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
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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
Given:
f is the parallel transfer to the vector
M
N
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
The point N is translated by the motion to the point N1 with the condition that the two vectors are equal: N
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
B
C
b)
the triangle ΔABC
A
B
C
Given:
the triangle ΔABC
AB = BC
the point D lies on AC: D
the point C lies on AD: C
BC
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
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
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)
ΔABC
EFPQ
Construction:
as it is shown on the picture.
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Definition:
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) A
OA = OA1
2) B
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
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) A
OA = OA1
2) B
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) B
BA = B1A
2) C
CA = AC1
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Given:
the point O is the center of the rotation
α = 120°
AB is the segment
AB→A1B1
A1B1
Construction:
1) A
OA = OA1
2) B
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) C
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:
ΔABC
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
Therefore
A
B
C
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
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
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,
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,
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 =
Then the area of the triangle
SΔABC =
= 30 (square units)
Answer: 30 square units.
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Given:
the triangle ΔABC is isosceles,
c = 4
Find: the area of the triangle SΔABC = ?
Solution:
SΔABC =
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
a2 = 16
a = 4 (unit)
Then the area of the triangle
SΔABC =
= 8 (square units)
Answer: 8 square units.
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Given:
the triangle ΔABC,
AH is the median
b = 8
Find: the radius of the circumscribed circle R = ?
Solution:
Since AH is the median, we see that CH =
By the Pythagorean theorem we obtain:
c2 = a2 + b2
c2 = 36 + 64
c = 10 (units)
Then CH =
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,
the ratio of the acute angles
AC = 4
Find: the radius of the circumscribed circle R = ?
Solution:
x + 2x + 90 = 180
3x = 90
x = 30
Then
Therefore, BC =
By the Pythagorean theorem we obtain:
AC2 + BC2 = AB2
AC2 +
AC2 =
AB2 =
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 =
Answer: 4.
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Given:
the triangle ΔABC,
tg A = 0.6
BC = 3
Find: AC = ?
Solution:
tg
0.6 =
Answer: 5.
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Given:
the triangle ΔABC,
AH = AC
Find:
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,
Answer: 30°.
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Given:
the triangle Δ ABC is regular, equilateral,
the area of the triangle
SΔABC =
Find: the length of the bisector BH = ?
Solution:
Consider Δ ABC is isosceles, where
Then BH is the median, the height.
Therefore, the segments are perpendicular: BH
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 =
SΔABH =
Consider the triangle Δ ABH.
Since BH is the bisector, we see that the angle
AH =
SΔABH =
AB • BH =
By the Pythagorean theorem we obtain:
AB2 = AH2 + BH2
AB2 =
BH2 =
BH =
Using the result (**) in equation (*), we obtain
AB •
AB2 =
AB =
Then AB • BH =
BH = 1 (unit)
Answer: BH = 1 unit
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Given:
the triangle Δ ABC is regular, equilateral,
R =
Find: the area of the triangle
SΔABC = ?
Solution:
Consider Δ ABO (AO = BO = R)
We draw the height OH from the vertex O to AB.
Consider Δ AOH, where
Since
OH =
By the Pythagorean theorem we obtain:
OH2 + AH2 = OA2
OH2 + AH2 =R2
=
AH2 =
Then the area of the triangle
SΔAOH =
Therefore, SΔABO = 2 • SΔAOH = 2 •
Then the area of the triangle
SΔABC = 3 • SΔABO = 3 •
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 =
Let x be the coefficient of the proportionality. Then
SABCD =
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 =
BO =
By the Pythagorean theorem we obtain:
AO2 + BO2 = AB2
The side of the rhombus is AB =
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 =
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
By the Pythagorean theorem we obtain:
AB2 = AH2 + BH2
BH =
Then the area of the triangle
SΔABD =
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
Then AB = (x – 15) unit (by the given condition).
By the Pythagorean theorem, we solve the equation:
(x – 15)2 = (
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:
CH is the bisector of the angle
Prove: the similarity of triangles
Δ BHC
Proof:
Since CH is the bisector of the angle
the angles are equal to
Consider Δ BHC
Then
Therefore the triangles are similar Δ BHC
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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
The angles are equal
Then the triangles are similar Δ BCO
Let BO = x, AO = 8 – x. Then 10x = 4 • (8 – x)
10x = 32 – 4x
14x = 32
x = 2
Therefore, BO = 2
Answer: BO = 2
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Given:
the similar triangles
ΔABC
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
Then the ratio of the perimeters of the triangles
From the equalities (*) and (**) it follows
B1C1 =
Then
A1B1 =
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:
Then, according to the first theorem of the similarity of the triangles:
the triangles are similar Δ AMD
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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