I remember vividly as an adolescent, I got thrilled and psyched seeing action and sci-fi movies (especially those involving large guns and grenades). The first day I ever heard the word 'polygon' in the class, as introduced by my Maths teacher, Mr. Finicky Spins, I was so elated. Do you know why? It sounded much like POLY-GUNS! I thought to myself, "Spins must have got lots of guns to distribute to us in class". My hopes were indeed dashed when he started drawing shapes although similar but not even close to bullets.
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Team Using Similar Polygons Teachers
Not to worry, nobody is getting shot here, rather you shall go beyond bullet points as the meaning, identification, application, formula as well as examples of similar Polygons would be addressed hereafter. Load, aim, and shoot 'polyguns'.
Similar Polygons can be described as two-dimensional Figures which have the same shape but vary in size.
Their vertices can be paired up so that the corresponding sides have lengths that are in proportion, and all of the pairs of corresponding angles are equal.
These two factors as a matter of fact define their shape.
You find these pairs of sides and angles by simply looking at the shape or by noting the labels of the vertices. The angles are congruent, and the directly proportional relationship of the sides is a consequence of this.
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Polygon-Polygone-Polygun!!! What is a polygon?
A polygon is a plane figure which is comprised of at least 3 straight lines or sides with no less than 3 angles. This means that a circle cannot be a polygon since it is curved.
All congruent polygons are similar as they have the same angles and their sides are equal, and so are in the same proportional relationship to each other. Also, if two regular polygons have the same number of sides then they will always be similar as they have the same angles. An example of this is how all Equilateral Triangles are similar, as they all have the same three angles and thus the same shape. It is clear to see that polygons will remain similar under reflection, translation, and enlargement.
Which among the following is a polygon?
Fig. 1. Identifying polygons.
Solution:
Figure \(i\) is obviously a polygon given the straight lines and having more than 3 sides. Likewise, \(iii\) is a polygon even though there are several sides; it does not matter how many sides so long as there are 3 or more sides. However, \(ii\) is not a polygon since its top right corner has a curve.
Therefore, only figures \(i\) and \(iii\) are polygons.
When making diagrams for similar polygons, careful attention must be given to the labeling of the vertices and the ordering of these labels. They help you to identify which sides of the polygons correspond to each other. The polygons will be named by letters going around the vertices in questions, the first two letters of each will represent the side between these vertices for each polygon and these sides have a proportional relationship. The same is true for all of the other pairs of letters in different positions in the names. For instance, consider the two similar polygons below:
Fig. 2. Diagrams of two similar quadrilaterals.
The similarity of the polygons \(ABCD\) and \(WXYZ\) can be written as \[ABCD \sim WXYZ\] The lengths of the corresponding sides of each polygon have the same proportional relationship to each other: a side on \(WXYZ\) is a constant multiplied by the length of the corresponding side on \(ABCD\):
\[\frac{AB}{WX}=\frac{BC}{XY}=\frac{CD}{YZ}=\frac{DA}{ZW}\]
As previously mentioned, this comes from how the angles at the corresponding vertices on each polygon are the same:
\[\angle A =\angle W ; \angle B =\angle X ; \angle C =\angle Y ; \angle D =\angle Z \]
There are several uses of the concept of similar polygons, and a major application is in finding missing sides and angles.
If it is known that two polygons are similar, this can be used to work out the values for unknown sides and angles. The sides must satisfy the proportional relationship for similar polygons and the angle on a vertex of one polygon must be equal to the angle at the equivalent vertex on the other polygon.
The two quadrilaterals \(ABCD\) and \(WXYZ\) in figure 1 are similar. The length of \(AB\) is \(5\, cm\), the length of \(WX\) is \(7.5\, cm\) and the length of \(CD\) is \(6\, cm\). What is the length of \(YZ\)?
