N lie on the sides of the parallelogram such that DM = 4 and DN = 2. We need to find the area of triangle BMN (Sbmn).

Solution

To find the area of triangle BMN, we can use the formula for the area of a triangle given the lengths of two sides and the included angle. The formula is:

Area = (1/2) * side1 * side2 * sin(angle)

In this case, we know the lengths of sides BM and BN, and the included angle is angle A. Let's substitute the given values into the formula and calculate the area.

Given: - DM = 4 - DN = 2 - Angle A = 60°

We need to find the area of triangle BMN (Sbmn).

Calculation

Using the formula for the area of a triangle, we have:

Area = (1/2) * BM * BN * sin(A)

Substituting the given values:

Area = (1/2) * 4 * 2 * sin(60°)

We know that sin(60°) = √3/2, so:

Area = (1/2) * 4 * 2 * (√3/2)

Simplifying:

Area = 4 * (√3/2)

Area = 2√3

Therefore, the area of triangle BMN (Sbmn) is 2√3.

Answer

The area of triangle BMN (Sbmn) is 2√3.