When We Talk About Similarities, We Can Easily Relate It to Our Daily Lives. Lets Take

SIMILARITIES ( )

When we talk about similarities, we can easily relate it to our daily lives. Lets take an example of cars of the same brand, Nissan Patrol and Nissan safari. According the photo below, they look similar.

Similarities in triangles.

P A

B C Q R

We now may say that triangle ABC and the triangle PQR are similar because they both have the same shape, although they are not congruent( discussed later).

Triangle ABC and triangle PQR would appear similar if we looked at it through a magnifying glass.

Thus we may state that two similar triangles have the three angles of one triangle respectively equal to the three angles of the other triangle, but the corresponding sides are NOT equal.

It follows that similar trangles are equiangular and hve theri corresponding sides propotional. Equiangular means the angles of the two triangles are equal.

Example:

A P

find

-PQ

-PR

X y

3cm

B 4cm C Q 8cm R

PQ AB=QRBC PRAC= QRBC

X3=84 Y5 = 84

X=8×3÷4 Y= 8×5÷4

X=6CM Y=10CM

Triangle ABC is similar to triangle PQR.

Why?

This is because the trinagles ABC and PQR are equiangular hence their corresponding sides are proportional.

TYPES OF SIMILARITIES OF TRIANGLES.(THEORUMS)

1. Side angle side triangles. (SAS)

In, this type of similarities, the triangles are similar when two sides are corresponding with the angle included, so ‘if one angle of a triangle is congruent to one angle of another triangle and the sides that include those angles are proportonal, then two triangles are similar.

For the triangles below, <YXZ and <PQR are equal and that is why these triangles are called to be similar.

c d

P R

X b Z

Example:

A X

6CM 4CM

B 10CM C Y x Z

FIND YZ

ACXZ=BCYZ

64=10X

X=4×10÷6 X=6.7CM

2.Angle angle side triangle (AAS)

In this type of similarities, the triangles are similar when they have two corresponding angles which are equal and has one side.

B a

Examplesno.1

B Q

Find the side labelled x.

∆ ABC is similar to ∆ QRP

ABQR=BCRP

X16=126

X=16×12÷6

X=32CM

EXAMPLE NO.2 Q

y 6CM

16CM

3CM

B 6CM C

FIND X AND Y

ABQR=BCPR CAPQ=BCPR

16Y=63 Y=16×3÷6

X6=63 Y=8CM

X=6×6÷3 X=12CM

3. side side side triangles (SSS)

In this type of similarities, the triangles are similar when three sides of the triangle are corespnding. So, ‘if all pairs of corresponding sides of two triangles are proportional, then triangles are similar.

NOTE: In side side side triangles, their corresponding sides are manified by a certain factor ,K.

a b ka kb

X Z P R

Thus these triangles above are similar because of their corresponding sides.

EXTENSION OF THIS CONCEPT

D X

4CM 3.8CM 6CM 5.7CM

E 5CM F

Y 7.5CM Z

DEXY= 46= 23

XFYZ=57.5= 23 THEREFORE:DEXY=XFYZ=DFXZ= 23

DFXZ= 3.85.7= 23

HENCE, <D=<X, <E=<Y, <F=<Z MEANS ANGLE

THUS, BOTH THE TRIANGLES ABOVE ARE SIMILAR.

EXAMPLE:

Y P

2CM 3.8CM 9CM 13CM

X 4CM Z Q x R

XYQP=XZQR

29= 4QR X=4×9÷2 X=12CM

EXERCISE (1) ON SIMILARITY OF TRIANGLES

FIND THE UNKNOWN VALUE.

m 6CM

M N

B C

18CM 9CM

x 6CM

4CM

B 6CM C

9cm

12cm 15cm

9CM

3cm

5cm

10cm

6cm

4cm

6cm

2cm

7. A tree of height 4m casts a shadow of lenght 6.5m. find the height of the house casting a shadow of 26m long.

8.A diagram shows theside view ofa swimming pool being filled with water. Calculate the length x.

15m

3.7m

1.2m

1.8m

9. A tree casts a 23cm shadow. At the same time, a 6cm peson casts a shadow 10cm long. What is the height of the tree?

SIMILAR POLYGONS

Suppose à triangular figure made of cardboard is held infront of a light bulb. If the triangle is held parallel to a screen, a shadow of the triangle apperas on the screen. (This concept Works with all polygons). The shadow on the screen will be larger than the cardboard triangle, but i twill be of the same shape.

