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F. Area of Projection of a Figure

Читайте также:
  1. A. Types of Projection
  2. B. Using Projection in Calculations
  3. Examples of Projection Using the Units and Products
  4. Facts and figures
  5. Finding the Distance Between Two Skew Lines by Projection
  6. I. Projection of a Point on a Line

Theorem: The area of the projection of a polygon on a plane is equal to the product of the area of the polygon and the cosine of the angle between the planes of polygon and its projection.

 
 
Areaprojection = Areafigure. cos q  

 

 


Proof:

We will prove this theorem for a triangle first and then generalize for all polygons.

 

Let us take a triangle ABC and a plane α. There are 3 possible cases:

1st case: If one side of ΔABC lies on plane α.

(Say side BC)(Figure 2.32).

 

Draw AA' ⊥ α where A' ∈ α.

 

We have ΔA'BC = Projα ΔABC.

 

Draw AH ⊥ BC.

 

By the three perpendiculars theorem, A'H ⊥ BC and therefore ∠AHA' = θ is the angle between planes of ΔABC and α.

 

In ΔA'BC,

In ΔAA'H,

.

 

2nd case: If one side of ΔABC (Say BC) is parallel to plane α then we take another plane β such that BC ∈ β and α // β (Figure 2.33).

 

ΔA'B'C' = Projα ΔABC

ΔNBC = Projβ ΔABC.

 

We know that ΔNBC ≅ ΔA'B'C'.

 

So A(ΔNBC) = A(ΔA'B'C') which means that A(ΔA'B'C') = A(ΔABC) ⋅ cos θ where θ is the angle

between planes of ΔABC and α.

 

3rd case: If ΔABC is not on plane α with no side parallel to plane α (Figure 2.34).

 

Let β be the plane of ΔABC.

 

Let α ∩ β = d.

 

Draw AN // d where N ∈ BC and A'N' = Projα AN.

 

Then A'N' // d.

 

So, for ΔABN and ΔANC, we can apply 2nd case.

 

Let θ be the angle between planes α and β.

 

A(ΔA'B'N') = A(ΔABN) ⋅ cos θ

A(ΔA'N'C') = A(ΔANC) ⋅ cos θ

+.

A(ΔA'B'C') = A(ΔABC) ⋅ cos θ

 

4th case: If we have a polygon, the area of its projection on a plane can be calculated easily by dividing this polygon into triangles and finding the sum of areas of projections of each triangle on the given plane.

 

Let A1B1C1D1E1 be the projection of polygon ABCDE on plane α (Figure 2.35).

 

Divide ABCDE into triangles ABC, ACD and ADE.

 

Let θ be the angle between plane of ABCDE and plane α.

 

 

Now, A(A1B1C1D1E1) = A(ΔA1B1C1) + A(ΔA1C1D1) + A(ΔA1D1E1)

= A(ΔABC) ⋅ cosθ + A(ΔACD) ⋅ cosθ + A(ΔADE) ⋅ cosθ

= (A(ΔABC) + A(ΔACD) + A(ΔADE)) ⋅ cosθ

= A(ABCDE) ⋅ cosθ

 

 

Example 59: Given two planes α and β. Equilateral triangle ABC whose one side length is a units lies in plane α. The angle between planes α and β is 60°. Find the area of projection of ΔABC on plane β.

Solution:

ΔABC is equilateral, so A(ΔABC) = .

 

Let ΔA'B'C' = Projβ ΔABC.

 

A(ΔA'B'C') = A(ΔABC) ⋅ cos 60°.

 

.

 

 

Example 60: Area of ΔABC is 30 cm2. The projection of ΔABC on a plane is ΔA'B'C' with side lengths 6 cm, 10 cm and 14 cm. Find the angle between the planes of these triangles.

Solution: Let θ be the angle between planes of ΔABC and ΔA'B'C'.

 

We know that A(ΔA'B'C') = A(ΔABC). cos α.

 

So, we just need to find A(ΔA'B'C').

 

We know all side lengths of ΔA'B'C' so by using the Heron’s Formula:

 

 

.

 

then α = 30°.

 

Example 61: What is the measure of the angle between the planes of a triangle and its projection if the area of its projection is half of the area of the triangle itself?

Solution: Let S be area of the triangle and S1 be area of its projection.

 

Let α be the angle between their planes.

 

Now, it is given that S = 2S1.

 

From the formula, S1 = S ⋅ cos α, S1 = 2S1 ⋅ cos α.

 

So cos α = and α = 60°.

Example 62: Given a regular hexagon with one side a = 8 cm. The angle between the plane of the hexagon and its projection plane is 30°. Find the area of the projection of this hexagon.

Solution: We know that the area of a regular hexagon with one side a is equal to .

 

So,

 

Here, the angle between two planes is 30°.

 

Therefore,

 

 

Example 63: If two lines intersect at an angle of 60° and each makes an angle of 45° with a plane, show that the projections of these lines on the plane are perpendicular to each other.

Solution: We are given two lines d and m, and projection plane α.

 

Let us take dm = P, Projα d = d 1,

Projα m = m 1, and Projα P = P1.

 

Now we take two points A and B on m and d, respectively such that PA = PB = k units.

 

Let Projα A = A1 and Projα B = B1.

 

ΔPAB will be equilateral because ∠APB = 60° and PA = PB. Therefore, PA = PB = AB = k units.

 

It is given that lines make 45° with plane α.

 

Now, P1A1 = PA ⋅ cos45° = k ⋅ and P1B1 = PB ⋅ cos45° = k ⋅ .

 

P1B1PB and P1A1AP are congruent trapezoids, so A1A = B1B and A1ABB1 is a rectangle.

 

It is obvious that AB = A1B1 = k.

 

So, in ΔP1A1B1 if we apply the Pythagorean theorem we will have,

P1A12 + P1B12= .

 

That means ∠A1P1B1 = 90°.

 

Therefore, projection lines are perpendicular.

 

Example 64: ABCD is a rectangle. Through vertex A, line AM is inclined to the

plane of rectangle ABCD. AM makes 60° with sides AD and AB.

Find the angle between AM and plane of rectangle.

Solution: Let us draw MM1 ⊥ (ABCD).

 

Point M1 will be on the bisector of ∠BAD.

 

Let us explain why.

 

Draw M1E ⊥ AD and M1F ⊥ AB, then

∠AEM = 90° and ∠AFM = 90°. (three perpendiculars theorem)

 

So, ΔAEM ≅ ΔAFM (A.A.A.).

 

Therefore, EM = FM which means their projections EM1 and FM1 are equal.

 

From here we can conclude that AFM1E is a square and AM1 is the angle bisector.

 

Now, let AM = a.

In ΔAME, we get AE = AM ⋅ cos60° = .

.

Finally, the angle between AM and plane of the rectangle, which is ∠MAM1, can be found as,

 

So ÐMAM1 = 45°.

 

 


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