Solution:
We know that the ratios of the corresponding sides on each polygon must be equal:
\[\frac{WX}{AB}=\frac{YZ}{CD}=1.5\]
\[YZ=1.5\times CD\]
knowing that \(CD\) is \(6\, \text{cm}\), hence,
\[YZ=1.5\times 6\, \text{cm}=9\, \text{cm}\]
This question refers again to figure 1. The polygon \(ABCD\) has angles at the vertices \(A\), \(B\) and \(C\) of \(80°\) , \(110°\) and \(105°\) respectively. What is the angle \(Z\) in the polygon \(WXYZ\)?
Solution:
We know that the interior angles of a quadrilateral sum up to \(360°\). This gives the angle at \(D\):
\[\angle D=360°-80°-110°-105°=65°\]
Also, the two polygons are similar and the vertex \(D\) corresponds to the vertex \(Z\) in \(WXYZ\) so they are equal:
\[\angle Z=65°\]
These two previous examples show how the properties of similar Triangles can be used to find missing sides and angles individually. The techniques above can be used together to find other properties of polygons such as their height, perimeter, and area.
Fig. 3. Diagrams showing two similar quadrilaterals. The angles at the vertices \(B\), \(C\), \(F\) and \(G\) are all right angles.
Two similar quadrilaterals are shown above. Find the length of \(FG\) and also find the area of \(EFGH\).
Solution:
We can see from the shape of the quadrilaterals that the side \(DA\) corresponds to \(HE\) and side \(CD\) to \(GH\). This means that the ratios of these pairs are equal:
\[\frac{HE}{DA}=\frac{GH}{CD}=\frac{6}{3}=2\]
This gives:
\[HE=2\times DA\]
Knowing that \(DA\) is \(2\, \text{cm}\), hence,
\[HE=2\times 2\, \text{cm}=4\, \text{cm}\]
The vertex \(D\) corresponds to the vertex \(H\) so their angles must be equal:\[\angle H=\angle D=30°\]
Using simple trigonometry, the length of \(FG\) can be worked out:
\[FG=4\, \text{cm} \times \sin(30°)=2\, \text{cm}\]
To find the area of the quadrilateral, it can be split in two by drawing a perpendicular line up from the base \(GH\) to point \(E\) to form a triangle on the left and a rectangle on the right.
Fig. 4. A diagram showing how the quadrilateral \(EFGH\) can be split up into a triangle and a rectangle by drawing a line down from point \(E\) which is perpendicular to the base \(GH\).
The side \(EH\) is known so the value of \(x\) can be computed:
\[x=4\, \text{cm} \times \cos(30°)=2\times\sqrt{3}\, \text{cm}\]
Note that \[\cos(30°)=\frac{\sqrt{3}}{2}\]
The area of the triangle, \(A_T\), is equal to half the base multiplied by the height, where the height is the length of \(FG\) and the base is equal to \(x\).
\[A_T=\frac{x\times FG}{2}\]
where \(x\) is \(2\sqrt{3}\, \text{cm}\) and \(FG\) is \(2\, \text{cm}\), then,
\[A_T=\frac{2\sqrt{3}\, \text{cm} \times 2\, \text{cm}}{2}=2\sqrt{3}\, \text{cm}^2\]
The area of the rectangle, \(A_R\), can be worked out by finding the length of the base minus \(x\) and then multiplying this quantity by the length of \(FG\). Therefore,
\[A_R=(GH-x)\times FG\]
this gives
\[A_R=(6-2\sqrt{3})\, \text{cm} \times 2\, \text{cm}=(12-4\sqrt{3})\, \text{cm}^2\]
The two areas can then be added to give the total area of the quadrilateral, \(A_Q\):
\[A_Q=A_T+A_R\]
thus,
\[A_Q=2\sqrt{3}\, \text{cm}^2+12-4\sqrt{3}\, \text{cm}^2=(12-2\sqrt{3})\, \text{cm}^2\]
therefore,
\[A_Q=8.53\, \text{cm}^2\]
Most times, it feels easier to have the formula of certain topics so that you can easily apply them when needed in problems. Unlike those topics, there is hardly any formula(s) one can attribute to finding a similar polygon. You are advised to take into consideration two principles - i.e., the proportion of corresponding sides and the size of corresponding angles.
