Download This topic was given by Mrs Kalyani, we were meant to understand

This topic was given by Mrs Kalyani, we were meant to understand the basic facts and thus
summarise it in our own words. We also had to demonstrate some sums. I have extracted
some information from the internet, understood it and then written it in my own words.
However, other shapes and figures were copied from many different sites, and also some
questions were extracted from other sites. I have also used some questions from the book
in order to demonstrate my ability.
Thank you:
Deip. D. Majithia
Grade 11.
In my project, I am assigned to write about Similarities. I will write everything that I have
understood in my own words, and will demonstrate certain sums. Most of the information I
have got is from the internet, I have also acknowledged some of the information from my
mathematics teacher and lastly I have used sources like my text book for extra sums and
more important information. I will demonstrate the sums and prove them.
Similarities: Similarities could mean two things which are the same. They could be similar
but not identical (NOT COMPLETELY SAME). Obviously this topic is to do with maths
(subtopic: Geometry). Two Figures are similar if they have the same shape. Shapes that are
similar have all corresponding (matching) sides equal in the same proportion. This means
they have the same shape but different sizes. If we say that a triangle A is similar to a
triangle B then we write :
A ~
thing.
-
B (~) this is the notation used to show that something is similar to another
In the above bolded statement, if shapes are similar their corresponding sides are
equal and in the same proportion.
6cm
e.g.1
3cm
Rectangle B
2cm Rectangle A
4cm
When I say corresponding sides, I mean sides 2cm and 4cm and sides 3cm and 6cm. Their
Ratio is the same. 4/2 = 6/3 or vice versa is equal to 2.
SIMILAR POLYGONS:
Polygons are said to be similar only if:
(1) Their corresponding angles are the same
(2) Their corresponding sides are in a ratio
(1) This above shape was extracted from a site, so that I could use it as an example for
my project. In the above shape. Quadrilateral ABCD ~ ( is similar to) EFGH. This
means that sides AB and EF, AD and EH, BC and FG and DC and HG are proportional.
In order to find out what is the ratio of their proportion. We can divide the second
(new) figure’s side by the first (old). If we take Quad ABCD as our first figure then:
Ratio= EH/AD is the ratio of proportion. 2/4. The ratio is ½.
Therefore we can say that: AB/EF = AD/EH= BC/FG = DC/HG
(2) And in the second theorem, I stated that their corresponding angles are the same.
This means angle A = angle E , angle B = angle F, angle D= angle H and lastly angle C =
angle G.
NOTE: In any figure, if corresponding angles are equal but corresponding sides are not
proportional then we cannot call that figure similar.
A
B
In the above shape, we cannot say Parallelogram A is similar to Parallelogram B, this is because the
corresponding angles are the same ( all are right angles) but sides are not proportional. And
therefore this parallelogram is not similar.
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SIMILAR TRIANGLES
Just like polygons, in Triangles if two angles of the triangle are the same, as the two angles in
another triangle then, the two triangles are therefore similar.
This above image was extracted from www.cliffsnotes.com .
In this image, we can see that angle B is equal to angle E. In figure ABC, if B is 100 degrees, C is 20
degrees then A obviously has to be 60 degrees, this is because:
Angle A + Angle B + Angle C = 180 degrees
Also in figure, DEF, we can easily claim that F is 20 degrees, the same theorem used above, can be
used in this shape, and we can prove that F is 20 degrees as:
Angle D + Angle E + Angle F = 180 degrees
All triangles add up to 180 degrees.
Let us use the same example as the above.
In TRIANGLE ABC AB= 3cm BC= 5 cm AC= 4 cm
In TRIANGLE DEF DE= 1.5cm EF= 2.5cm DF= 2 cm
Taking in the Ratio, AB/DE = BC/EF = AC/DF
Let us assume that we haven’t got DF, therefore DF= x , if we take a common ratio, then in order to
find x, we can take AB/DE = AC/x
SOLN: 3cm/1.5cm = 4cm/ x – If we cross multiply, then 3x is equal to (1.5 * 4): Therefore 3x= 6.
X= 6/3 and x which is DF= 2cm, this proves that the ratio analysis I have used is correct.
