Consider two given triangles PQR and ABC. They are said to be similar (~) if their corresponding angles are equal and corresponding sides are proportional.

__Similar
Triangles__

Consider
two given triangles PQR and ABC. They are said to be similar (~) if their corresponding
angles are equal and corresponding sides are proportional. That is ∠P = ∠A, ∠Q = ∠B and ∠R = ∠C and also PQ/AB = PR/AC = QR/BC. This
is denoted as ∆PQR ~ ∆ABC .

There are
4 ways by which one can prove that two triangles are similar.

**(i) AAA (Angle – Angle
– Angle) or AA (Angle – Angle) Similarity**

Two triangles
are similar if two angles of one triangle are equal respectively to two angles of
the other triangle. In the given figure, ∠*A* = ∠*P* , ∠*B* = ∠*Q* .

Therefore, ∆ABC ~ ∆PQR.

**(ii) SAS (Side – Angle
– Side) Similarity**

Two triangles
are similar if two sides of one triangle are proportional to two sides of the other
triangle and the included angles are equal. In the given figure, AC/PQ = AB/PR and
∠A = ∠P and hence ∆ACB ~ ∆PQR.

**(iii) SSS (Side-Side-Side)
Similarity**

Two triangles
are similar if their corresponding sides are in the same ratio. That is, if AB/PQ
= AC/PR = BC/QR , then ∆ABC ~ ∆PQR.

**(iv) RHS (Right Angle –
Hypotenuse – Side) Similarity**

Two right
triangles are similar if the hypotenuse and a leg of one triangle are respectively
proportional to the hypotenuse and a leg of the other triangle. That is, if ∠*B =
*∠*Q* = 90º and AC/PR = BC/QR then, ∆ABC ~ ∆PQR.

If ∆ABC ~ ∆PQR, then the corresponding
sides to AB, BC and AC of ∆ABC are PQ, QR and PR respectively and the corresponding
angles to A, B and C are P, Q and R respectively. Naming a triangle has a significance.
For example, if ∆ABC ~ ∆PQR then, ∆BAC is not similar to ∆PQR.

** **

__Example 5.5__

In the Fig.
5.16, if ∆PQR ~ ∆XYZ , find *a* and *b*.

*Solution:*

Given that
∆PQR
~ ∆XYZ

Their corresponding
sides are proportional.

∴
b = 22.4 *cm*

** **

__Note__

• All circles and squares are similar to each other.

• Not all rectangles need to be similar always.

• If two angles are both congruent and supplementary then, they are
right angles.

• All congruent triangles are similar.

• The symbol ~ is used to denote similarity.

** **

__Example 5.6__** **(Illustrating AA Similarity)

In** **the Fig. 5.17 , ∠*ABC* ≡
∠*EDC* and the** **perimeter of ∆*CDE* is 27 *units*,** **prove that *AB* ≡
*EC*.

*Proof:*

⇒ 8/6
= 16/EC

⇒
EC =12 *units*

Given, the
perimeter of ∆CDE = 27 *units*,

ED + DC + EC = 27

⇒
ED + 6 + 12 = 27

⇒
ED = 27–18 = 9 units

AB/9 = 8/6 ⇒ AB = 12 units and hence AB = EC.

** **

** Example 5.7** (Illustrating AA Similarity)

In the given
Fig. 5.18, if ∠1 ≡ ∠3 and ∠2 ≡ ∠4
then, prove that ∆*BIG* ~ ∆*FAT* . Also find FA.

*Proof:*

Also, their
corresponding sides are proportional

** **

__Example 5.8__** **(Illustrating SAS Similarity)

If A is the
midpoint of *RU* and T is the midpoint of
*RN*, prove that ∆*RAT* ~ ∆*RUN* .

*Proof:*

** **

__Example 5.9__** **(Illustrating SSS similarity)

Prove that
∆PQR
~ ∆*PRS* in the given Fig. 5.20.

*Solution:*

That is,
their corresponding sides are proportional.

By SSS Similarity,
∆PQR
~ ∆*PRS*

** **

__Example 5.10__** **(Illustrating RHS similarity)** **

The** **height of a man and his shadow form** **a triangle similar to that formed by a
nearby tree** **and its shadow. What is the
height of the tree?

*Solution:*

Here, ∆ABC
~ ∆*ADE* (given)

Their corresponding
sides are proportional (by RHS similarity).

⇒
h = [5×96] / 12 = 40 feet

The height of the tree is 40 *feet.*

** **

__Activity__

The teacher cuts many triangles that are similar or congruent from
a card board (or) chart sheet. The students are asked to find which pair of triangles
are similar or congruent based on the measures indicated in the triangles.

Tags : Geometry | Chapter 5 | 8th Maths , 8th Maths : Chapter 5 : Geometry

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8th Maths : Chapter 5 : Geometry : Similar Triangles | Geometry | Chapter 5 | 8th Maths

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