Similar Triangles - GCSE Maths
Summary
TLDREste video educativo explora el concepto de triángulos similares, presentando dos tipos de problemas: triángulos conectados y triángulos superpuestos. Se explica cómo determinar la relación de escala constante entre los lados correspondientes de los triángulos y cómo usar la similitud para resolver problemas. Se destacan técnicas como la división de lados en el mismo triángulo y la utilización de fracciones equivalentes. Además, se abordan problemas prácticos con triángulos conectados y superpuestos, y se muestra cómo aplicar la similitud y las proporciones para encontrar longitudes desconocidas en triángulos.
Takeaways
- 🔍 El vídeo trata sobre triángulos similares y cómo resolver problemas donde los triángulos están conectados o se superponen.
- 🔄 Para que dos triángulos sean similares, deben tener una relación de escalado constante entre sus lados correspondientes.
- 📏 El factor de escala se puede encontrar dividiendo los lados correspondientes de los triángulos, y siempre dará el mismo valor.
- ↔️ El factor de escala también se puede encontrar al dividir los lados dentro del mismo triángulo, siempre que se dividan de manera consistente.
- 🔼 Al resolver problemas con triángulos similares, se pueden utilizar ángulos verticalmente opuestos y ángulos alternos para demostrar la similitud.
- 📐 La similitud de triángulos se puede demostrar al superponer triángulos y comparar los ángulos correspondientes.
- 🔢 Para encontrar valores desconocidos en triángulos similares, se establecen fracciones de lados correspondientes y se resuelven ecuaciones.
- 📏 En problemas donde los triángulos se superponen, se identifican las longitudes de los lados correspondientes y se resuelven ecuaciones para encontrar longitudes desconocidas.
- 💡 Es importante ser consistente al establecer fracciones al inicio de un problema, ya que afecta la resolución de la ecuación.
- 📋 En problemas con ratios dados, se pueden asignar longitudes ficticias a los lados en proporción para facilitar el cálculo de longitudes desconocidas.
Q & A
¿Qué son los triángulos similares y cómo se definen?
-Los triángulos similares son dos triángulos que tienen las mismas proporciones y las mismas formas, pero no necesariamente el mismo tamaño. Se definen por tener un factor de escala constante para la ampliación o reducción de uno a otro, y por tener todos sus ángulos correspondientes iguales.
¿Cómo se determina si dos triángulos son similares?
-Para determinar si dos triángulos son similares, se verifica que sus ángulos correspondientes sean iguales y que la proporción de sus lados correspondientes sea constante. Esto se puede hacer dividiendo los lados correspondientes de uno de los triángulos por los lados correspondientes del otro triángulo y ver si el resultado es el mismo para todos los pares de lados.
¿Qué es el factor de escala de amplificación y cómo se calcula?
-El factor de escala de amplificación es el número por el cual se multiplica el lado de un triángulo para obtener el lado correspondiente en el otro triángulo similar. Se calcula dividiendo un lado del triángulo más grande por el lado correspondiente del triángulo más pequeño.
Si dos triángulos son similares, ¿qué relación tienen sus lados?
-Si dos triángulos son similares, los lados correspondientes están en la misma proporción, lo que significa que si se toma cualquier par de lados en uno de los triángulos y se divide el lado más grande entre el lado más pequeño, el resultado será el mismo que para cualquier otro par de lados en los triángulos.
¿Cómo se resuelven problemas donde los triángulos están conectados?
-Para resolver problemas donde los triángulos están conectados, se identifican los lados correspondientes y se establecen las relaciones de proporción utilizando el factor de escala de amplificación. Luego, se resuelven las ecuaciones para encontrar los valores desconocidos, asegurándose de que los factores de escala se apliquen consistentemente.
En problemas donde los triángulos se solapan, ¿cómo se demuestra que son similares?
-En problemas donde los triángulos se solapan, se demuestran similares al mostrar que tienen todos los ángulos correspondientes iguales, ya sea por ser ángulos alternos o por tener lados paralelos que forman ángulos verticales opuestos.
