The contents of this webpage:
- 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.
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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Given:
the segment AB
Build:
AB → A1B1
A → A1 : =
B → B1 : =
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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
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.
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a)
the triangle ΔABC
A A1:
=
B B1:
=
C C1:
=
b)
the triangle ΔABC
A A1: =
B B1:
=
C C1:
=
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
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
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b)
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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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.
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Definition:
MOM1 = α and OM1 = OM.
O is the center of the rotation
α is the angle of the rotation
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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
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Theorem:
The rotation is the motion.
Given:
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..
***
Given:
the point O is the center of the rotation
AB is the segment
AB→A1B1
Build:
A1B1
Construction:
1) AA1;
AOA1 = 180°
OA = OA1
2) BB1;
BOB1 = 180°
OB = OB1
***
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
***
Given:
the point O is the center of the rotation
α = 120°
AB is the segment
AB→A1B1
A1B1
Construction:
1) AA1;
AOA1 = 120°
OA = OA1
2) BB1;
BOB1 = 120°
OB = OB1
***
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
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.
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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
AB
BC
CA
I.e. ΔABC → ΔBCA.
So, Δ ABC is mapped to itself.
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Revising.
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°
Answer: 80°.
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Given:
the triangle ΔABC is the isosceles triangle,
one angle is greater than the other:
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°.
Answer: 40°.
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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)
Answer: 24 cm.
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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)
Answer: 30 square units.
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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)
Answer: 8 square units.
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Given:
the triangle ΔABC,
A = 90°
AH is the median
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.
Answer: 5 units.
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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:
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)
Answer: 4 units.
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Given:
the triangle ΔABC,
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)
Answer: 4.
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Given:
the triangle ΔABC,
C = 90°
tg A = 0.6
BC = 3
Find: AC = ?
Solution:
tgA =
0.6 = ; AC = 3 • = 5 (units)
Answer: 5.
***
Given:
the triangle ΔABC,
A = 90°
AH = AC
Find: ABC = ?
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°.
Answer: 30°.
***
Given:
the triangle Δ ABC is regular, equilateral,
the area of the triangle
SΔABC = square units
Find: the length of the bisector BH = ?
Solution:
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)
Answer: BH = 1 unit
***
Given:
the triangle Δ ABC is regular, equilateral,
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)
Answer: 2.25 square units
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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
Answer: 20.
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Given:
the triangle Δ ABD is isosceles,
the side AB = 10
Find: the area of the triangle
SΔABD = ?
Solution:
the area of the triangle
SΔABD = AD • BH
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
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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.
Answer: AC = 40 units
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Given:
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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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.
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Given:
the similar triangles
ΔABC ΔA1B1C1 ,
AB = 12 dm,
BC = 16 dm,
AC = 20 dm,
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.
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Given:
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:
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)
Answer: 6.5 cm, 6 cm.
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