F. Area of Projection of a Figure

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 = Proj_α 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' = Proj_α ΔABC

ΔNBC = Proj_β 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' = Proj_α 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' = Proj_β 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 d ∩ m = P, Proj_α d = d ₁,

Proj_α m = m ₁, and Proj_α P = P₁.

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

Let Proj_α A = A₁ and Proj_α 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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