Download similar poly similar polygons olygons

Prepared by: Alia. Ayub.
Ayub. 11B
SIMILAR POLYGONS
What does the word similar mean?
It means not exactly the
he same but pretty close.
Congruent polygons
In order to be congruent, two figures must be the same size and same shape
shape.. Congruent
figures are exact replicas of each other. They have the same shape and the same size.
≅
Now consider figures that have the same shape but not the same size. Such
Such figures look 'similar' but
in essence are simply proportionate to each other
What are similar polygons?
Similar polygons are ones that have the same shape, but their sizes may be different.
~ is the symbol that means “similar.” It is a portion of the “is congruent to” sign
Two or more polygons are similar only if:
The ratios of all pairs of corresponding sides are equal
All pairs of corresponding angles are equal
Consider the polygons ABCD and LMNO in the figure 1
Figure 1
≅
Their corresponding angles are equal but their sides are not proportional. Hence they are
not similar.
Now the sides may be proportional but the angles may not be congruent. For instance we
have polygons like PQRS and HIJK (figure 2)
Figure 2.Again
Again they are not similar.
Thus to be similar polygons must satisfy both, the condition of congruent angles and that of
proportionate sides. Figure 3 shows some similar polygons.
Figure 3
Sizes:
Although the size of the two shapes can be different, the sizes of the two shapes
must differ by a factor
In this case, the factor is x 2
These dodecagons differ by a factor of 6.
Enlargements:
•
When you have a photograph enlarged, you make a similar photograph.
Reduction:
A photograph can also be shrunk to produce a slide
Similar Triangles
Two triangles are similar if their corresponding angles are congruent and their
corresponding sides are proportional. It is however not essential to prove all 3 angles of
one triangle congruent to the other, or for that matter all three proportional to the other
sides. Out of these if some particular conditions get satisfied the rest automatically get
satisfied. Those particular conditions would be sufficient to ensure similarity. These are the
two conditions: 1) In 2 triangles if the corresponding angles are congruent, their corresponding sides are
equal.
2) If the sides of 2 triangles are proportional then the corresponding angles are congruent.
The following mentioned below, are tests used to show that two triangles are similar.
1)Postulate
1)Postulate 17 (Angle Angle similarity Postulate):
Postulate): If two angles of one triangle are congruent
congruent
with the corresponding two angles of another triangle, the two triangles are similar.
similar.
HOW? The sum of all three angles of a triangle is 180º. Therefore if two angles are
congruent the third is automatically congruent. Therefore the sufficient condition requires
only two angles to be congruent.
2)S
2)Side Angle side Test: If two sides of one triangle are proportionate to the two
corresponding sides of the second triangle and the angles between the two sides of each
triangle are equal the two triangles are similar.
HOW? If two sides of a triangle are equal, the angles opposite the sides have equal
measures.
3)S
3)Side Side Side Test: If the three sides of one triangle are proportional to the three
corresponding
corresponding sides of another triangle,
triangle, then two triangles are similar.
ABC and
DEF have a one to one correspondence such that
AB =6 DE= 3
AC = 8 DF = 4
BC= 10
EF = 5.
Are the two triangles similar ? If so, why ?
Solution: AB/DE = AC/DF= BC/EF = 2/1. Yes. The two triangles are similar because their
corresponding sides are proportionate. And the measure of their corresponding lenths
varies by a factor, 2.
Example 1 : Are the triangles shown in the figure
similar?
Solution:
Find the ratios of the corresponding sides.
XY 9 3 XZ 15 3
-- = -- = - -- = -- = TS 12 4 SU 20 4
The sides that include angle X and angle S are proportional.
Angle X and angle S are congruent (the information is given in the figure).
Triangle YXZ and triangle TSU are similar by SideSide-AngleAngle-Side.
Side
Properties of Similar triangles
Perimeters of similar triangles:
Theorem 60: Perimeters of similar triangles are in the same ratio as their corresponding
sides and this ratio is called the scale factor. If the two triangles have a scale factor of a:
a: b,
b,
then the ratio of their perimeters is a:
a: b aswell.
Example 1:
In figure 3 there are two similar triangles . LMN and PQR.
Figure 3
LM/PQ=MN/QR=LN/PR= 1.33
The scale factor is therefore 1.33
1.33
Perimeter of LMN = 8cm + 7cm + 10cm = 25cm
Perimeter of PQR = 6cm + 5.25cm + 7.5cm = 18.75cm
Perimeter of LMN/ Perimeter of PQR= 25cm/18.75cm = 1.33
Thus, the perimeters of two similar triangles are in the ratio of their scale factor.