¿Cómo se utilizan las proporciones para encontrar lados desconocidos en triángulos similares?
-Se establecen fracciones con los lados correspondientes y se resuelven las ecuaciones para encontrar los lados desconocidos. Es importante ser consistente en la selección de lados para las fracciones, ya sea siempre poniendo el triángulo más pequeño en la parte superior o siempre en la parte inferior.
¿Qué sucede si se tiene una relación de longitudes dada en un problema de triángulos similares?
-Si se tiene una relación de longitudes dada, como 5 a 4, se puede asumir una longitud para cada lado en proporción y luego se utilizan estas longitudes para establecer fracciones y resolver las ecuaciones para encontrar los lados desconocidos.
¿Cómo se abordan los problemas donde los triángulos tienen longitudes totales conocidas y partes desconocidas?
-Se identifican las longitudes totales y se establecen ecuaciones para las partes desconocidas, utilizando las proporciones de los lados correspondientes de los triángulos similares. Se resuelven estas ecuaciones para encontrar los valores de las longitudes desconocidas.
¿Cómo se pueden usar las fracciones para resolver problemas de triángulos similares?
-Las fracciones se usan para establecer relaciones de proporción entre los lados correspondientes de los triángulos similares. Se forman fracciones con los lados y se resuelven las ecuaciones resultantes para encontrar los valores desconocidos, asegurándose de que las fracciones sean consistentes en su estructura.
Outlines
📐 Introducción a los Triángulos Similares
Este primer párrafo introduce el concepto de triángulos similares, explicando que para que dos triángulos sean similares, deben existir una relación constante de escalado entre sus lados correspondientes. Se ilustra con un ejemplo de dos triángulos donde la relación de similitud se demuestra al multiplicar los lados de uno por dos para obtener los lados del otro. Además, se mencionan otras observaciones, como el cálculo del factor de escala y la consistencia en los fracciones equivalentes que se pueden obtener al dividir los lados dentro de los mismos triángulos o entre triángulos similares.
🔍 Triángulos Similares y Puntos en Paralelo
Este segmento se enfoca en problemas donde los triángulos están conectados a través de líneas paralelas. Se destaca la importancia de los ángulos verticalmente opuestos y los ángulos alternos, que son iguales cuando las líneas están paralelas. Se resuelve un ejemplo donde se encuentran los valores perdidos de los lados de los triángulos, utilizando tanto la multiplicación directa como la resolución de fracciones para encontrar las soluciones. Se enfatiza la consistencia en la forma de establecer las fracciones al inicio del problema.
🧩 Solución de Problemas con Triángulos Overlapped
En este apartado, se aborda cómo resolver problemas donde los triángulos se solapan. Se demuestra la similitud de los triángulos a través de la comparación de sus ángulos correspondientes. Se presenta un ejemplo donde se identifican los lados correspondientes y se resuelven los valores perdidos de los lados X e Y utilizando la técnica de las fracciones y la consistencia en la elección de los pares de lados. Se resalta la importancia de la consistencia en la selección de lados para establecer las fracciones y se resuelven los valores de X e Y.
🔢 Utilizando Proporciones en Triángulos Similares
Finalmente, se presenta un desafío más complejo que involucra proporciones en triángulos similares. Se describe un escenario donde se conoce la proporción de lados de un triángulo y se debe encontrar la longitud de uno de sus lados. Se utiliza una estrategia de resolución donde se asume una longitud para los lados en proporción y se resuelve el problema como se habría hecho con longitudes reales. Se resalta la utilidad de esta técnica para resolver problemas en exámenes, y se resuelve un ejemplo específico donde se encuentra la longitud de un lado desconocido.
📚 Conclusión y Recomendaciones para Práctica
El último párrafo ofrece una conclusión del video y recomienda al espectador explorar más videos relacionados, suscribirse para no perderse futuros contenidos y practicar con preguntas de exámenes en el tema. Se invita a los espectadores a intentar resolver problemas similares para fortalecer su comprensión y habilidad en el uso de triángulos similares.
