Similar Polygons are Polygons which have the same size but may differ in size. We do use similar smaller figures to represent larger real life objects to solve problems related to it. A rectangular field is indicated by a small rectangle in your book to describe the problem. The actual field and the Book-figure are thus similar polygons. (The Book rectangle is nevertheless labeled with original dimensions). Proportions of corresponding sides are used to solve problems related to similar polygons.

Polygons with similar shape but may differ in size are called similar polygons.
Two Polygons with one to one correspondence between the vertices are similar if
  1. Each pair of corresponding angles are congruent.
  2. The measures of corresponding sides are proportional.

Similar Polygons Definitions


The symbol ∼ reads as 'similar to'. In the above diagram ABCD ∼ PQRS.
The order of vertices on either side of the similarity statement determines the corresponding vertices and hence the corresponding angles and sides. Using the definition of similarity for the above situation
∠A  ∠P,   ∠B  ∠Q,   ∠C  ∠R,   ∠D  ∠S.                    (Corresponding angles are congruent)

$\frac{AB}{PQ}$ = $\frac{BC}{QR}$ = $\frac{CD}{RS}$ = $\frac{DA}{SP}$         (Lengths of corresponding sides are proportional).

If each of the ratio given in the extended proportion above = k, then k is known as the scale factor of similarity or constant of proportionality.

The similar polygons may be repositioned by sliding, flipping or turning in order to identify the corresponding parts easily.

The ratio of areas of similar polygons is related to the ratio of the lengths of the corresponding sides.
The ratio of areas of two similar polygons is the square of the ratio of the lengths of the corresponding sides.

If k is the scale factor of similarity, then the ratio of areas = k2.

Solved Example

Question: The blueprint of a building is made to a scale factor of 1 cm : 1 meter. If the actual area of a rectangular corridor is 300 sq.meters, what is the area of the portion on the blue print that represents the corridor?
Solution:
 
The ratio of corresponding sides of the polygons = 1 cm : 1 meter = $\frac{1}{100}$.      (1 meter = 100 cm)

Ratio of the areas of the similar rectangles = ($\frac{1}{100}$)2 = $\frac{1}{10000}$

let the area of the portion of the corridor on blue print = x sq cm.

The proportion for areas can be written as

$\frac{x}{300\times 10000}$ = $\frac{1}{10000}$.

x = 300 sq.cm.

Hence the area of the corridor represented on blue print = 300 sq.cm

The ratio of perimeters of two similar polygons is same as the ratio of their corresponding sides.
 


Solved Examples

Question 1: The dimensions of three quadrilaterals are given. Determine whether they are all similar or two of them are similar. Also find the scale factor wherever the similarity exists.
    Similar Polygons Examples
Solution:
 
When we consider the polygons ABCD and PQRS, their sides are congruent, but the angles are not congruent.
     Between polygons PQRS and UVWX we can find a congruent correspondence between the angles, but the second
     polygon has to be named in a different order as the corresponding congruent angles are,
     ∠P  ∠W,   ∠Q  ∠ X,  ∠R   ∠U  and ∠S   ∠V
     Considering the ratios of corresponding sides of polygons PQRS and WXUV,

     $\frac{PQ}{WX}$ = $\frac{6}{4.5}$ = 1.33

     $\frac{QR}{XU}$ = $\frac{6}{4.5}$ = 1.33

     $\frac{RS}{UV}$ = $\frac{4}{3}$ = 1.33

     $\frac{SP}{VW}$ = $\frac{5.6}{4.2}$ = 1.33

     The ratios of corresponding sides are also proportionate.
     Hence PQRS ∼ WXUV
     The scale factor of similarity = 1.33
 

Question 2: Find the value of  if the two triangles ABC and DEF are similar.

     Similar Polygons Examples
Solution:
 
Given ABC ∼ DEF
     Hence we can write the extended proportion as,

     $\frac{AB}{DE}$ = $\frac{BC}{EF}$ = $\frac{CA}{FD}$

     $\frac{4}{x+2}$ = $\frac{3}{6}$ = $\frac{4}{x+2}$

     $\frac{4}{x+2}$ = $\frac{1}{2}$                                Proportion simplified

           2 (4) = x + 2                                                      Cross multiplication

           x + 2 = 8  ⇒ x = 6
 

Question 3: In the given diagram Polygon ABCDE ∼ PQRST.

    Examples of Similar Polygons

    a. Find the scale factor of similarity.
    b. Find the perimeter of the polygon PQRST
    c. Find the ratio of the areas ABCDE and PQRST.
Solution:
 
a. Side AB in polygon ABCDE and side PQ in Polygon PQRST are the corresponding sides.

        Hence the scale factor of similarity = $\frac{5}{8}$ = 0.625

    b. The ratio of perimeters of similar polygons = Scale factor of similarity.

        Perimeter of polygon ABCDE = 5 + 2 + 4 + 4 + 2 = 17 units

        $\frac{Perimeter\ of\ ABCDE}{Perimeter\ of\ PQRST}$ = 0.625

        $\frac{17}{Perimeter\ of\ PQRST}$ = 0.625

        Perimeter of PQRST = $\frac{17}{0.625}$ = 27.2 units

   c. $\frac{Area\ ABCDE}{Area\ PQRST}$ = ($\frac{AB}{PQ}$)2 = (0.625)2 = 0.390625