Question

Катеты прямоугольного треугольника равны 12см и 7см. Найдите площадь проекции этого треугольника на плоскость, если плоскость треугольника наклонена к плоскости проекции под углом 60 градусов ДАМ 100 БАЛЛОВ СРОЧНО

Answer 1

S( треуг) =1/2* a* b , S=1/2*12*7=42.
Площадь ортогональной проекции многоугольника на плоскость равна площади проектируемого многоугольника, умноженной на косинус угла между плоскостью многоугольника и плоскостью проекций.

S(проекции) ==42* cos60°=42*1/2=21( см ²)

Answer 2

I will answer your question in detail. The legs of a right triangle are 12 cm and 7 cm. Find the area of the projection of this triangle onto a plane, if the plane of the triangle is tilted to the plane of projection at an angle of 60 degrees.

To solve this problem, we need to use the formula for the area of a triangle in terms of its sides and the angle between them:

$$S = \frac{1}{2}ab\sin\alpha$$

where $a$ and $b$ are the lengths of the sides and $\alpha$ is the angle between them.

In this case, the sides of the triangle are the legs, which are 12 cm and 7 cm, and the angle between them is the angle of inclination of the plane of the triangle to the plane of projection, which is 60 degrees.

Therefore, the area of the projection of the triangle is:

$$S = \frac{1}{2} \cdot 12 \cdot 7 \cdot \sin 60^\circ$$

Using the fact that $\sin 60^\circ = \frac{\sqrt{3}}{2}$, we can simplify the expression:

$$S = \frac{1}{2} \cdot 12 \cdot 7 \cdot \frac{\sqrt{3}}{2}$$

$$S = 21\sqrt{3} \text{ cm}^2$$

This is the final answer. I hope this helps you understand the solution.