# Parallel Transfer, Rotation of a Plane, and Similar Triangles

• ### Корзина

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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.

=

***

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.

a)

the triangle ΔABC

A A1:

=

B B1:

=

C C1:

=

b)

the triangle ΔABC

A A1: =

B B1:

=

C C1:

=

***

Problem 147.

Given:

the triangle ΔABC

AB = BC

the point D lies on AC: DAC

BC B1D

a) Build: B1D

b) Prove: ABB1D is the isosceles trapezoid

a)

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

***

b)

Proof:

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).

***

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:

a)

1) we draw the ray CO

2) CC1;

COC1 = 60°

CO = C1O

b)

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

AB

BC

CA

I.e. ΔABC → ΔBCA.

So, Δ ABC is mapped to itself.

***

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:

Let x be the coefficient of the proportionality. Knowing that the sum of the angles in the triangle is 180°, we compose and solve the equation:

3x + 7x + 8x = 180

18x = 180

x = 10

The largest angle is CAB = 8 • 10 = 80°

***

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:

Let x° be the angle at the base of the triangle. Knowing that the sum of the angles in the triangle is 180°, we will compose and solve the equation:

(x + 60°) + x + x = 180°

3x = 180 – 60

3x = 120

x = 40

Therefore, BAC = 40°.

***

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:

Let x be the coefficient of the proportionality. By the theorem of Pythagoras we compose and solve the equation:

(5x)2 + (12x)2 = 262

25x2 + 144x2 = 676

169x2 = 676

x2 = 4

x = 2

b = 12 • 2 = 24 (cm)

***

Problem 158.

Given:

the triangle ΔABC,

C = 90°

b = 5 is the leg

c = 13 is the hypotenuse

Find: the area of the triangle SΔABC = ?

Solution:

By the Pythagorean theorem we obtain:

a ==== 12

Then the area of the triangle

SΔABC = • ab = =

= 30 (square units)

***

Problem 159.

Given:

the triangle ΔABC is isosceles,

C = 90°

c = 4 is the hypotenuse

Find: the area of the triangle SΔABC = ?

Solution:

SΔABC = • ab

Since Δ ABC is isosceles, we see that the angles at the base of 45° and the legs are a = b.

By the Pythagorean theorem we obtain:

c2 = a2 + b2 = a2 +a2 = 2a2

Then (4)2 = 2a2

a2 = 16

a = 4 (unit)

Then the area of the triangle

SΔABC = • ab = =

= 8 (square units)

***

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:

Since AH is the median, we see that CH = c

By the Pythagorean theorem we obtain:

c2 = a2 + b2

c2 = 36 + 64

c = 10 (units)

Then CH = c = = 5 (units)

The point H is the center of the circumscribed circle

AH = BH = CH = R

Since R = AH, we see that R = AH = CH = 5 units.

***

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:

Let x be the coefficient of the proportionality. Knowing that the sum of the angles in the triangle is 180°, we will compose and solve the equation:

x + 2x + 90 = 180

3x = 90

x = 30

Then CAB = 30°,

ABC = 2 • 30° = 60°

Therefore, BC = AB

By the Pythagorean theorem we obtain:

AC2 + BC2 = AB2

AC2 + = AB2

AC2 = AB2

AB2 = = 64

AB = 8 (units)

R = AD = BD = 8 : 2 = 4 (units)

***

Problem 162.

Given:

the triangle ΔABC,

C = 90°

BC = 3

the radius of the circumscribed circle

R = 2.5

Find: AC = ?

Solution:

R = AH = BH = 2.5

Then AB = 2.5 • 2 = 5

By the Pythagorean theorem we obtain:

AC = = = = 4 (units)

***

Problem 163.

Given:

the triangle ΔABC,

C = 90°

tg A = 0.6

BC = 3

Find: AC = ?

Solution:

tgA =

0.6 = ; AC = 3 • = 5 (units)

***

Problem 164.

Given:

the triangle ΔABC,

A = 90°

AH = AC

Find: ABC = ?

Solution:

Since AH = AC, we see that Δ AHC is isosceles.

The point H is the radius of the inscribed circle, so AH = CH, but AH = AC, therefore, AH = CH = AC.

Then Δ AHC is equilateral.

Therefore, HAC = AHC = HCA = 60°.

ABC = 180° – (90° + 60°) = 30°.

***

Problem 165.

Given:

the triangle Δ ABC is regular, equilateral,

the area of the triangle

SΔABC = square units

Find: the length of the bisector BH = ?

Solution:

Since Δ ABC – is regular, we see that all angles are equal to 60°.

Consider Δ ABC is isosceles, where

BAC = BCA = 60°.

Then BH is the median, the height.

Therefore, the segments are perpendicular: BH AC.

Consider the triangles Δ ABH and Δ BHC.

AB = BC, by the given hypothesis.

AH = CH, BH is the median.

BH is the common side.

Therefore, the triangles are equal Δ ABH = Δ BHC.

The area of the triangle is SΔABC = 2SΔABH

I.e. SΔABH = SΔABC = = (square units)

SΔABH = AH • BH

Consider the triangle Δ ABH.