However, while carrying out certain tasks which involve the application of similar polygons, you may be required to form equations.
There can be questions when all of the sides on one out of two similar polygons are unknown. However, if the length of the sides is given by quantities that are related, then the directly proportional relationship between the sides of the polygons can be used to form an equation and find the unknown lengths.
Fig. 5. Two similar Triangles in which the smaller triangle has unknown sides.
Solution:
The diagram above shows the similar triangles \(ABC\) and \(XYZ\). Find the value of \(x\).
The ratio between the corresponding sides of each triangle must be equal. From the names of the triangles (try to look closely at them), side \(XY\) will be proportional to \(AB\) and side \(YZ\) will be proportional to \(BC\):
\[\frac{XY}{AB}=\frac{YZ}{BC}\]
This equation can be rearranged so that, if the values are put in, one side will be completely in terms of \(x\):
\[XY\times BC=AB\times YZ\]
this gives
\[12x=8x+8\]
Bring like terms together and solve for \(x\) and get
\[x=2\]
There can also be cases where all of the sides of two similar polygons are unknown. This can still be solved if all of the sides are in terms of the same unknown quantity. The ratios of the corresponding sides of each polygon can be set equal to each other to form a quadratic equation of this unknown quantity.
This will give two solutions, but it is always important to remember that a length cannot be negative and the solutions which result in this for any of the sides must be ignored.
Take a look at the example below.
Fig. 6. Diagrams showing two similar quadrilaterals with sides of unknown lengths in terms of \(x\).
The diagram above shows two similar quadrilaterals \(ABCD\) and \(PQRS\). Find the value of \(x\).
Solution:
The same method that was used previously can again be applied to this problem. The order of the letters in the names gives us the corresponding sides. Sides \(BC\) and \(QR\) make a pair and so do sides \(CD\) and \(RS\). Equating the ratios for these pairs of sides gives:
\[\frac{BC}{QR}=\frac{CD}{RS}\]
A quadratic equation for \(x\) can be found by multiplying the denominator on one side of the equation by the numerator on the other side and vice versa:
\[BC\times RS=QR\times CD\]
which is
\[2x(x-1)=(x+3)(x+1)\]
\[2x^2-2x=x^2+4x+3\]
giving you the quadratic equation
\[x^2-6x-3=0\]
This equation does not factorize easily, so, you are advised to solve for \(x\) by applying to complete the square method or using the quadratic formula.
In this case, we shall apply the quadratic formula.
Recall that in the quadratic formula, for a quadratic equation in the form
\[ax^2+bx+c=0\]
then,
\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
Therefore
\[x=\frac{6\pm\sqrt{6^2-(4\times 1 \times (-3))}}{2(1)}\]
which gives
\[x=\frac{6\pm\sqrt{48}}{2}\]
hence,
\[x=3-2\sqrt{3}\]
or
\[x=3+2\sqrt{3}\]
The solution with the minus sign must be discarded as it results in sides \(BC\) and \(RS\) being negative, this leaves:
\[x=3+2\sqrt{3}\]
Problems can come in any form about similar polygons, it is advisable to try several question types.
A 4-sided polygon has two identical sides such that one side is twice the other. If its perimeter is \(36\, cm\), and a similar polygon which is a one-third of this polygon is to be cut out, find the length of the smallest side of the new polygon.
Solution:
Do not fret about the word problem.
Interpret the question word for word from the first sentence to the next.
Since the polygon has four sides with a pair identical, you can say let one side be \(a\), while the other could be \(b\). That way the four sides becomes \(a\), \(a\), \(b\) and \(b\).
