Proportionality of Sides Shapes in Similar Figures
The lengths of the corresponding sides of two figures will be proportional when they are similar. The ratio of their areas is equal to the square of the ratio of their respective sides.
In the above figure, each of the shapes have been changed in size by a scale factor. Since they all have the same shape, they are similar to one another, hence their sides will be proportional.
Area of Similar Shapes Formulas
The formula for the area of similar shapes is given below:
\(\frac{Area~of~figure~A}{Area~of~figure~B}=\left(\frac{a}{b}\right)^2\)
To find the area of two similar shapes we can use the knowledge that the ratio of their areas is equal to the ratio of the square of their respective sides.
Let us take an example; △ABC \(\sim\) △DEF
So their corresponding parts must be proportional
\(\frac{AB}{DE}=\frac{AC}{DF}=\frac{BC}{EF}\)
\(\frac{Area~of~△ABC}{Area~of~△DEF}=\left(\frac{AB}{DE}\right)^2=\left(\frac{AC}{DF}\right)^2=\left(\frac{BC}{EF}\right)^2\)
Two squares are similar. In the figure below, let the sides of square 1 and 2 be s1 and s2 respectively.
\(\frac{Area~of~square~1~(A_1)}{Area~of~square~2~(A_2)}=\left(\frac{s_1}{s_2}\right)^2\)
Solved Examples
Example 1: If the corresponding sides of two similar triangles are in the ratio 4:5, what is the ratio of their areas?
Solution:
We know that in similar figures the ratio of areas is equal to the ratio of the squares of their respective sides.
Let the ratio of sides of two triangles be \(\frac{s_1}{s_2}=\frac{4}{5}\)
\(\frac{Area~of~triangle~1}{Area~of~triangle~2}=\left(\frac{s_1}{s_2}\right)^2\)
\(=\left(\frac{4}{5}\right)^2\)
\(=\frac{16}{25}\)
Therefore, the ratio of area of two triangles is 16 : 25.
Example 2: Find the area of a smaller figure.
Solution:
Since the shapes of both the figures are the same, they are similar. We know that in similar figures the ratio of their areas is equal to the ratio of the squares of their respective sides.
\(\frac{Area~of~bigger~figure}{Area~of~smaller~figure}=\left(\frac{Side~of~bigger~figure}{Side~of~smaller~figure}\right)^2\)
⇒ \(\left(\frac{196}{Area~of~smaller~figure}\right)=\left(\frac{7}{9}\right)^2\) [Substitute the given values]
⇒ \(\left(\frac{196}{Area~of~smaller~figure}\right)=\left(\frac{49}{25}\right)\)
⇒ Area of smaller figure × 49 = 196 × 25 [Cross multiplication]
⇒ Area of smaller figure = \(\frac{196~\times~25}{49}\) [Simplify]
∴ Area of smaller figure = \(100~m^2\)
Therefore, the area of smaller figure is \(100~m^2\)
Example 3: There are two circular pools in the school. The bigger pool has a diameter of 21 m and the area is \(2205m^2\) and the area of the smaller pool is \(1125m^2\). Find the radius of the smaller pool.
Solution:
We know that in similar figures the ratio of their areas is equal to the ratio of the squares of their respective sides. We also know that all circles are similar; hence, the ratio of the areas is the ratio of squares of their respective radii or diameters.
\(\frac{Area~of~bigger~pool}{Area~of~smaller~pool}=\left(\frac{Diameter~of~bigger~pool}{Diameter~of~smaller~pool}\right)^2\)
⇒ \(\frac{2205}{1125}=\left(\frac{21}{Diameter~of~smaller~pool}\right)^2\)
⇒ \(\frac{2205}{1125}=\frac{441}{\left(Diameter~of~smaller~pool\right)^2}\)
⇒ \(\left({Diameter~of~smaller~pool}\right)^2=\frac{441~\times~1125}{2205}= 225\)
⇒ \(\left({Diameter~of~smaller~pool}\right) = \sqrt{225}=15~ m\)
Now, Radius \(=\frac{Diameter}{2}\)
∴ Radius of smaller pool \(=\frac{15}{2}\) = 7.5 m
Hence, the radius of the smaller pool is 7.5 m.
Example 4: Find the missing length
Solution:
The ratio of areas of two similar figures is the ratio of the squares of their respective sides.
\(\frac{Area~of~figure~A}{Area~of~figure~B}=\left(\frac{a}{b}\right)^2\)
Let the missing length be x
\(\frac{49}{256}=\frac{14^2}{x^2}\) [Substitute the values]
\(\frac{7}{16}=\frac{14}{x}\) [Taking square root]
x = \(\frac{14~\times~16}{7}\)
x = 32 cm
Hence the missing length is 32 cm.
Example 5: Are the given figures similar?
Solution:
Two figures are similar if the dimensions of the corresponding sides have the same ratio.
\(\frac{Length~of~figure~A}{Length~of~figure~B}=\frac{7.5}{6}=\frac{75}{60}=\frac{5}{4}\) [Substitute the value and Simplify]
\(\frac{Length~of~figure~A}{Length~of~figure~B}=\frac{5}{4}\) [Substitute the values]
Here we can see that the ratios of corresponding dimensions of both the figures are the same.
Therefore, the given figures are similar.
A scale factor is the ratio of the corresponding sides of two similar objects.
Considering two two figures A and B
\(Scalar~factor=\frac{Dimension~of~figure~A}{Dimension~of~figure~B}\)
Similar triangles are similar in shape but differ in size, whereas congruent triangles are identical in shape and size.
Similar figures have similar shapes but they are not identical and that is why they do not have equal areas. Congruent figures, on the other hand, will have equal areas.
If two figures are similar, then the ratio of their volumes is the ratio of the cubes of their respective dimensions.
\(\frac{Volume~of~figure~A}{Volume~of~figure~B}=\left(\frac{a}{b}\right)^3\)
The term ‘similar figure’ refers to two figures that have the same shape. If two figures are similar, then the lengths of their corresponding dimensions will have the same ratio.