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Parallel Transfer, Rotation of a Plane, and Similar Triangles

The contents of this webpage:

– the lesson "Parallel transfer" is presented in the example of solving problems 145 - 148;
– in the worksheets under numbers 149 - 154 of this workbook for mathematics, the rotation of the plane around the point by an angle is considered;
– revising of the geometry of grade 9 in solutions is given in the examples 155 - 173:

- the angles of the triangle,

- the area of the triangle through the legs and the hypotenuse,

- the calculation of the radius of the circumscribed circle,

- the side of the rhombus,

- the similar triangles.

Parallel transfer

Definition:

The parallel transfer onto a vector is a mapping of the plane onto itself if the following conditions hold: each point M is mapped to the point M₁ such that two vectors are equal.

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Problem 145.

Given:

the vector

the segment AB

Build:

AB → A₁B₁

A → A₁ : =

B → B₁ : =

***

Theorem:

With the parallel transfer to the vector the distance between the points is kept, i.e. the parallel transfer is the motion.

Given:

f is the parallel transfer to the vector

M M₁

NN₁

Prove:

f is the motion (MN = M₁N₁)

Proof:

The point M is translated by the motion to the point M₁ with the condition that the two vectors are equal: M M₁: = MM₁

The point N is translated by the motion to the point N₁ with the condition that the two vectors are equal: NN₁: = NN₁

Therefore, the obtained segments are parallel MM₁ || NN₁ and the constructed segments are equal MM₁ = NN₁

Therefore, the quadrilateral MM₁N₁N is the parallelogram.

Therefore MN = M₁N₁, then f is the motion.

***

Problem 146.

the triangle ΔABC

A A₁:

B B₁:

C C₁:

the triangle ΔABC

A A₁: =

B B₁:

C C₁:

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Problem 147.

Given:

the triangle ΔABC

AB = BC

the point D lies on AC: DAC

the point C lies on AD: CAD

BC B₁D

a) Build: B₁D

b) Prove: ABB₁D is the isosceles trapezoid

Construction:

1) draw the line a from the point B parallel to the vector : a ||

2) the point B is translated by the motion to the point B₁

3) draw the line B₁D parallel to the segment BC:

B₁D || BC

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Proof:

Consider the quadrilateral BB₁DC.

Since the bases of BB₁ || CD and the lateral sides BC || BD are parallel, we see that BB₁DC is the parallelogram (by the definition)

By the property of the parallelogram:

the bases BB₁ = CD and the lateral sides BC = BD are equal, but AB = BC, then AB = B₁D

Since BB₁ || AD are parallel and AB B₁D are not parallel, therefore ABB₁D is the trapezoid (by the definition).

Since AB = B₁D, we see that ABB₁D is the trapezoid (by the definition).

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Problem 148.

Given:

the triangle ΔABC

EFPQ is the trapezoid

the circle (O; R)

the vector

Build:

circle (O;R) circle (O₁;R₁)

ΔABC ΔA₁B₁C₁

EFPQ E₁F₁P₁Q₁

Construction:

as it is shown on the picture.

***

Rotation of the plane around the point by the angle

Definition:

Rotating the plane around the point O by an angle α is a mapping of the plane onto itself such that each point M is mapped onto the point M₁ such that the rotation angle

MOM₁ = α and OM₁ = OM.

O is the center of the rotation

α is the angle of the rotation

***

Problem 149.

Given:

α = 75° (counter-clockwise)

O is the center of the rotation

AB is the segment

Build:

A₁B₁

Construction:

1) AA₁;

AOA₁ = 75°

OA = OA₁

2) BB₁;

BOB₁ = 75°

OB = OB₁

***

Theorem:

The rotation is the motion.

Given:

f is the rotation

α is the angle of the rotation (counter-clockwise)

the point O is the center of the rotation

MN is the segment

Prove:

f is the motion, MN = M₁N₁

Proof:

Let M→M₁; N→N₁

Then the triangles are equal ΔOMN = ΔOM₁N₁ by two sides and the angle between them:

OM = OM₁

ON = ON₁

MON = M₁ON₁

Then MN = M₁N₁, therefore, f is the motion..

***

Problem 150.

Given:

the point O is the center of the rotation

α = 180°

AB is the segment

AB→A₁B₁

Build:

A₁B₁

Construction:

1) AA₁;

AOA₁ = 180°

OA = OA₁

2) BB₁;

BOB₁ = 180°

OB = OB₁

***

Problem 151.

Given:

the triangle ΔABC

the point A is the center of the rotation

α = 160° (counter-clockwise)

ΔABC→ΔAB₁C₁

Build: ΔAB₁C₁

Construction:

1) BB₁;

BAB₁ = 160°

BA = B₁A

2) CC₁;

CAC₁ = 160°

CA = AC₁

***

Problem 152.

Given:

the point O is the center of the rotation

α = 120°

AB is the segment

AB→A₁B₁

Build:

A₁B₁

Construction:

1) AA₁;

AOA₁ = 120°

OA = OA₁

2) BB₁;

BOB₁ = 120°

OB = OB₁

***

Problem 153.

Given:

the point C is the center of the circle (C; R)

the point O is the center of the rotation

the rotation angle α = 60° (counter-clockwise)

a) the point C and the point O do not coincide

b) the point C and the point O coincide

Build:

circle (C₁; R)

Construction:

1) we draw the ray CO

2) CC₁;

COC₁ = 60°

CO = C₁O

Since the point O as the center of rotation and the point C as the center of the circle coincide, we see that the circles (C;R) and (C₁;R) will also coincide.