Another detail is that one side is twice the other. You can interpret this by saying, \(a\) is the bigger side, so that
\[a=2b\]
The next detail says the perimeter is \(36\, cm\). This implies that the perimeter of the polygon, \(P_p\) is given as
\[P_p=a+a+b+b\]
Recall that
\[a=2b\]
and \[P_p=36\, cm\]
Thus
\[36\, \text{cm}=2b+2b+b+b\]
by making \(b\) the subject of the formula, hence,
\[b=6\, \text{cm}\]
That means that the biggest side, \(a\), is
\[a=2\times 6\, \text{cm}=12\, \text{cm}\]
However, a similar polygon is to be cut out which is one-third of the polygon. This means that the sides of this new polygon can be gotten by finding the product of the sides of the old polygon an \(\frac{1}{3}\).
Apparently, we need just the smallest side; thus, because these polygons are similar, the smallest side of the bigger polygon corresponds with the smallest side of the smaller polygon.
Therefore, the smallest side of the new smaller polygon \(S_s\) is
\[S_s=6\, \text{cm} \times \frac{1}{3}\]
Hence,
\[S_s=2\, \text{cm}\]
A hanger is to be made from a triangular copper sheet of height \(12\, \text{cm}\) and base length \(30\, \text{cm}\). If the hanger is formed by cutting a smaller but similar shaped triangular sheet from the original triangle, find the area of the hanger if the internal base length is \(24\, \text{cm}\).
Solution:
Step 1: Make a sketch of the information provided so as to get a clear picture of the problem.
Fig. 7. Using similar polygons in making a hanger.
Step 2: Since you have been asked to find the area of the hanger, you need to find the area of the original triangle, \(A_o\), so that
\[A_o=\frac{1}{2}\times 30\, \text{cm} \times 12\, \text{cm}\]
Recall that the area of a triangle is \(\frac{1}{2} \times base \times height\)
\[A_o=180\, \text{cm}^2\]
Step 3: Find the area of the smaller triangle cut out. This is where you would apply what you have learned about a similar angle. Knowing that both triangles are similar you can find the height of the cut-out triangle. But you would need to know the ratio between both angles. So, you should use the base length of both triangles to determine their ratio.
\[b_s:b_b=\frac{24}{30}\]
where \(b_s\) is the base length of the small triangle (cut-out triangle), and \(b_b\) is the base length of the big triangle.
\[b_s:b_b=\frac{4}{5}\]
Now, you should multiply the height of the big triangle, \(12\, \text{cm}\) by \(\frac{4}{5}\) to get the height of the smaller triangle, \(h_s\):
\[h_s=12\, \text{cm} \times \frac{4}{5}\]
\[h_s=9.6\, \text{cm}\]
Therefore, the area of the small triangle, \(A_s\), is:
\[A_s=\frac{1}{2}\times 24\, \text{cm} \times 9.6\, \text{cm}\]
\[A_s=115.2\, \text{cm}^2\]
Step 4: Find the area of the hanger. You can achieve this by subtracting the area of the smaller triangle, \(A_s\), from the area of the bigger triangle, \(A_o\). Thus the area of the hanger, \(A_h\) is calculated to be:
\[A_h=A_o-A_s\]
\[A_h=180\, \text{cm}^2-115.2\, \text{cm}^2\]
Hence, the area of the hanger is:
\[A_h=64.8\, \text{cm}^2\]
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What are similar polygons?
Similar polygons have the same shape but a different size.
What is an example of polygons being similar?
All equilateral triangles are similar.
How do you work out the length of unknown sides and the values of angles on similar polygons?
You can find the lengths of sides by using the fact that the length of corresponding sides in two polygons are directly proportional to each other and you can find unknown angles by noting that corresponding vertices have the same angle.
How do you identify similar polygons?
Similar polygons have the same number of sides, the same angles at corresponding vertices and the lengths of their corresponding sides have a directly proportional relationship.
How to solve a similar polygon?
You solve similar polygons by verifying that those polygons are of the same shape and proportionate corresponding sides.
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