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SEMEJANZA DE TRIÁNGULOS Y TEOREMA DE PITÁGORAS
REALIZÓ: Q.F.B María Consuelo Villanueva Magdaleno
ALUMNA
UNIDAD IV
Actividad 2
Mi Recurso Especializado
TUTORA: Mtra. Elia Astorga Mendoza
Febrero-2021
SEMEJANZA DE TRIÁNGULOS
¿Qué son los
Triángulos
Semejantes?
SEMEJANZA DE TRIÁNGULOS
SEMEJANZA DE TRIÁNGULOS
➢La igualdad de dos razones se llama: PROPORCIÓN
➢Los Elementos homólogos son los Ángulos y lados que se corresponden
en ambas figuras.
➢El cociente de dos lados homólogos se llama razón de semejanza.
3 cm
3 cm
1.5 cm
6 cm
CRITERIOS DE SEMEJANZA DE
TRIÁNGULOS
Identifica cuáles proporciones son correctas en
los siguientes ejercicios:
a)
3
4
=
15
20
b)
2
13
=
4
24
c)
1
7
=
7
49
d)
27
18
=
3
2
e)
40
5
=
8
1
Respuesta: a, c, d y e son
correctas.
Solo b, es incorrecta.
En los siguientes ejercicios encuentra el término
desconocido de cada proporción.
1)
𝑥
3
=
20
12
2)
7
𝑥
=
63
18
3)
1
8
=
𝑥+1
10
4)
2𝑥
5
=
10
𝑥
5)
3
4
=
10𝑥 −5
𝑥
RESPUESTAS
1)𝑥 = 5
2) 𝑥 = 2
RESPUESTAS
3)
1
8
=
𝑥+1
10
10 (
1
8
)= 𝑥 + 1
10
8
-
8
8
= 𝑥
2
8
=
1
4
=𝑥
𝑥 =
1
4
10
5
=
10
5
3
4
=
555
740
1
8
=
1/4+4/4
10
=
5/4
10
=
1
8
=
5
40
RESPUESTAS
4)
2𝑥
5
=
10
𝑥
x (
2𝑥
5
)= 10
2𝑥²
5
=10
2x² = 50
𝑥² = 25
𝑥= 25
𝑥= 5
2(5)
5
=
10
5
10
5
=
10
5
RESPUESTAS
5)
3
4
=
10𝑥 −5
𝑥
x (
3
4
)= 10x-5
3𝑥
4
= 10x-5
3𝑥=4 (10𝑥 -5)
3𝑥 = 40𝑥 − 20
3𝑥 − 40𝑥= -20
- 37𝑥 = -20
37𝑥= 20
𝑥=
20
37
3
4
=
10
20
37
−5
20
37
3
4
=
200
37
−
5
1
=
200−185
37
−
15
37
20
37
3
4
=
15
37
20
37
=
3
4
=
555
740
Aplicando el Teorema de Pitágoras,
encuentra en cada caso el lado
desconocido:
𝑐2 = 𝑎2 +𝑏2
Aplicando el Teorema de Pitágoras,
encuentra en cada caso el lado
desconocido:
𝑐2 = 𝑎2 +𝑏2
a) a= 7m, b= 24m y c=?
a=7m
b=24m
c=?
Nota: Una vez que se
resuelve de manera
analítica el problema, se
pasa a construir de
forma gráfica con
medidas a escala o
reales.
C= 25m
b) c= 41cm, b= 40cm y a=?
c = 41 cm
b= 40 cm
a=?
a= 9 cm
c) c= 17cm, a= 15cm y b=?
c= 17 cm
b= ?
a=15 cm
b= 8cm
d) a=4, b=3 , c=?
e) a=?, b=24cm, c=26cm
f) a=12cm, b=5, c=?
TEOREMA DE TALES DE MILETO
“Las rectas paralelas dividen a dos transversales en
segmentos proporcionales”
a
b d
c
𝑎
𝑏
=
𝑐
𝑑
TEOREMA DE TALES DE MILETO
En un viaje a Egipto Tales de Mileto, determinó la altura de
la Gran Pirámide basándose en la sombra de los objetos.
Para ello utilizó sus conocimientos sobre triángulos
semejantes y proporciones.
