SIMILAR POLYGONS
Introduction
Let us begin by examining the two words; Similar and Polygons
Similar means to have properties that are alike one another. NB. Similar is not ‘exactly the same.’
A polygon is a closed figure consisting of sides and angles. Different polygons have different sides and angles in terms of their numbers and sizes.
Collectively, similar polygons refer to 2 (or more) figures with characteristics that are alike one another, but not entirely the same. What make them similar are their shapes, but they differ in sizes.
For two shapes to be similar, they must abide by the following criteria:
· Corresponding angles should be equal
· The ratios of pairs of corresponding sides should be equal
The symbol used to denote two (or more) similar figures is “∼”. For example, ∆ABC ∼ ∆DEF.
Similar Quadrilaterals
Example 1 Consider the quadrilaterals ABCD and EFGH below in which ABCD ∼ EFGH:
The above pair of figures can be classified as similar polygons because they abide by the criteria mentioned above, i.e.:
· ∠A = ∠E, ∠B = ∠F, ∠ C= ∠G and ∠D = ∠H
·
Non-similar polygons
Remember, that for polygons to be similar, they must abide by both rules in the criteria. It is possible to have polygons that abide by only one of them. These polygons cannot be considered similar polygons.
Example 2 Consider the following quadrilaterals:
Though the ratios of pairs of corresponding sides are the same, the corresponding angles are different, so the above quadrilaterals are not similar polygons.
i.e.
but 90°≠ 110°, 90° ≠ 70°
Example 3 Consider the following quadrilaterals:
Though the corresponding angles are the same, the ratios of pairs of corresponding sides are different, so the above quadrilaterals are not similar polygons.
i.e. 90° = 90° but 4/3 ≠ 2/2
Question
The above quadrilaterals are similar, calculate the value of x.
Using the formula stated above, we know that for 2 quadrilaterals to be similar polygons, the ratios of corresponding should be the same
i.e. AD/CD = EH/GH
8/6 = 4/X
X = (6*4)/8
X = 3
Similar triangles
Generally, to prove that 2 polygons are similar, you must show that all pairs of corresponding angles are equal and that all ratios of pairs of corresponding sides are equal, but with triangles, this is not necessary.
Postulate 17 (AA similarity Postulate) states that if two angles of one triangle are equal to two angles of another triangle, then the pair of triangles is similar.
Example 1 Consider the triangles below:
∠B = ∠E = 80°
∠A + ∠B + ∠C = 180°
∠C = 180° - 80° - 40°
∠C = 60°
But since ∠F = 60°, ∠C =∠F
By Postulate 17 (AA Postulate), the above triangles are similar; the ratio of the pairs of corresponding sides must be the same.
i.e.
Example 2 Consider the figure below
The figure above consists of two similar triangles. How can we prove they are similar? As mentioned above, for two triangles to be similar, they must have 2 corresponding angles equal to each other.
∠x = ∠y (Because they are opposite angles)
∠A = ∠C = 50°
Or
∠B = ∠D = 60°
(When a transversal meets a pair of parallel lines, the alternate interior angles are equal)
This proves that by Postulate 17 (AA Postulate), the above figure consists of 2 triangles which are similar to one another, (∆AOB ∼ ∆COD).
Examples 3 Consider the following isosceles triangles:
In ∆ABC, AB = AC, and in ∆DEF, DE=DF
Therefore, B=C and E=F (The angles opposite 2 equal sides of a triangle are always equal)
In ∆ABC, A + B + C = 180°
In ∆DEF, D + E + F = 180°
Because B=C and E=F
A + 2B = 180° D + 2E = 180°
30° + 2B = 180° 30° + 2E = 180°
2B = 150° 2E = 150°
B = 75° E = 75°
Therefore, A=D=30° and B=E=C=F=75°. By Postulate 17 (AA Postulate); ∆ABC ∼ ∆DEF
Example 4 Consider the following right angled triangles:
∠B = ∠E = 90°
∠A = ∠D (denoted on the figure)
Therefore, by Postulate 17 (AA Postulate); ∆ABC ∼ ∆DEF
Perimeters and areas
When 2 polygons are similar, the reduced ratio of any two corresponding sides is the scale factor of the similar polygons.
Perimeters
Consider the following triangles:
The reduced factor of all sides will be equal (i.e. 8/4 = 6/3 = 4/2). All of these ratios reduce to 2/1. Therefore it is said that the scale factor of these similar polygons is 2: 1.
The perimeters of ∆ABC and ∆DEF are 18cm and 9cm respectively. By comparing the ratios of the perimeters, you will find that it is also 2: 1. This brings us to our next theorem.
Theorem 60: if two similar polygons have a scale factor of a: b, then the ratio of their perimeters will also be a: b.
Example 1: consider the following similar triangles (NB. Units are in centimetres)
ABC ∼ DEF, find the perimeter of DEF.
By theorem 60;
Areas
Consider the following similar rectangles
The scale factor of these rectangles = 6/4 = 3/2
Area of ABCD = L*W
= 2*4
=8cm2
Area of EFGH = L*W
= 6*3
=18cm2
The ratios of the area = 18/8
= 9/4
9/4 = 32/22
Therefore, when two similar polygons’ corresponding sides have the ratio of a: b, then the ratio of their areas will be a2: b2
This is called theorem 61
Example 1: Consider the following similar triangles:
The reduced factor of the sides =DE / AB
= 6 / 3
= 2
Therefore, EF = 2BC
=8
Area of ∆ABC = ½ (b*h)
=½ * 12
=6 cm2
Area of ∆DEF =½ (b*h)
= ½ * 24
=24 cm2
Ratio of areas = 24/6
= 4/1
4/1= 22/12
Question 1: the areas of two similar rectangles are 12cm2 and 48cm2 respectively. The sum of their perimeters is 42. Find the perimeter of each rectangle.