Thus the cardboard triangle and its shadow on the screen are said to be similar. Note that similar figures have the same shape, but not necessarily the same size.

Definition: similar polygons are polygons for which all the corresponding angles are congruent and all the corresponding sides are proportional.

To say that corresponding sides are similar polygons are proportional means that the ratios of their measures are equal.

Lets look at these pairs of polygons that are not similar. 3cm

2cm 3cm

3cm

The rhombus and the square are NOT similar. All the corresponding sides are propotional, but it is not true that all corresponding angles are congruent.

8cm

5cm

4cm 4cm

8cm 4cm 4cm

The two rectangles are not similar. All the

corresponding angles are congruent, but it is not true

that all the corresponding sides are proporonal.

NOTE: For polygons to be similar, it requires that all corresponding sides are proportional and all the corresponding angle must be conguent.

Example:1.

Find the value of x and y and measure <p.

4 6 y

S V

x P R

86°

∆STV ~∆PQR

46=X9

46=7Y Y=6×7÷4

X= 4×9÷6

X=6

To find angle p, note that <s and <p are corresponding angles thus <p=<s <p=60°

2. Z R

X Y P Q 1. <x=<?

2.<p=<?

3.xzqr=zy?

Solutions: 1) Q 2) Y 3) RP

EXERCISE(2) ON SIMILAR POLYGONS.

1. An airplane has a length of 24m and a wingspan of 32m. A scale model is made with a wingspan of 12cm. Calculate the model’s length.

2. A tree casts a shadow 24m long on horizontal ground. A vertical post 3m high casts a shadow 4 m long at the same time of the day. Calculate the height of the tree.

3. A building 25m high and 20m wide is shown on a tv program. If its image on the screen is 12cm wide, what is the height of the image on the screen?

4. A swimming pool and the concrete walk around it form similar rectangles. The swimming pool is 12 feet wide by 30 feet long. The width of the rectangle formed by the outer edge of the walk is 18 feet. What is the length of the outer rectangle?

5. A photographic slide is 34mm wide and 22mm high. Projected on a screen, the image of the slide is 85cm wide. How high is the image?

EXAMPLE 2:

A tree casts a 23m shadow. At the same time, a 6m person casts a shadow 10m long. What is the height of the tree?

N A E R

10M 23M

MART=ANER

6X=1023

X=13.8m

AREAS AND PERIMETER OF SIMILAR POLYGONS(TRIANGLE)

PERIMETER OF SIMILAR TRIANGLES

a kb

perimeter1=(a+b)2

perimeter2=(ka+kb)2

P2P1=2K(A+B)2(A+B)

P2P1=K

A X

kr kq

B C Y Z

P kp

P1=r+q+p

P2=kr+kq+kp p2p1=k(p+q+r)p+Q+r =k

The two triangles above are similar, the ratio of the coresponding sides being k.

Perimeter of triangle ABC= p+q+r

Perimeter of triangle XYZ=kp+kq+kr

Perimeter of ∆XYZperimeterof ∆ABC=k(p+q+r)(p+q+r)

This illustrates to us an important general rule for all similar shapes.

If two figures are similar and the ratio of their corresponding sides is k, then the ratio of the perimeter is k.

Example:

The ratio of the perimeters of triangles ABC to triangle PQR is 1:3. Find the side of ∆ABC is two of its sides are 9cm and 15cm and the perimeter of triangle PQR is 90cm given that ∆ABC ~ ∆PQR?

P190 =13

90÷3=30CM

9+15+X=30

X=6CM.

P A

6CM 9CM 18CM 27CM

Q 15CM R B C

PERIMETER= 30CM 45CM PERIMETER=90CM

AREAS OF SIMILAR TRIANGLES

A B W X

D a C Z ka Y

The two rectangles are similar, the ratio of the coresponding sides being k.

AREA OF ABCD= ab

AREA OF WXYZ= ka×kb=k2ab

AREA WXYZAREA ABCD=k2abab

=k2

This illustrates an important general rule for all similar shapes:

If two figures are similar and the ratio of the corresponding sides is k, then the ratio of their areas is k2.

NOTE: k is sometimes called the linear scale factor.

Examples:

1. Two similar trinagles have areas of 18cm2 and 32cm2 respectively. If the base of the smaller triangle is 6cm, find the base of the larger triangle.

6cm

X cm

Ratios of areas (k2)=3218=169

Ratio of corresponding side= √169 =4 base=43×6=8cm.

2. A

XY is parallel to BC.

ABAX=32

If the area of ∆AXY is 4cm2, find the area of triangle ABC.