Mindmap
Keywords
💡Triángulos similares
💡Escala de ampliación
💡Lados correspondientes
💡Ángulos verticalmente opuestos
💡Ángulos alternos
💡Fracciones equivalentes
💡Relación de proporcion
💡Problemas de triángulos conectados
💡Problemas de triángulos superpuestos
💡Ecuaciones de dos pasos
Highlights
Exploración de triángulos similares y dos tipos de problemas: triángulos conectados y triángulos superpuestos.
Definición de triángulos similares basada en una constante de escala constante entre ellos.
Demostración de similitud triangular mediante la multiplicación de longitudes de lado correspondientes.
Observaciones sobre cómo encontrar la constante de escala a través de la división de pares de lados correspondientes.
Ejemplo práctico de cómo dividir lados dentro del mismo triángulo para encontrar la constante de escala.
Importancia de las fracciones equivalentes en los triángulos similares para resolver problemas más complejos.
Análisis de triángulos conectados con líneas paralelas y cómo esto afecta la similitud.
Uso de ángulos verticalmente opuestos y ángulos alternos para demostrar similitud en triángulos conectados.
Técnica para encontrar longitudes desconocidas en triángulos similares conectados mediante la multiplicación y la división.
Enfoque alternativo para resolver problemas de triángulos similares utilizando fracciones y la consistencia en la selección de lados.
Solución de problemas de triángulos superpuestos utilizando la similitud y la correspondencia de ángulos.
Estrategia para manejar longitudes de lados en triángulos superpuestos y cómo encontrar longitudes desconocidas.
Uso de la consistencia en la formación de fracciones al resolver problemas de triángulos similares superpuestos.
Resolución de problemas con triángulos similares y ratios dados, utilizando la técnica de asumir longitudes para los ratios.
Ejemplo de cómo asumir longitudes para ratios ayuda a encontrar la longitud de un lado desconocido en un triángulo.
Conclusión del video con una revisión de los conceptos y la recomendación de intentar problemas de exámenes relacionados.
Transcripts
[Music]
in this video we're going to look at
similar triangles we're going to look at
two types of problem one's where the
triangles are connected like this one
and ones where they overlap like this
one here before we do this we're going
to take a closer look at what it means
for two triangles to be similar take
these two triangles here if these two
triangles are similar there must be a
constant scale for out of enlargement
from one to the other so if we match up
the corresponding sides on the base here
we have 12 and 24 on the left we have 5
and 10 for the height and the hypotenuse
of these are 13 and 26 to get from 12 to
24 we'd multiply by two this is the same
as going from 5 to 10 on the height and
also for the hypotenuse from 13 to 26
since this is always the same number
these shapes must be similar it works in
the other direction as well so if we go
from 24 to 12 we would multiply by 1/2
and and this is the same for the other
sides as well so we know these two
shapes are similar there are some other
observations we can make we can find
that scale factor of enlargement by
dividing the pairs of corresponding
sides so if we take the purple sides 24
and 12 and divide them you'll see we get
two this works for the other pairs of
sides as well so 10id 5 that's also 2
and 26id 13 that's once again two but it
also works the other way around so if we
take those fractions and do the
reciprocal like
this then these will always give you the
same value as well which this time is
1/2 in these fractions here we divided
one side from one of the triangles by
the corresponding side on the other
triangle but we can actually just divide
sides within the same triangle as well
so if we take the green side in the
small triangle and divide it by the
purple side we get 5 over2 and then if
we do the same thing in the larger
triangle the green side divid by the
purple side that's 10id 24 these two
fractions actually give the same value
and it will be the same for any of the
pairs of sides as long as you're
consistent in what you do for instance
if I take the blue side and divide it by
the green side and then do the same on
the second triangle blue divide by Green
I get the same number once again these
two fractions give the same value or I
could have done the purple side divide
by the blue side so we get 12 over 13 on
this one and on the other one 24/ by 26
and these also give the same value and
so do their reciprocals so if we take
all of these fractions and take the
reciprocal of them those will also be
vehicle as well so there's actually lots
and lots of equivalent fractions that