Ratio of area = 48/12
= 4/1
=a2/b2
Therefore a/b =√4/√1 (theorem 61)
=2/1
Ratio of perimeters = a/ b = 2/ 1 (theorem 60)
2x + x = sum of perimeters
2x + x = 42
3x = 42
x= 14
If x= 14cm, then 2x= 28cm
The perimeters of the two rectangles are 14cm and 28cm respectively.
Question 2: the ratio of the perimeters of two rectangles is in the ratio 5:4. The sum of their areas is 20.5cm2. Find the area of each rectangle
Scale factor of the rectangles is the ratio of the perimeters = a/b = 5/4 (theorem 60)
Therefore, the scale factor of their areas = a2/b2 = 25/16 (theorem 61)
25x + 16x = 20.5cm2
41x = 20.5cm2
x = 0.5cm2
25x= 0.5cm2 *25 = 12.5cm2
16x= 0.5cm2 *16 = 8cm2
The areas of the two rectangles are 12.5cm2 and 8cm2 respectively.
3 Dimensional figures
Similarities also occur with 3-dimensional figures. The conditions are similar to those of 2 dimensional figures:
· Corresponding angles should be equal
· The ratios of pairs of corresponding sides should be equal
In other words, the shape of the two figures should be exactly the same, but their sizes may vary.
Consider the following cube:
The above cubes are similar because the ratios of their lengths, widths and heights are the same.
I.e. 3/2 = 3/2 = 3/2
Question: find the length of the sides marked ‘a’ and ‘b’ in the following cuboids:
3a=12cm (cross product property) 3b=24cm (cross product property)
a=4cm b=8cm
Surface areas
Example 1: Consider the following similar cuboids:
Just like with 2-dimensional figures, the scale factor of any two figures is determined by the ratio of its corresponding sides.
In the above example, the scale factor is 6/3= 4/2= 2/1
Surface area of figure 1 = 2(LW) + 2(LH) + 2(HW)
= 2(6cm2) + 2(3cm2) +2(2cm2)
= 12cm2 + 6cm2 + 4cm2
= 22cm2
Surface area of figure 2 = 2(LW) + 2(LH) + 2(HW)
= 2(24cm2) + 2(12cm2) +2(8cm2)
= 48cm2 + 24cm2 + 16cm2
= 88cm2
Ratio of surface areas = 88/22
= 4/1
But the scale factor = 2/1
Therefore 4/1 =22/12
Just like with 2-dimesional figures, if the scale factor of 2 similar 3-dimensional figures is a: b, then the ratio of its surface area is a2: b2
Example 2: consider the following similar figures as proof for the theory above:
Surface area of figure 1 = 2(LW) + 2(LH) + 2(HW)
= 2(25cm2) + 2(50cm2) +2(50cm2)
= 50cm2 + 100cm2 + 100cm2
= 250cm2
Surface area of figure 1 = 2(LW) + 2(LH) + 2(HW)
= 2(4cm2) + 2(8cm2) +2(8cm2)
= 8cm2 + 16cm2 + 16cm2
= 40cm2
Ratio of surface area = 250/40
=25/4
But scale factor =5/2
Therefore, 25/4 = 52/22
Volumes
Example 1: consider the following similar cubes:
Their scale factor = 3/2
Volume of cube1 = (2cm)3 volume of cube = (3cm)3
=8cm3 = 27cm3
Ratio of their volumes = 27/8
This can be expressed in terms of the scale factor, i.e. 27/8 = 33=23
If the scale factor of a pair of 3-dimensional figures is a: b, then the ratio of their volumes is a3: b3
Example 2: consider the following cuboids:
Scale factor= 6/4 = 2/3
Volume of cuboid1 = LWH
= 3cm*3cm*6cm
= 54cm3
Volume of cuboid2 = LWH
= 2cm*2cm*4cm
= 16cm3
Ratio of volumes = 54/16
= 27/8
But scale factor = 3/2
Therefore, ratio of volumes = 33/23= 27/8
Question 1: The ratio of the surface areas of 2 cubes is 16/9. The sum of their volumes is 91cm3. Find the volume of each cube.
Scale factor (a: b) = √16/√9
= 4/3
Ratio of volumes (a3: b3) = 43/33
= 64/27
64x + 27x = 91cm3
91x =91cm3
x =1cm3
64x=64cm3, 27x=27cm3
Therefore the volumes of the cubes are 64cm3 and 27cm3 respectively.
Question 2: the ratio of the volumes of two cuboids is 27: 8. The sum of their surface areas is 65cm2. Find the surface area of each cuboid.
Ratio of volumes = a3: b3
Therefore, scale factor = 3√27 / 3√8
= 3/2
Ratio of surface area = a2: b2
= 9: 4
9x + 4x = 65cm2
13x = 65cm2
x = 5cm2
Therefore, 9x= 45cm2, 13x= 20cm2
The surface areas of the two similar cuboids are 45cm2 and 20cm2 respectively.