P
5
3
Q
2
T
3
R
S
Z cm.
In the above question, extracted from Maths Text book. We have to find side Z, in this
figure. We know that PQT ~ PRS. For that reason, the ratios will be equal.
PQ/PR = QT/RS = PT/PS. So in order to get RS, we can use either of the ratios. For that
reason PT/PS = QT/RS
SOLN: PT/PS = QT/RS
5/8 = 3/Z
Cross Multiplication: 5 x Z = 8 x 3
5Z = 24
Z= 24/5 = 4.8cm
Question 3:
C
In question 3, we have to find X cm and W cm. We
know that Triangle ABE ~ Triangle ACD. And therefore:
AB/AC= AE/AD= BE/CD. In order to get w that is BC, we
can find AC as that is in the ratio and subtract 3cm
from AC.
Soln: AB/AC = BE/CD
3/w = 4/6
w= (6 x 3) divided by 4
w
6cm
B
3cm
4
A
X cm
E
2
D
In the figure on the left, such figures are often called Side
Splitters, we can say that AB/DB = AC/DE = BC/BE
however, AB/DB becomes AB-DB/ DB so therefore actually
DB/AD =EB/EC = AC/DE.
Question 1.
In the figure on left, we are assigned to find X, in this
figure, i can say the ABC ~ DBE. Therefore AB-DB /DB =
BC-BE/BE = AC/DE. x/12 = 15/10
10x=180
Therefore, x= 18
PERIMETER AND AREA OF SIMILAR TRIANGLES
In similar triangles, that is corresponding angles are the same and corresponding sides are
proportional then, ratio of corresponding sides = ratio of their perimeter.
So lets take an example that a triangle, ABC with sides 3cm , 6cm and 5 cm respectively and another
triangle DEF, with sides 6 cm , 12 cm, and 10 cm, then the proportion between DEF and ABC is
6/3=12/6=10/5 which gives us 2/1. Therefore the ratio of corresponding sides is 2:1, so according to
my theorem this would also mean that the perimeter ratio will be 2:1. Perimeter of Triangle ABC=
14cm ( 3cm +6cm + 5cm) While Perimeter of Triangle DEF= 28 cm ( 6cm+12cm+ 10cm).
Perimeter of Triangle DEF
-----------------------------------Perimeter of Triangle ABC
28 cm / 14 cm
=
2/1
This theorem proves to us that in a similar triangle, the ratio between their corresponding sides is
the same ratio between its perimeter. Therefore if the ratio between corresponding sides is z:x
then the ratio between their perimeter will also be z:x.
Examples
ABC ~ DEF, so according to the theorem, this means
that:
Perimeter of ABC
Perimeter of DEF
=
AB
DE
24 inches / x (perimeter of DEF) = 6/9
6x = 216
ABC ~ DEF
x= 36
In this figure, GHI ~ JKL, therefore one can prove that
the ratio between the corresponding sides is the same
as the ratio between the figures’ perimeter.
Perimeter of GHI
Perimeter of JKL
=
HG
KJ
24/36 = 6/9
Ratio simplifies to 2/3=2/3.
Well, now we have reached in the areas’ section. We
will first try to find the relationship between the
areas of triangle (right angled) and corresponding
sides.
AREAS:
GHI ~ JKL.
Corresponding Sides ratio = ( GH/JK or any other can
be chosen i.e. GI/JL or HI/KL) Ratio= 6/9 simplified
2/3. Ratio of area = Area of GHI divided by AREA of
KJL, GHI =( ½ (6)(8) = 24) KJL= ( ½ (12) (9)= 54 )
Area of GHI = 24
Area of JKL
54
= 4
9
Corresponding Sides = HG/ KJ= 6/9 = 2/3
4/9 and 2/3 . Relationship (2/3)2
----- Therefore, relationship of the Area of Similar
Triangle is such that, if ratio of corresponding sides
are a:b then relationship between area of the
triangle will be a2 : b2
Examples:
4cm2
36cm2
6 cm
y cm
Now, 4/36 is the ratio between the area of the two triangle. a2 : B2 this means we have to square
root. By square rooting 4/36 it is simplified to 2/6. ( 2 x 2) / ( 6 x 6). Ratio is 2/6 therefore solution
will be as follow:
2/6 = 6/y
2y = 36
Y= 18 cm.