you can write down using similar
triangles this is going to be helpful in
solving some of the more complicated
questions in this
video now let's go back to the original
type of question the first one we said
was when we have two triangles that are
connected and notice we have a parallel
line at the top and at the bottom the
first thing to notice here is that this
angle here is the same as the one below
it because these are vertically opposite
angles then if we draw in this angle
here and draw on some lines like this we
find that the green angle is the same as
the green angle down here and we call
these alternate angles you can see we've
got a zed shape there the same idea
works for this angle as well so if we
draw in a zed shape here this time we
find that this blue angle is the same as
the blue angle at the bottom down here
once again due to alternate angles in
the previous video on similar shapes we
learned that when the angles are all the
same the shapes must be similar it's
even more obvious if I move this
triangle down to the bottom here and
flip it round you can see the blue
angles on the left are the same so are
the red ones at the top and the green
ones on the right but I'd also need to
move down these sides like this notice
how the sides have moved to a different
position this time if I put the original
diagram back you can see the eight was
on the right hand side but now it's on
the left hand side the 10 was on the
left and now it's on the right and the
13 was on the top and now it's on the
bottom so let's go ahead and try and
find these missing values now that we've
put this information on so to get from
10 to 20 we multiply by two so we must
multiply the 13 by two on the bottom to
get to Y and 13 * 2 is
26 then of course we must multiply the 8
by two to get to the x value and 8 * 2
is
16 now this uses the technique that we
did in the previous video on similar
shapes but I'm also going to show you
this by using fractions you don't need
to use this approach to solve these ones
but it's going to help us to practice it
especially for the questions later in
this video so what we're going to do is
pair up the corresponding sides so on
the left here we have X and 8 and on the
right we have 20 and 10 we're going to
use these pairs of sides to find the
value of X so if we divide the green
ones by doing xide by 8 this must be the
same as when we divide the blue ones so
20 over 10 so we form an equation like
this we can solve this equation by
multiplying both sides by eight if you
multiply the left side by 8 the 8 will
cancel and then you multiply the right
side by 8 we get this 20 * 8 on the top
there is 160 so we have 160 over 10
which was 16 and we knew that was the
answer to the question because we worked
it out before so so X is 16 and let's do
a similar idea to work out y so y pairs
up with the 13 on the bottom here so if
we divide the purple Sid we get y over
13 and this must be equal to any of the
other pairs divided I'm going to go for
the blue pair again so 20 over
10 here we just multiply both sides by
13 and we get 20 lots of 13 over 10 20 *
13 is 260 and divide this by 10 and you
get
26 which again we knew this was the
answer from before
now let's take a look at how a question
like this could be worded so if we take
some triangles like this and we're told
to work out the values of X and Y notice
they haven't even told us that these two
triangles are similar they don't need to
because they've marked on the parallel
lines so we should already know that
information so let's try with X first
you can see that X matches up with a 15
on the bottom then if we remember that
if we flip that triangle from the top
upside down the8 is actually going to
match up with a 12 so even though the
eight is on the right it matches up with
the 12 on the left left and the four
matches up with the Y so let's divide
the purple sides so x / 15 and this is
going to be equal to one of the other
pairs divided now since we don't know
why I'm going to avoid the blue pair and
go for the green one so the green one
would be eight IDE by 12 then we can
just solve this like we did for the
previous two we just multiply both sides
by 15 so on the left we get X and on the
right we get 8 over2 multiplied 15 which
we could combine to one fraction like
this 8 * 15 is 120 so we get x = 120 /
12 and 120 / 12 is just 10 so we found
the value of x it's 10 one thing that's
really important is you need to be
consistent in the way you set up the
fractions at the start of this question
the X here was from the smaller triangle
the eight was also from the smaller
triangle the 15 was from the larger
triangle and so was the 12 it's
important that you're consistent in
