According to fundamental operations of plane geometry, a plane figure can be moved from one point o another point (called Translation), can be rotated about the fixed point (called Rotation) or can be expanded or contracted such that the distance between the points in the plane figure increases or decreases in the same ratio (called Expansion). IF any figure is subjected to any of these operations in succession will result to similar figure or one figure can be brought to coincide with another by using the above properties. We have noticed similar figures like the map of the world or country, model of a house or a dam, in all these cases the similar figures bear a constant ratio with the original. In this section let us discuss about the similarity and few problems based on similar figures.

## Similarity Definition

Similarity Definition: Two figures are similar, if they have the same figures. Similar figures differ in size.

The following diagrams show pairs of similar figures.

Uses of similarity: Similarity is very much useful to make models of buildings, roads, factory etc. We have seen that maps which show the geographical location of the place. We easily find the distance between any two places in the word using the scale factor shown in the map. The housing plan also help us to construct buildings.

Similar Triangle: Two triangles are said to be similar if they are equiangular and the corresponding sides are proportional.

The following diagram shows the pair of triangles which are similar and the corresponding proportional sides.

Similarity Ratio: In the above two triangles, $\Delta ABC$ and $\Delta PQR$,
$\angle A$ = $\angle P$
$\angle B$ = $\angle Q$
$\angle C$ = $\angle R$

$\frac{BC}{QR}$ = $\frac{AC}{PR}$ = $\frac{AB}{PQ}$, this ratio is called Similarity Ratio.

1. If two triangles are equiangular then their sides opposite to equal angles are proportional.
2. If the sides of two triangles are proportional, then the angles opposite to proportional sides are equal

## Principle of Similarity

The following properties will help us to identify if the given triangles are similar or not.

1. AAA similarity Criteria: If two triangles are equiangular, then they are similar.

In the above two triangles, $\Delta ABC$ and $\Delta PQR$,
$\angle A$ = $\angle P$
$\angle B$ = $\angle Q$
$\angle C$ = $\angle R$
Therefore, $\Delta ABC$ $\sim$ $\Delta PQR$, according to AAA similarity criteria.

2. SSS similarity Criteria (AA similarity): If the corresponding sides of two triangles are proportional, then they are similar.

In the above two triangles, $\Delta ABC$ and $\Delta PQR$,

$\frac{AB}{PQ}$ = $\frac{BC}{QR}$ = $\frac{AC}{PR}$

Therefore, $\Delta ABC$ $\sim$ $\Delta PQR$, according to SSS similarity criteria.

3. SAS Similarity Criteria: If in two triangles, one pair of corresponding sides are proportional and the included angles are equal then the two triangles are similar.

In the above two triangles, $\Delta ABC$ and $\Delta PQR$,

$\frac{AC}{PR}$ = $\frac{BC}{QR}$

and $\angle C$ = $\angle R$

Therefore, $\Delta ABC$ $\sim$ $\Delta PQR$, according to SAS similarity criteria.

## Similar Triangles Problems

### Solved Examples

Question 1: In the figure given below, ABCD is a trapezoid in which AB // DC, and the diagonals AC and BD intersect at O.

Prove that,
a. $\Delta OCD$ $\sim$ $\Delta OAB$
b. If OA = ( 2x + 1 ) cm, OB = ( 5x - 3 ) cm, OC = ( 6 x - 5 ) cm and OD = ( 3 x - 1 ) cm, find the value of x.
Solution:

a. In $\Delta OCD$ and $\Delta OAB$
we have, $\angle COD$ = $\angle AOB$ [ vertically opposite angles ]
$\angle OCD$ = $\angle OAB$ [ alternate angles as AB // DC and CA is a transversal ]
Therefore, $\Delta OCD$ $\sim$ $\Delta OAB$ [ by AA Similarity ]

b. $\Delta OCD$ $\sim$ $\Delta OAB$

=>  $\frac{OC}{OA}$ = $\frac{OD}{OB}$ [ since the triangles are similar, their sides are proportional ]

=> ($\frac{( 6 x - 5 )}{( 2x + 1 )}$ = $\frac{( 3 x - 1 )}{( 5x - 3 )}$ [ substituting for OA, OB, OC and OD ]

=>                   ( 6 x - 5 ) ( 5 x - 3 )   = ( 3 x - 1 ) ( 2 x + 1 ) [ cross multiplying ]
=>                   30 x2 - 18 x - 25 x + 15  = 6 x2 - 2 x + 3 x - 1
=>                    30 x2     - 43 x    + 15 = 6 x2 + x - 1
=>    30 x2 - 43 x + 15 - ( 6 x2 + x - 1 ) = 0
=>        30 x2 - 6 x2 - 43 x - x + 15 + 1 = 0
=>                          24 x 2 - 44 x + 16 = 0
=>                               6 x2 - 11 x + 4 = 0  [ Dividing both sides by 4 ]
=>                    6 x2 - 8 x - 3 x + 4    = 0  [ by splitting the middle term ]
=>             2 x ( 3 x - 4 ) - 1 ( 3 x - 4 ) = 0
=>                    ( 3 x - 4 ) ( 2 x - 1 )    = 0
=>                                          3 x - 4 = 0 ( or ) 2 x - 1 = 0

=>                                                  x = $\frac{4}{3}$ or x = $\frac{1}{2}$

Substituting x = $\frac{1}{2}$, we get OB = $\frac{5}{2}$ - 3 = 2. 5 - 3 = - 0.5 < 0 [ negative ]

Therefore, x $\neq$ $\frac{1}{2}$

Therefore, the value of x is $\frac{4}{3}$

Question 2: A vertical pole 12m long casts a shadow 5 m long on the ground. At the same time a tower casts a shadow 30 m on the ground. Calculate the height of the tower.

Solution:

From the above figures, AB = 12 m, AC = 5 m and PR = 30 m
In $\Delta$ ABC and $\Delta PQR$, we have,
$\angle A$ = $\angle P$ = 90
and $\angle C$ = $\angle R$   [ angle of  inclination of the Sun's rays ]
Therefore      $\Delta ABC$ $\sim$ $\Delta PQR$ [ by AA similarity ]

$\frac{AB}{PQ}$ = $\frac{AC}{PR}$

=>      $\frac{12}{x}$ = $\frac{5}{30}$

=>                    5 x = 30 x 12
=>                    5 x = 360

=>                      x = $\frac{360}{5}$

= 72 m

The height of the tower is 72 m