Math Problem Statement

They are given in mm. The ground line is parallel and the small paper below the axis is 2cm. Stump 0 from the left of the 4cm paper. They give A (0.70,20) B (60, -10.60).

  1. Draw a right triangle ABC ABC Top C Horizontal e (x = 60, y = 90)
  2. Find the orthogonal Ω of triangle ABC and the vertex D of the rectangle ABCD (D.ABC) which is a right triangle.
  3. Find the cross section of the ABCD rectangle, the plan is parallel to the plan B1, the seal is 45.

Solution

Let's break down your problem and solve it step by step.

1. Draw a Right Triangle ABC

Given:

  • Points A(0.70,20)A(0.70, 20) and B(60,10.60)B(60, -10.60)
  • The top vertex CC is horizontally aligned at C(60,90)C(60, 90).

We can now determine the structure of triangle ABC.

  1. Vertices of Triangle ABC:
    • A(0.70,20)A(0.70, 20)
    • B(60,10.60)B(60, -10.60)
    • C(60,90)C(60, 90)

This triangle is right-angled at vertex BB, as CC is vertically above BB (same x-coordinate).

Diagram:

You can visualize this triangle, where:

  • Line AB represents one side,
  • Line BC is the vertical height from point BB to CC,
  • Line AC is the hypotenuse.

2. Find the Orthogonal Ω and Vertex D

Orthogonal Ω:

The orthogonal Ω\Omega of triangle ABC refers to the line that is perpendicular to the plane of triangle ABC.

  • Since we know that BC\overrightarrow{BC} is vertical, we can focus on the orthogonal that is perpendicular to the plane defined by the points A,B,CA, B, C.
  • The orthogonal line passes through a point in space perpendicular to the surface.

Finding Vertex D:

  • DD forms a rectangle ABCD. Since triangle ABC is right-angled at BB, we need to find a point DD such that ABCDABCD is a rectangle, and DD is positioned such that DD makes a right angle with lines ABAB and BCBC.

We know:

  • ABAB is diagonal along the base.
  • BCBC is vertical.

Thus, DD is located horizontally from point AA and vertically aligned with CC.

3. Find the Cross Section of the Rectangle

  • The cross section refers to the intersection between the plane parallel to the rectangle and another plane.
  • The plane B1B1 and a seal at 45° would define the angle for this cross-section.

Given that the plane B1 is parallel and the seal is inclined at 45°, the intersection of the ABCD rectangle with the seal will give you a section of the rectangle. You can compute the coordinates of the points of intersection by projecting along the angle of the seal.


Would you like a more detailed breakdown of any of these points? Here are some related questions you could explore:

  1. How do we calculate the lengths of sides of triangle ABC?
  2. What is the area of triangle ABC?
  3. How do we determine the slope of the hypotenuse ACAC?
  4. How do we find the equation of the plane containing triangle ABC?
  5. What does the 45° seal mean geometrically for the cross-section?

Tip:

Always start by analyzing the given points in relation to each other on the coordinate plane before solving more complex geometric properties.

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Math Problem Analysis

Mathematical Concepts

Geometry
Coordinate Geometry
Right Triangles
Cross Sections

Formulas

Pythagorean theorem (a^2 + b^2 = c^2)
Area of a triangle (1/2 * base * height)
Slope formula (m = (y2 - y1) / (x2 - x1))
Equation of a plane

Theorems

Pythagorean Theorem
Properties of Rectangles and Triangles

Suitable Grade Level

Grades 10-12