having the small on the top and the
large on the bottom or of course you
could have the large on the top and the
small on the bottom it doesn't matter
which way around you do it as long as
you're consistent I put the small on the
top here because we were trying to find
X so let's replace the X with a 10 and
let's go and find y so this time we're
going to divide the blue pair so y over
4 and this is going to be equal to one
of the other pairs I'm going to go for
the green one once again but this time
it will be 12 over 8 notice y was on the
larger triangle so 12 must also be on
the larger triangle and on the bottom
we've got four and eight which are from
the smaller triangle to solve this then
we just multiply both sides by four on
the left we'd have y and on the right 12
8 multiplied by 4 we can combine that to
one fraction so it's 12 * 4 over 8 12
fours are just 48 so it's 48 over 8 and
48 / by 8 is 6 so the value of y is
6 now let's have a look at the other
type of problem so a question that looks
like this this time the triangles are
both there but they're overlapping each
other we we can show that they're
similar Again by looking at their angles
so if we draw in these lines here then
this green angle on the left must be the
same as this green angle up here this is
because they're corresponding angles the
same idea works on the right side so if
we draw in this line and then these two
this angle here that's blue must be the
same as this one here that's blue once
again because they're corresponding
angles and then finally we have the
angle at the top here well that's
actually the same angle for both of the
triangles so if I separate them off like
this we've got the small triangle and
then the large triangle you can see all
of the angles match again so they must
be similar shapes but this one can be a
bit more tricky to do on the top
triangle I actually have lengths for all
of the sides so I've got the 15 on the
bottom the 18 on the left and the 12 on
the right on the larger triangle though
it's a bit more difficult I can see I
have 25 on the base so I can mark that
on but the left and right side can be
tricky so for the left side here it's
the whole length from here to here so we
have a y and an 18 so the total length
must be y + 18 it's similar on the right
hand side so we need to go all the way
from here to here which we have X and 12
so the total length is x + 12 now that
we form these triangles we're ready to
find the missing values of X and Y to do
this we're going to divide corresponding
pairs of sides again so I'm going to
start by trying to find X so I can see
I've got X on the larger triangle in the
x + 12 side so x + 12 and I'm going to
divide that by the side that matches on
the smaller shape which is 12 so I have
x + 12 over 12 then this is going to be
equal to a second fraction and I need to
pick another pair of sides now I'm not
going to pick the sides on the left
because they have a y in them if I pick
the base of the triangle then I can see
I have both of those sides I've got 15
and 25 I need to be consistent and since
I started with the larger triangle on
the last fraction I'll do the same here
so it's 25 on the top and 15 on the
bottom now we can go ahead and solve
this so for this one we would multiply
both sides by 12 that would give us x +
12 on the left and on the right we'd
have 25 multiplied by 12 all over 15 we
could also multiply both sides by 15 now
so if we multiply the left side by 15 we
need to use a bracket so we'd have 15
lots of x + 12 and on the right hand
side the 15s will no cancel so it's just
25 * 12 on the left I would now expand
out this bracket so 15 lots of X is 15 x
and 15 lots of 12 is 180 on the right
side I need to do 25 multip by 12 which
is 300 this is now just a nice two-step
equation to solve we can subtract 180
from both sides so we get 15x = 120 and
then divide both sides by 15 we get x =
8 so we found the value of x it's 8
cm now let's do the same idea and work
out why so we'll start with the y + 18
and divide it by the side that matches
on the other triangle which is
18 and this will be equal to another
fraction and it makes sense to choose
the sides 25 and 15 again so 25 over 15
you can see I've been consistent again
and I have the larger triangle on the
top and the smaller triangle on the
bottom of those fractions we'll solve
this in the same sort of way so we'll
multiply both sides by 18 on the left
this will give us y + 18 and on the
right 25 * by 18 all over
15 then multiply both sides by 15 so we
get 15 lots of y + 18 and on the right
the 15 cancel so it's just 25 * 18 if we
expand out the brackets here we get 15
lots of y That's Just 15 Y and 15 * 18
is