2
Eg 2: The areas of two similar triangles are 45cm and 80 cm
Find the perimeter of each triangle.
2.
In the figure below, the first triangle has an area of
45cm square while the second triangle has an area of 80
The sum of their perimeters is 35cm.
cm square. First of all we’ll try to find the ratio between
these triangles.
Soln:
45cm2
80cm
2
1st Part: 45/80 if simplified by 5, then it results 9/16:
therefore, by square rooting, ratio will be ¾ (3:4)
2nd Part: If we take sides of the triangles as X, as they
are corresponding. Then the equation will be as follows,
3x + 4x = 35, 7x= 35
Therefore x=5.
Perimeter of triangle 1 will be 3x5= 15cm
Perimeter of triangle 2 willobviously be 20cm, as they
add up to 35 cm, but even then to prove it we can take
4 x (X) which is 5cm, resulting to 20cm.
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VOLUMES OF SIMILAR FIGURES
When three dimensional figures are similar, the shapes are the same but the sizes are often
different. A line has one dimension, and the scale factor is used once, An area has two dimensions
and thus the scale factor is used twice. Lastly volume has three dimensions and the scale factor is
used thrice.
:- this means, in a one dimension figure ratio of corresponding side a:b is corresponding to
perimeter a:b
a:b = a:b
:- In a two dimension figure, ratio of corresponding side a : b is corresponding to area of figure a2 : b2
:- In a three dimension figure, ratio of corresponding side a : b is corresponding to area of figure
a3 :b3.
However, when considering the surface area, if ratio of corresponding sides is “a” then surface area
will be “a” square.
3cm
6cm
6cm
12
12cm
6cm
3cm
6cm
5cm
10cm
In the following shape we can find that the ratio between the corresponding sides first, and then we
can find the surface area of the Cubes, and then we can work out the ratio.
Soln: 3cm/6cm = 5cm/10cm= 6cm/12cm
Therefore, ratio between sides is 1:2 .
Surface area of figure one is:
Surface area of figure two is:
2(LW) + 2(LH) + 2 ( HW)
2(LW) + 2(LH)+ 2(HW)
= 2 (15) + 2(30) + 2(18)
= 2(60) + 2 (120) + 2(72)
= 126cm square.
= 504 cm2
126/504= By simplifying this,
¼, if we square root one divided by 4 or ( 0.25) answer will be 1/2 . This Proves that the surface area
of a figure corresponds with its sides. If ratio of sides are a:b then ratio of surface area will be a2 : b2
Now using the same figure as above, we will calculate the volume. Volume of a cuboid is L x W x H(
length x width x height). However, as we have already got the ratios between their sides that is 1:2.
Soln:
If we compare the two volumes, in ratio. It gives us 90/720 this is 1/8. While our ratio between
corresponding sides is ½. Therefore, the relationship between the ratio of corresponding sides and
its volume is as follows:
If ratio of corresponding sides is a:b then ratio between volume will be a3 : b3
= 3cm x 6cm x 5cm is equal to
90 cubic centimetres.
= 6cm x 12cm x 10cm is equal to
720 cubic centimetres.
Questions:
Two similar jugs have heights of 4cm and 6cm respectively. If the capacity of the smaller jug is
50cm3. Find the capacity of the larger jug.
Soln. First of all, we have to find the ratio between jug A and jug B
a:b
= 4 : 6 which is simplified to 2 : 3
In this question we can use cross multiplication, so thus, if ratio is 2:3 then 50cm/x.
2x = 150cm,
X=75 cm3
Question 2:
Two similar spheres are made of the same material having weights of 32kg and 108kg respectively. If the radius of the
larger sphere is 9cm, find the radis ouf the smaller sphere.
A:b = a3 : b3 , therefore 32/108= 8/27, if we cube root this, our answer will be 2/3. If the larger sphere is 9 cm, that mean
2/3 x 9 = x
Therefore x= 6 cm.