Areas of similar triangles:
The ratio of the areas of two similar triangles is equal to the ratio of the squares of the
i.e. the square of the scale factor.
corresponding sides,
ABC ~ PQR
Ratio of corresponding sides (k) = 6/4=3/2
Area of
ABC = 1/2ab SinC
½ ×6×3 SinC
=Area of ABC/Area of
Area of PQR = 1/2pq SinR
½ × 4 ×2 SinQ
PQR =18/8 = 9/4= k²
If the two figures are similar and the ratio of their corresponding sides is k, then the ratio of
their areas will be k²
Example 1
Areas of two similar triangles are 144 cm². and 81cm ². If one side of the first triangle is 6
cm then find the corresponding side of the second triangle.
Solution :
= 4.5 c.m
Example 2
The side of an equilateral triangle ABC is 5 cm. Find the length of the side of another
equilateral
PQR whose area is four times area of ABC.
Solution:
Since both triangles are equilateral they are similar.
Example 3
The corresponding sides of two similar triangles are 4 cm and 6 cm. Find the ratio of the
areas of the triangles.
Solution:
Example 4
ABC ~ PQR such that AB : PQ is 8 : 6. If area of
smaller triangle?
Solution:
ABC is 48cm² what is the area of the
3D Figures
For 3D shapes such as Cubes, cuboids, cylinders etc, to be similar, the following conditions
apply:
1} The shape must be the same. Eg/ the two shapes must be only cubes or only cylinders, but
not a combination of a cube and a cylinder.
2}Their corresponding sides must be in proportion
3}All
3}All pairs
pairs of corresponding angles must be equal
Surface area:
S.A = 2lb + 2lh + 2hb
S.A of
A = 2 (4cm×3cm) + 2(4cm×2cm) +2(3cm×2cm) = 52cm²
S.A of
B = 2(12cm×9cm) + 2(12cm×6cm) +2(9cm×6cm) = 468 cm²
Ratio of corresponding sides(k) :
k = 12cm/4cm = 9cm/3cm = 6cm/2cm = 3
Ratio of S.A’s(x)
x = 468 cm²/52cm² = 9
x= k²
x= k². Therefore, given two similar 3D shapes, if the ratio of corresponding sides is k, then
the ratio of their surface areas is k²
Example 1:
1) The ratio of the surface area of two spheres is 3:1. If the surface area of the larger sphere is 540 cm².
Find the surface area of the smaller one.
Solution:
S.A of larger
/ S.A of smaller
Ratio of larger
/ Ratio of smaller
= 540 cm²/ X
= 3/1
540/X= 3/1 (cross product multiplication)
Therefore X = 180 cm²
Example 2:
2 )Water is put into two similar cylinders . The larger one has a surface area of 256cm2 and the ratio of
the sides of the two cylinders is 6:3. Find the surface area of the second cylinder.
Solution:
As we know, the ratio of the surface areas is the square of the ratio of the sides.
Ratio of sides : 6/3
Ratio of surface areas : 256 /Y
Therefore, {6/3}² = 256/Y
36/9 = 256/Y ( cross product multiplication)
Y = 64 cm
To prove the theory we can check: (Ratio of sides)² should be equal to Ratio of surface areas.
(Ratio of sides)² : Ratio of surface areas = { 6/3}² : 256/64
36/9 :4
4:4. The theory is therefore proved right.
Volume:
Volume = Base area × height
Volume of
A = (4cm×3cm)×2cm = 24cm³
Volume of
B = (12cm×9cm) ×6cm = 648cm³
Ratio of corresponding sides (k) :
k = 12cm/4cm = 9cm/3cm = 6cm/2cm = 3
Ratio of Volume’s(y)
y = 648 cm³/24cm³ = 27
y= k². Therefore, given two similar 3D shapes, if the ratio of corresponding sides is k, then
the ratio of their volumes is k³
Example 1:
Two similar cylinders have a height of 2 cm and 4cm respectively. If the volume of the
smaller cylinder is 20 cm², find the volume of the larger cylinder?
1. {4/2}³ = x/20
2. 64/8 = x/20
3. 64 ×20
8
x= 160 cm ³
Example 2:
Two similar cones have a surface area in the ratio 4:16. Find the ratio of :
a) their lengths
As we know, the surface areas of similar 3D figures are proportionate to the square of the
ratio of corresponding lengths.
Therefore, let the sides be x and y rexpectively.
4/16 = {x/y}²
4/16 =2/4. The lengths of the two similar cones are 2cm and 4cm respectively.
b) their volumes
As we know, the volumes of similar 3D figures are proportionate to the cube of the ratio of
corresponding lengths.
Ratio of sides : 2/4
Ratio of volumes : {2/4}³ = 8/64