Similar Triangles - GCSE Maths

Summary

This 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

00:00

📐 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.

05:01

🔍 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.

10:02

🧩 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.

15:03

📏 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.

Keywords

💡Similar Triangles

Similar triangles are a fundamental concept in geometry where two triangles are said to be similar if their corresponding angles are equal and the ratios of their corresponding sides are equal. In the video, this concept is central as it is used to solve various problems involving triangles that are either connected or overlapping. The script explains that if two triangles are similar, there must be a constant scale factor relating their corresponding sides, which is demonstrated through examples where sides are multiplied by a constant number to match corresponding sides of another triangle.

💡Scale Factor

The scale factor is a ratio that describes the relationship between corresponding sides of similar triangles. It is a key concept in the video as it is used to determine how sides of one triangle relate to the sides of another. The script illustrates this by showing that if one side of a triangle is twice as long as the corresponding side of another, the scale factor is two, and this relationship holds for all corresponding sides.

💡Corresponding Sides

Corresponding sides in the context of similar triangles refer to the sides that are matched up when the triangles are compared. The video emphasizes that to determine if two triangles are similar, one must look at the relationship between these corresponding sides. The script provides examples where specific sides are matched and used to find the scale factor or to solve for unknown side lengths.

💡Vertically Opposite Angles

Vertically opposite angles are a pair of angles that are formed when two lines intersect. The video mentions these angles in the context of establishing similarity between triangles that are connected. When a parallel line is drawn, the angles on either side of the intersecting line are vertically opposite and thus equal, which is a clue that the triangles formed may be similar.

💡Alternate Angles

Alternate angles are a pair of angles that are on opposite sides of a transversal crossing two other lines (the parallel lines). The video script uses the concept of alternate angles to explain why certain angles in connected triangles are equal, which is a necessary condition for the triangles to be similar. This is demonstrated when the script describes drawing lines to reveal the 'Z' shape, which helps identify equal alternate angles.

💡Reciprocal

The reciprocal of a number is another number which, when multiplied with the original number, results in a product of one. In the video, the concept of reciprocal is used to explain that the scale factor can be found not only by dividing the larger side by the smaller side but also by taking the reciprocal of this division, which will yield the same scale factor. This is shown in the script when the presenter calculates the scale factor by dividing sides and then taking the reciprocal of those divisions.

💡Fractions

Fractions are used throughout the video to represent the relationships between the sides of similar triangles. The script explains that by forming equivalent fractions with the sides of the triangles, one can solve for unknown side lengths. This is demonstrated when the presenter sets up equations using fractions to find the values of missing sides in the triangles.

💡Parallel Lines

Parallel lines are lines in a plane that do not meet; they are always the same distance apart and never touch. In the video, parallel lines are used to identify similar triangles that are connected. The script mentions that when triangles are connected with parallel lines at the top and bottom, the angles formed are vertically opposite or alternate, which helps to establish similarity.

💡Ratio

A ratio is a quantitative relationship between two numbers, showing the number of times one value contains or is contained within the other. In the video, the concept of ratio is applied to the sides of triangles to establish similarity and to solve for unknown side lengths. The script gives an example where a ratio of sides is given, and the presenter uses this ratio to find the length of an unknown side.

💡Two-Step Equation

A two-step equation is a linear equation that typically requires two algebraic steps to solve. In the video, two-step equations are derived from setting up proportions based on the similar triangles' properties. The script illustrates how to solve these equations by first isolating the variable and then solving for its value, as seen when the presenter manipulates equations to find the lengths of sides labeled 'X' and 'Y'.

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

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think you should watch next subscribe so

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go and try the exam questions on this

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