Download Two geometrical objects are called similar if they both have the

Two geometrical objects are called similar if they both have the same shape.
More precisely, one is congruent to the result of an enlarging or shrinking of the
other. Corresponding sides of similar polygons are in proportion, and
corresponding angles of similar polygons have the same measure. One can be
obtained from the other by uniformly "stretching" the same amount on all
directions, possibly with additional rotation and reflection, i.e., either have the
same shape, or one has the same shape as the mirror image of the other. For
example, all circles are similar to each other, all squares are similar to each other.
SIMILAR POLYGONS
Similar Polygons: Two polygons containing vertices that can be paired so that the
corresponding angles are congruent and the corresponding sides are in
proportion.
Example:
If you are given two similar polygons ABCD and JKLM, we can write
ABCD ~ JKLM which is read as "polygon ABCD is similar to polygon JKLM". The
wavy line symbol means 'similar to'.
PROPERTIES OF SIMILAR POLYGONS
1. Corresponding angles are the same
2. Corresponding sides are all in the same proportion
3. Corresponding diagonals are in the same proportion. In each polygon the
corresponding diagonals are in the same proportion. Their ratio is the
same as the ratio of the sides.
4. Area ratio: the ratio of the areas of the two polygons is the square of the
ratio of the sides. So if the sides are in ratio 3:1 then the area will be in the
ratio 9:1.
Although it could be possible that the quadrilaterals are not similar to one
another as maybe their corresponding sides are not equal or their corresponding
angles.
Scale factor: The ratio of the lengths of two corresponding sides of similar
polygons
Example 1:
/ A is congruent to / E
/ B is congruent to / F
/ C is congruent to / G
/ D is congruent to / H
AB/EF = BC/FG= CD/GH = AD/EH
The scale factor of polygon ABCD to polygon EFGH is 10/20 or ½
REFLECTION AND ROTATION
Polygons can still be similar even if one of them is rotated, and/or mirror image
of the other. In the figure below, all three polygons are similar.
Starting with the original polygon on the left, the center polygon is rotated
clockwise 90°, the right one is flipped vertically. Mark the angles in the two right
polygons that correspond with the angle P on the left. This is illustrated in more
depth for triangles in "Similar Triangles", but is true for all similar polygons, not
just triangles.
TYPE OF QUESTIONS IN WHICH POLYGONS COULD HELP…
1. Problem: Find the value of x, y, and the measure of angle P.
Solution: To find the value of x and y, Write proportions involving
corresponding sides. Then use cross products to solve.
4/6 = x/9
6x = 36
x=6
4/6 = 7/y
4y = 42
y = 10.5
To find angle P, note that angle P and angle S are corresponding angles.
By definition of similar polygons, angle P = angle S = 86o.
SIMILAR TRIANGLES
The triangle has a couple of special rules dealing with similarity. They are
outlined below.
1. Angle-Angle Similarity - If two angles of one triangle are congruent to two
angles of another triangle, then the triangles are similar.
Example: Prove triangle ABE is similar to triangle CDE.
Solution: Angle A and angle C are congruent (this Information is given in the
figure).
Angle AEB and angle CED are congruent because vertical angles are
congruent.
Triangle ABE and triangle CDE are similar by Angle-Angle.
2. Side-Side-Side Similarity - If all pairs of corresponding sides of two triangles
are proportional, then the triangles are similar.
3. Side-Angle-Side Similarity - If one angle of a triangle is congruent to one
angle of another triangle and the sides that include those angles are proportional,
then the two triangles are similar.
Example: Are the triangles shown in the figure similar?
Solution: Find the ratios of the corresponding sides.
UV/KL=9/12 VW/LM=15/20
The sides that include angle V and angle L are proportional.
Angle V and angle L are congruent (the information is given in the figure).
Triangle UVW and triangle KLM are similar by Side-Angle-Side.
PARALLEL LINES AND TRIANGLES
You can create similar triangles by drawing a segment parallel to one side of a
triangle in the triangle. This is useful when you have to find the value of a
triangle's side.
If a segment is parallel to one side of a triangle and intersects the other sides in
two points, then the triangle formed is similar to the original triangle. Also,
when you put a parallel line in a triangle, the sides are divided proportionally.
Example: Find PT and PR
Solution: 4/7=x/12
because the sides are divided proportionally when you draw a parallel line to
another side.
7x = 48 Cross products
x = 48/7
PT = 48/7
PR = 12 + 48/7 = 132/7
PERIMETRES AND AREAS OF SIMILAR TRIANGL
Perimeters of similar triangles are in the same ratio as their corresponding sides and this
ratio is called the scale factor.
Example: There are two similar triangles.  LMN and  PQR.
Seg LM/ segPQ = seg MN/seg QR= Seg NL/Seg RP= 1.33
This ratio is called the scale factor.
Perimeter of  LMN = 8 + 7 + 10 = 25
Perimeter of  PQR = 6 + 5.25 + 7.5 = 18.75
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 corresponding sides, i.e. the square of the scale
factor.
ABC   PQR
Area of triangle ABC/Area of triangle PQR= line (AB)2/line (PQ)2= line (BC)2/Line
(QR)2=Line(CA)2/Line (RP)2.
Prove Check;
Draw perpendicular from A and P to meet seg.BC and seg.QR at D and S respectively.
Since  ABC ~  PQR
Line AB/ line PQ= line BC/line QR= line CA/Line RP . . . . . . . 1
also  B   Q
In  ABD and  PQS
also  B   Q and  ADB   PSQ
  ABD   PQS by Angle to Angle method.
Line AB/PQ= line BD/ Line QS= Line DA/line SP………2
= Line BC/ Line QR x Line BC/ line QR
= line BC/ Line (QR)2 = Line AB/ Line (PQ)2
= line AC/ Line (PR)2
=Line BC/ Line (PR)2 => (Scale Factor)
Thus the areas of two similar triangles are in the same ratio as the square of their scale
factors.
2) Areas of two similar triangles are 144 sq.cm. and 81 sq.cm. If one side of the first
triangle is 6 cm then find the corresponding side of the second triangle.
Solution:
= 4.5 c.m.
3) The corresponding sides of two similar triangles are 4 cm and 6 cm. Find the ratio of
the areas of the triangles.
Solution:
3D shapes
Three-dimensional figure is a figure that has depth in addition to width and
height. Everyday objects such as a tennis ball, a box, a bicycle, and a redwood
tree are all examples of space figures. Some common simple space figures
include cubes, spheres, cylinders, prisms, cones, and pyramids. A space figure
having all flat faces is called a polyhedron. A cube and a pyramid are both
polyhedrons; a sphere, cylinder, and cone are not.
VOLUMES
Ratio of volumes of similar figures; if the ratio of corresponding sides of figure
A and B is a: b, then the ratio of their volume is a3: b3.
The volumes of two similar solids are proportional in the ratio k3 when k is the
line ratio of the solids.
Example:
B
A
5
12
4
3
Volume of A = 5x3x4
=60
Ratio of volumes: 480/60
:8
= 23
Therefore:
Ratio of volume= (ratio of sides)3
5555
10
Volume of B= 12x10x4
=480
4
SURFACE AREAS
Ratio of surface areas of similar figures; if the ratio of corresponding sides of
figure A and B is a: b, then the ratio of their surface area is a2: b2.
Example:
B
A
4
20
2
6
Scale factor is 1:5
30
10
Surface area of A= 2(lw)+2(lh)+2(wh)
= 48+24+16
= 88
Surface area of B=2(lw)+2(lh)+2(wh)
= 1200+600+400
=2200
Therefore ratio of surface area = (ratio of sides) 2