Similar Triangles - GCSE Maths
Summary
TLDRThis educational video explores the concept of similar triangles, focusing on two types of problems: connected triangles and overlapping ones. It explains that similar triangles have a constant scale factor, and this is used to solve problems involving unknown side lengths. The video demonstrates how to find missing values by setting up equations based on the properties of similar triangles, including corresponding angles and proportional sides. It also covers how to handle more complex scenarios with given ratios and multiple triangles, providing a step-by-step approach to solving such geometric problems.
Takeaways
- 🔍 The video discusses the concept of similar triangles, focusing on two types of problems: triangles that are connected and triangles that overlap.
- 📏 It explains that for two triangles to be similar, there must be a constant scale factor when comparing their corresponding sides.
- 📐 The script demonstrates how to find the scale factor by dividing pairs of corresponding sides and shows that the reciprocal of these fractions also yield the same value.
- 📝 It highlights that the scale factor can be determined by comparing sides within the same triangle, emphasizing the consistency in the method used for comparison.
- 🔄 The video uses the concept of alternate angles to show that triangles connected by parallel lines are similar, as these angles are equal.
- 📈 It provides a method to solve for missing side lengths in similar triangles by setting up equations based on the proportionality of corresponding sides.
- 📚 The script introduces the technique of using fractions to solve problems involving similar triangles, which is particularly useful for more complex questions.
- 📋 The video gives examples of worded problems involving similar triangles and demonstrates step-by-step solutions, emphasizing the importance of consistency in setting up fractions.
- 🔢 It tackles a more challenging problem involving the calculation of side lengths in overlapping triangles, showing how to handle different scenarios where some side lengths are known and others are not.
- 📘 Lastly, the video includes a problem with a given ratio of sides, illustrating how to use this ratio along with the properties of similar triangles to find unknown lengths.
Q & A
What is the definition of similar triangles?
-Similar triangles are defined as two triangles that have the same shape but not necessarily the same size, meaning their corresponding angles are equal and their corresponding sides are in proportion.
How do you determine the scale factor between two similar triangles?
-The scale factor between two similar triangles is determined by dividing the corresponding sides of the triangles. If the sides of one triangle are all twice as long as the corresponding sides of another, the scale factor is 2.
What is the relationship between the sides of similar triangles?
-The sides of similar triangles are in proportion. This means that if you take any two corresponding sides from each triangle, the ratios of those sides will be equal.
How can you use the concept of similar triangles to solve for missing side lengths?
-You can set up a proportion using the known sides of the triangles and the unknown side you're trying to find. By solving the proportion, you can determine the length of the missing side.
What is the significance of alternate angles when dealing with similar triangles?
-Alternate angles are equal when the lines forming the angles are parallel. This is significant in similar triangles because it helps establish that the triangles are similar when the corresponding angles are the same.
How does the presence of parallel lines affect the similarity of triangles?
-The presence of parallel lines indicates that certain angles are alternate and therefore equal, which is a key factor in determining that two triangles are similar.
What is the reciprocal of the scale factor and how does it relate to similar triangles?
-The reciprocal of the scale factor is the scale factor inverted. For similar triangles, if the scale factor is 2 (meaning one triangle is twice as large as the other), the reciprocal would be 1/2, indicating the scaling down from the larger to the smaller triangle.
Can you find the scale factor by dividing sides within the same triangle?
-Yes, you can find the scale factor by dividing sides within the same triangle, as long as you are consistent with the sides you choose for comparison.
What is the importance of being consistent when setting up fractions to solve for unknowns in similar triangles?
-Consistency is crucial when setting up fractions to ensure that the smaller triangle's side is always on top or always on the bottom, relative to the larger triangle's side, to avoid errors in solving for unknowns.
How can you use the concept of similar triangles to solve problems with overlapping triangles?
-Even with overlapping triangles, you can still identify corresponding sides and angles, and use these to set up proportions to solve for unknown side lengths, taking into account the overlapping parts as part of the triangle sides.
What additional information might be provided in a problem involving similar triangles, and how does it help?
-Problems might provide ratios of certain sides, like a 5:4 ratio in the script, which helps establish the relationship between unknown sides by assuming arbitrary lengths that maintain the given ratio.
Outlines
📐 Introduction to Similar Triangles
This paragraph introduces the concept of similar triangles, explaining that similarity is determined by a constant scale factor between corresponding sides. It demonstrates how to identify this scale factor by comparing side lengths of triangles and how this factor remains consistent across different pairs of corresponding sides. The paragraph also discusses how to find the scale factor by dividing corresponding sides or by using the reciprocal of these divisions. It emphasizes the importance of consistency in choosing which sides to compare when determining similarity.
🔍 Problem Solving with Connected Triangles
This paragraph delves into solving problems involving connected triangles, where parallel lines indicate similarity. It highlights the use of vertically opposite angles and alternate angles to confirm similarity. The paragraph illustrates how to find missing side lengths by setting up equations using the scale factor of the triangles. It also demonstrates the use of fractions to solve for unknowns, emphasizing the importance of consistency in the setup of fractions when dealing with similar triangles.
🧩 Tackling Overlapping Triangles
The focus of this paragraph is on solving problems with overlapping triangles, where similarity is established by showing that corresponding angles are equal. It guides through the process of identifying corresponding sides and setting up equations to find unknown side lengths. The paragraph also covers the method of using fractions to solve for unknowns in overlapping triangles, maintaining the consistency in the setup of fractions as seen in previous examples.
📏 Applying Ratios to Similar Triangles
This paragraph introduces the concept of using given ratios to solve problems involving similar triangles. It explains how to work with the ratio of side lengths to find unknown sides, even when the actual lengths are not directly provided. The paragraph demonstrates a step-by-step approach to solving such problems, including setting up equations based on the ratio and the known side lengths, and then solving for the unknowns. It concludes with a practical example that shows how to apply these methods in exam-type questions.
Mindmap
Keywords
💡Similar Triangles
💡Scale Factor
💡Corresponding Sides
💡Vertically Opposite Angles
💡Alternate Angles
💡Reciprocal
💡Fractions
💡Parallel Lines
💡Ratio
💡Two-Step Equation
Highlights
Exploring similar triangles and their properties.
Understanding the concept of constant scale factor for similar triangles.
Demonstrating how to find the scale factor by dividing corresponding sides.
Using reciprocals to confirm the scale factor in similar triangles.
Observing that consistent side division within a triangle yields the scale factor.
Applying the concept of similar triangles to connected triangles with parallel lines.
Identifying vertically opposite angles and alternate angles in connected triangles.
Solving for missing side lengths using multiplication based on similar triangles.
Using fractions to solve for unknown side lengths in similar triangles.
Consistency in setting up fractions is key when solving similar triangle problems.
Solving word problems involving similar triangles with overlapping sides.
Using corresponding angles to prove similarity in overlapping triangles.
Addressing the challenge of identifying corresponding sides in overlapping triangles.
Solving for unknowns in overlapping triangles using consistent fraction setup.
Applying ratios to find unknown side lengths in triangles with given ratios.
Practical example of using assumed lengths based on ratios to solve triangle problems.
Final walkthrough of solving a complex problem involving similar triangles and ratios.
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
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