***

Problem 154.

Given:

Δ ABC is the isosceles, equilateral triangle

D is the point of intersection of the bisectors

D is the center of the rotation

the rotation angle α = 120°

Prove:

ΔABCΔABC

Proof:

Since Δ ABC is the regular triangle, we see that all angles of it are equal to 60°.

Since the point D is the center of the circumscribed and inscribed circle, we see that

AD = BD = DC = R.

Δ ABD = Δ BDC = Δ DAC (by three sides).

Therefore that ADB =BDC =CDA

Therefore

I.e. ΔABC → ΔBCA.

So, Δ ABC is mapped to itself.

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Revising.

Problem 155.

Given:

the triangle ΔABC

the ratio of the angles

ABC :BCA :CAB = 3 : 7 : 8

Find: the largest angle of the triangle

Solution:

Answer: 80°.

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Problem 156.

Given:

the triangle ΔABC is the isosceles triangle,

one angle is greater than the other:

ABC >

BAC by 60°

Find: the angle at the base of the triangle

Solution:

Answer: 40°.

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Problem 157.

Given:

the triangle ΔABC is right

c = 26 cm is the hypotenuse

the ratio of the legs:

a : b = 5 : 12

Find: the greater leg b

Solution:

Answer: 24 cm.

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Problem 158.

Given:

the triangle ΔABC,

C = 90°

b = 5 is the leg

c = 13 is the hypotenuse

Find: the area of the triangle S_Δ_ABC = ?

Solution:

Answer: 30 square units.

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Problem 159.

Given:

the triangle ΔABC is isosceles,

C = 90°

c = 4 is the hypotenuse

Find: the area of the triangle S_Δ_ABC = ?

Solution:

Answer: 8 square units.

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Problem 160.

Given:

the triangle ΔABC,

A = 90°

AH is the median

a = 6

b = 8

Find: the radius of the circumscribed circle R = ?

Solution:

Answer: 5 units.

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Problem 161.

Given:

the triangle ΔABC,

C = 90°

the ratio of the acute angles

ABC : CAB = 1 : 2

AC = 4

Find: the radius of the circumscribed circle R = ?

Solution:

Answer: 4 units.

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Problem 162.

Given:

the triangle ΔABC,

C = 90°

BC = 3

the radius of the circumscribed circle

R = 2.5

Find: AC = ?

Solution:

Answer: 4.

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Problem 163.

Given:

the triangle ΔABC,

C = 90°

tg A = 0.6

BC = 3

Find: AC = ?

Solution:

Answer: 5.

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Problem 164.

Given:

the triangle ΔABC,

A = 90°

AH = AC

Find: ABC = ?

Solution: Answer: 30°.

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Problem 165.

Given:

the triangle Δ ABC is regular, equilateral,

the area of the triangle

S_Δ_ABC = square units

Find: the length of the bisector BH = ?

Solution:

Answer: BH = 1 unit

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Problem 166.

Given:

the triangle Δ ABC is regular, equilateral,

the radius of the circumscribed circle

R =

Find: the area of the triangle

S_Δ_ABC = ?

Solution:

Answer: 2.25 square units

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Problem 167.

Given:

the rhombus area is S_ABCD = 384

the ratio of the diagonals of the rhombus:

AC : BD = 3 : 4

Find: the rhombus side AB = ?

Solution:

Answer: 20.

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Problem 168.

Given:

the triangle Δ ABD is isosceles,

the base AD = 16

the side AB = 10

Find: the area of the triangle

S_Δ_ABD = ?

Solution:

Answer: the area of the triangle is S_Δ_ABD = 48 square units

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Problem 169.

Given:

the triangle Δ ABC is isosceles,

the height BH = 15

the base AC is greater than the height BH by 15: AC > BH by 15

Find: the base AC = ?

Solution:

Answer: AC = 40 units

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Similar triangles

Problem 170.

Given:

the triangle Δ ABC, two angles

A = 54°

B = 18°

CH is the bisector of the angle C

Prove: the similarity of triangles

Δ BHC Δ ABC

Proof:

C = 180° – (A +B)

C = 180° – (54° + 18°) = 108°

Since CH is the bisector of the angle C, we see that

the angles are equal to

BCH = HCA = 108° : 2 = 54°

Consider Δ BHC

HBC = B = 18°

BCH = A = 54°

Then CHB = C = 108°

Therefore the triangles are similar Δ BHC Δ ABC.

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Problem 171.

Given:

ABCD is the trapezoid,

the top base BC = 4 cm

the bottom base AD = 10 cm

the diagonal BD = 8 cm

Find:

the part of the diagonal BO = ?

the ratio of the perimeters of the triangles

= ?

Solution:

Answer: BO = 2

cm,

= 0.4.

***

Problem 172.

Given:

the similar triangles

ΔABC ΔA₁B₁C₁,

AB = 12 dm,

BC = 16 dm,

AC = 20 dm,

the perimeter of the triangle:

P (ΔA₁B₁C₁) = 60 dm

Find:

the sides of the triangle ΔA₁B₁C₁

A₁B₁, A₁C₁, B₁C₁= ?

Solution:

Answer: A₁C₁ = 25 dm, A₁B₁ = 15 dm, B₁C₁ = 20 dm.

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Problem 173.

Given:

ABCD is the trapezoid,

the sides of the trapezoid intersect at the point M:

AB ∩ CD = M,

BC = 5 cm,

AD = 8 cm,

AB = 3.9 cm,

CD = 3.6 cm

Find:

MB, MC = ?

Solution:

Answer: 6.5 cm, 6 cm.

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