The triangles ABC and AXY are similar.

Ratio of corresponding sides (k)= 32

Ratio of areas (k2)= 94

Area of ABC= 94×4=9cm2

EXERCISE(3) ON AREA AND PERIMETER OF SIMILAR POLYGONS.

Find the unknown area.

3cm 6cm

5CM 3CM

20CM 10CM

8CM 12CM

FIND THE LENGTHS OF THE SIMILAR SHAPES.

4CM x

D E

B C

Given that AD= 3cm , AB=5cm and the area of triangle ADE=6cm2.

Find:

a) The area of∆ABC

b) The area of DECB

6. What is the linear scale factor of perimeter of similar triangles?

7. What happens to the perimeter, if the sides have been doubled?

CONGRUENCE (≅)

Two figures are congruent if they have the same size and shape. When two triangles are congruent, you can fit one on top of the other so that the two figures match exactly. The sides and angles that match are corresponding parts. The vertices that match are corresponding vertices.

A E

B C D

Corresponding vertices corresponding parts

A corresponds to D A corresponds to <D AB corresponds to DE

B corresponds to E <B corresponds to <E BC corresponds to EF

C corresponds to F <C corresponds to <F CA corresponds to FD

When you write a congruence statement, list the corresponding vertices in the same order. Here are all your ways this can be done for the triangles above. Note that all these congruence statements imply the same correspondence.

∆ABC≅∆DEF ∆CAB≅∆FDE

∆ACB≅∆DFE ∆CBA≅∆FED

∆BAC≅∆EDF

∆BCA≅∆EFD

Definition: Congruent triangles are triangles whose vertices can be made to correspond in such a way that the corresponding parts of the triangle are congruent.

Every triangle has got six parts- three sides and three angles. By the definition, if two triangles are congruent, then the six pairs of corresponding parts are congruent.

Also, if the vertices can be matched so that all six pairs of corresponding parts are congruent, then the triangles are also congruent.

CONGRUENCE THEORUMS

1. SIDE ANGLE SIDE (SAS)

Suppose 2 sticks are attached so that you can vary the angle formed at A. set an angle of 60°. There is now only one way to complete the triangle. The distance from B to C has been fixed.

Triangle ABC so formed is congruent to triangle JKL. Both the triangles have 4 and 6cm sides that form a 60°

4CM

A C J

6CM

6CM K

The angle formed by the two sides of a triangle is said to be included by those two sides. In both triangles, the angle is included by the sides measured 4 and 6.

So, ‘If two sides and the included angle of one triangle are congruent to the two sides and the included angle of a second triangle, then the two triangles are congruent.’

In thes situations, we cannot use SAS.

5 5

This cannot use SAS because the two pairs of sides are congruent but the included angle are not congruent.

5 3 5 3

This cannot use SAS because two sides of one triangle are congruent to two sides of the other triangle. However, the congruent angles are not included angles for those congruent sides.

Example: P S

R 5 Q V 5 T

100°

IS ∆PQR≅∆STV BY SAS? EXPLAIN

PQ≅ST

<QQR=<TTV

∆PQR≅∆STV by SAS.

2. SIDE SIDE SIDE (SSS)

Suppose you form a quadrilateral from four stiks. You will able to change the measures of the angle without disconnecting the sticks. The figure forme dis not rigid. Howeer, triangles are rigid figures.

So this theorum states ‘ If three sides of one triangle are congruent to the three sides of another triangle, then the two triangles are congruent.

Example:

Y Q

SHOW THAT ∆QYN≅∆QYP

QN≅QP

YN≅YP

YQ≅YQ

THUS ∆QYN≅∆QYP BY SSS.

3. ANGLE SIDE ANGLE (ASA)

The side that is a common side of two angles of a triangle is said to be included by those angles. In the figure below, the 5cm side is included by the 30° and 70° angles.

Example: B

P 5 D

A 5

SHOW THAT ∆BAP≅∆CDP

<AAP=<DDP

<BPA≅<CDP

∆BPA≅∆CDP BY ASA

4. ANGLE ANGLE SIDE (AAS)

Consider these two triangles. The parts with the given measures form a combination AAS- that is two angles and a side not included by those angles. Are they congruent?

AAS is a theorum and not a postulate.

Thus this theorum states ‘ If two angles and a nonincluded side of one triangle are congruent to two angles and the corresponding nonincluded side of another riangle, then the two triangles are congruent.

Example:

For each of the congruent triangles below, write ASA or AAS as the reason for the congruence.

A F