𝑎
𝑏
=
𝑐
𝑑
Resuelve aplicando el Teorema de Tales de Mileto
Resuelve aplicando el Teorema de
Tales de Mileto
https://www.bing.com/images/search?view=detailV2&ccid=SnqFTi4W&id=1A657CE36F679AD90F8AD558DA75C083
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=5A27C9540C53041B556BB2D9C42178BB&selectedindex=11&form=IRPRST&ajaxhist=0&pivotparams=insightsTok
en%3Dccid_SnqFTi4W*cp_35F24A1DCE1D0B421255A33B5E92477E*mid_1A657CE36F679AD90F8AD558DA75C083C
CA18861*simid_608037777775789322*thid_OIP.SnqFTi4WSpIz1PJ3AVVx8wHaFj&vt=0&sim=11&iss=VSI
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E9431D&thid=OIP.0aNzNaS7AgLoRkvcsD42EQHaEK&mediaurl=https%3a%2f%2fi.ytimg.com%2fvi%2fWQAHqRVMK
Fo%2fmaxresdefault.jpg&exph=720&expw=1280&q=criterios+de+semejanza+de+triangulos&simid=60802968599960
0590&ck=7F7CD6341D70827938557E3AA6415D31&selectedIndex=3&FORM=IRPRST&ajaxhist=0
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6tA%2fmaxresdefault.jpg&exph=720&expw=1280&q=planteamiento+de+problemas+matem%c3%a1ticos&simid=608
001201831216473&ck=FADD68A288E8DB5C312EAA57840B2CBA&selectedIndex=10&FORM=IRPRST&ajaxhist=0
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AC593EC612F51&thid=OIP.-
98MR_ux1XrTOOzeVjDeZAHaEK&mediaurl=https%3A%2F%2Fi.ytimg.com%2Fvi%2FoeHYvjgYbAY%2Fmaxresdefau
lt.jpg&exph=720&expw=1280&q=teorema+de+tales+de+mileto&simid=608002193983016313&ck=60EC50893DBB24D
BD604338B8520CB19&selectedindex=37&form=IRPRST&ajaxhist=0&vt=0&sim=11
https://www.bing.com/images/search?view=detailV2&ccid=a4LW99nF&id=35CB95EA2184234791D9FCB8A2945C6E
43F20A26&thid=OIP.a4LW99nFsHXbMYaQ5iLznAAAAA&mediaurl=https%3a%2f%2fth.bing.com%2fth%2fid%2fR6b8
2d6f7d9c5b075db318690e622f39c%3frik%3dJgryQ25clKK4%252fA%26riu%3dhttp%253a%252f%252f2.bp.blogspot.c
om%252f-pxu8pwfSFxo%252fUX_uxv7tJsI%252fAAAAAAAAFVs%252fjSuAD-BcNzo%252fs1600%252ftales-de-
mileto.jpg%26ehk%3dAOw%252fKhKld551enrkdPXO%252b9r1b%252fBeKYUjK%252foLg13nW98%253d%26risl%3d
%26pid%3dImgRaw&exph=477&expw=330&q=tales+de+mileto&simid=608011226306840162&ck=D36F6937C2C3E01
ED262657F7451543E&selectedIndex=17&FORM=IRPRST&ajaxhist=0
https://www.bing.com/images/search?view=detailV2&ccid=ewV30deR&id=B4F7840BA1E92F52144F602925CCC6BF
A1B548D5&thid=OIP.ewV30deRo0m80qe4iNPVFQHaDj&mediaurl=https%3a%2f%2fcienciamatematica.com%2fwp-
content%2fuploads%2fimagen.jpg&exph=540&expw=1125&q=pitagoras&simid=607998809469291459&ck=2ABB04C
CFAF590F37373609110BF86FA&selectedIndex=84&FORM=IRPRST&ajaxhist=0
FUENTES CONSULTADAS
18-feb-2021
FUENTES CONSULTADAS
- Programa de Estudios 2do Semestre DGB/DCA/06-2017
- Villanueva M.C Programación Semestral de Matemáticas II
FUENTES CONSULTADAS
- Programa de Estudios 2do Semestre DGB/DCA/06-2017
- Villanueva M.C Programación Semestral de Matemáticas II
https://www.bing.com/images/search?view=detailV2&ccid=CwLbxF4L&id=739DF6F4298
A864FED23F560405EC000E25F80B7&thid=OIP.CwLbxF4Lda1P2KKiIipwNAHaEU&mediaurl=ht
tps%3a%2f%2fwww.lifeder.com%2fwp-content%2fuploads%2f2017%2f08%2ffuentes-de-
consulta.jpg&exph=525&expw=900&q=fuentes+consultadas&simid=607987771441418019
&ck=8C1AE0FF039009BA8F7B738B5FFD8A33&selectedIndex=3&FORM=IRPRST&ajaxhist=0
Consultada el 18-feb-2021

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