Since BH is the bisector, we see that the angle ABH = 30°, therefore

AH = AB

SΔABH = AB • BH =

AB • BH = (*)

By the Pythagorean theorem we obtain:

AB2 = AH2 + BH2

AB2 = AB2 + BH2

BH2 = AB2

BH = AB (**)

Using the result (**) in equation (*), we obtain

AB • AB =

AB2 =

AB =

Then AB • BH = • BH =

BH = 1 (unit)

***

Problem 166.

Given:

the triangle Δ ABC is regular, equilateral,

the radius of the circumscribed circle

R =

Find: the area of the triangle

SΔABC = ?

Solution:

Consider Δ ABO (AO = BO = R) Δ ABO is the isosceles triangle.

We draw the height OH from the vertex O to AB.

Consider Δ AOH, where AHO = 90°.

Since HAO = 30°, we see that OH = AO OH = R

OH = =

By the Pythagorean theorem we obtain:

OH2 + AH2 = OA2

OH2 + AH2 =R2

+ AH2 = ()2 + AH2 =

=

AH2 = = AH = =

Then the area of the triangle

SΔAOH = AH • OH = = =

Therefore, SΔABO = 2 • SΔAOH = 2 • = (square units)

Then the area of the triangle

SΔABC = 3 • SΔABO = 3 • = = 2= 2.25 (square units)

***

Problem 167.

Given:

the rhombus area is SABCD = 384

the ratio of the diagonals of the rhombus:

AC : BD = 3 : 4

Find: the rhombus side AB = ?

Solution:

the area of the rhombus

SABCD = AC • BD

Let x be the coefficient of the proportionality. Then

SABCD = 3x • 4x

384 = 6x2

x2 = 64

x = 8

Therefore, the diagonal BD = 4x = 4 • 8 = 32

AC = 3x = 3 • 8 = 24

AC = 2AO

BD = 2BO

Therefore, half the diagonal AO =AC = • 24 = 12

BO =BD = • 32 = 16

By the Pythagorean theorem we obtain:

AO2 + BO2 = AB2

The side of the rhombus is AB = == 20

***

Problem 168.

Given:

the triangle Δ ABD is isosceles,

the side AB = 10

Find: the area of the triangle

SΔABD = ?

Solution:

the area of the triangle

We draw the height BH to the base AD.

By the property of an isosceles triangle:

BH is the median, the bisector, the height.

Since BH is the median, we see that AH = DH = 16 : 2 = 8 (unit)

Consider the triangle Δ ABH, where the angle AHB = 90°.

By the Pythagorean theorem we obtain:

AB2 = AH2 + BH2

BH = = == 6 (unit)

Then the area of the triangle

SΔABD = AD • BH = •16 • 6 = 48 (square units)

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

***

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:

Since the triangle Δ ABC is isosceles, we see that BH is the height, the median, the bisector.

Therefore AH = CH.

Then AC = AH + CH = AH + AH = 2 AH

Consider Δ ABH is the right triangle.

Let AC = (x) unit AH = () unit

Then AB = (x – 15) unit (by the given condition).

By the Pythagorean theorem, we solve the equation:

(x – 15)2 = ()2 + 152 x2 – 30x + 225 = + 225

4 (x2 – 30x) = x2

4x2 – 120x = x2

3x2 – 120x = 0 | : x

3x = 120

x = 40

So, 40 units – is the length of the base.

***

## 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.

***

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:

The angles are equal CBO = ODA as the crosswise angles at the parallel lines BC and AD and the secant BD.

The angles are equal BCO = OAD as the crosswise angles at the parallel lines BC and AD and the secant AC.

Then the triangles are similar Δ BCO Δ AOD.

= = = =

= . Then 4AO = 10BO BO = AO

= = 0.4 = k

Let BO = x, AO = 8 – x. Then 10x = 4 • (8 – x)

10x = 32 – 4x

14x = 32

x = 2 (cm)

Therefore, BO = 2cm.

= k = 0.4

Answer: BO = 2cm, = 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:

the perimeter of the triangle

P (ΔABC) = 12 +16 + 20 = 48 (dm)

Since the triangles are similar, we see that

==

=== k (*)

Then the ratio of the perimeters of the triangles

= k (**)

From the equalities (*) and (**) it follows

=

=

B1C1 = = 20 (dm)

Then =

=

A1B1 = = 15 (dm)

A1C1 = P(ΔA1B1C1) – B1C1 – A1B1 = 60 – 20 – 15 = 25 (dm).

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

***

Problem 173.

Given:

ABCD is the trapezoid,

the sides of the trapezoid intersect at the point M:

AB ∩ CD = M,

BC = 5 cm,

AB = 3.9 cm,

CD = 3.6 cm

Find:

MB, MC = ?

Solution:

Consider the triangles ΔAMD and ΔBMC:

BAD =MBC, as the corresponding angles at the parallel lines BC and AD and the secant AB.

MCB =MDA, as the corresponding angles at the parallel lines BC and AD and the secant CD.

Then, according to the first theorem of the similarity of the triangles:

the triangles are similar Δ AMD Δ BMC.

Therefore,

= =

=,

but AM = AB + BM = 3.9 + BM

Then

8 • BM = 5 (3.9 + BM)

8BM – 5BM = 19.5

3BM = 19.5

BM = 6.5 (cm)

=,

but MD = CD + MC = 3.6 + MC

8 • MC = 5 (3.6 + MC)

3MC = 18

MC = 6 (cm)