270 on the right 25 lots of 18 is 450
once again we just have a two-step
equation to solve if you subtract 270
from both sides we get 15 y = 180 and
then divide both sides by 15 and we'll
get y = 12 now let's try a second
example of this this a little bit more
difficult so this time we've got this
diagram and we're just trying to find X
so if we separate these off into two
triangles we have a small triangle that
looks like this and we can see the side
on the left there at the top is X and on
the right we have a nine then if we do
the big triangle here the side on the
top left is actually all the way from
here to here so we need to do x + 5 and
on the right this time is
12 now that we have these two triangles
we can go ahead and find X so we need to
divide the sides that match so I'm going
to divide x + 5 on the big triangle by X
on the small
triangle and this will equal another
fraction and since I started with the
larger one I'll do 12 over 9 this one's
a little bit different in its structure
but we use the same idea to solve it
let's multiply both sides by X this time
on the left this will cancel the X on
the bottom so I get x + 5 and on the
right I'll get 12 lots of X over
9 then we multiply both sides by 9 on
the left hand side this will give us 9
lots of x + 5 and on the right the N9
will cancel so we just have 12 lots of X
expanding the bracket on the left gives
us 9x + 45 and on the right 12 lots of X
is just 12 x to solve this equation we'
subtract 9x from both sides
if you subtract it from the left it will
cancel so we've just got 45 and if you
subtract 9x from 12x you get
3x then finally divide both sides by
three and you find that X is equal to 15
now I want to look at one more of these
questions but there's going to be some
ratio involved so if we take this
diagram here and notice we actually have
labels for the sides this time this is
quite common in exams especially for Ed
Exel and this time they're going to tell
us that the ratio of a d to be is 5 to 4
and we're going to be tasked with
finding the length of De
now at first glance it looks like we
don't have very much information on this
diagram we only have the 15 cm so it
might feel like we don't have enough to
answer the question but we do we're
going to start with this ratio here A D
to be being at 5: 4 now the easiest way
to think about this is just to pretend
that a d is 5 cm and that be is 4 cm
even if it doesn't look like they're
that long in the diagram it's still
going to work for the question so I'm
going to label be as four and AD as five
from here we just proceed with the
question as normal it's important that
you understand that those aren't
actually that length it's just their
ratio but for this question it will help
us solve the problem finally since we're
trying to find de I'm going to label
that as X so what we'll do now is
separate them off into the two triangles
so we have the small triangle that looks
like this that's triangle EBC and for
this triangle we know some of its
lengths from E to C is 15 and E to B is
four or at least we're pretending it's
four and for the large triangle which is
triangle da we know the height on the
left from a to d is 5 or at least we're
pretending it's five and we also know
the length all the way along the
diagonal there is x + 15 now we just go
ahead and set up some fractions again so
I'm going to start with x + 15 and
divide it by the same side on the other
triangle which is 15 then I have my
second fraction that this is equal to
and it will be 5 over
4 now we just solve this equation like
we have done all of the previous ones
multiply both sides by 15 that's x + 15
on the left and on the right 5 * 15 all
over 4 then multiply both sides by four
on the left we've got four lots of x +
15 and on the right the four will cancel
so 5 lots of
15 expand out the bracket on the left
we've got 4x + 60 and on the right 5 *
15 is
75 now solve this two-step equation by
subtracting 60 from both sides 4X will
be 15 and then finally divide both sides
by four and we'll find that X is
3.75
thank you for watching this video I hope
you found it useful check out the one I
think you should watch next subscribe so
you don't miss out on future videos and
go and try the exam questions on this
topic that are in this video's
description
Ver Más Videos Relacionados
2416 Razón entre los lados de un triángulo 45 45 90
GeoGebra Triángulo de Napoleón
FUNCIONES TRIGONOMETRICAS Super facil | Para principiantes | Encontrar medida del angulo
Cómo y cuándo usar el Teorema del Coseno - Parte 1
RAZONES TRIGONOMÉTRICAS DE ÁNGULOS DE CUALQUIER MEDIDA || ÁNGULOS EN POSICIÓN NORMAL #14
Teorema de Pitágoras | Ejercicio de práctica
5.0 / 5 (0 votes)