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Areas of Similar Triangles

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Areas of Similar Triangles - Lesson Summary

Two triangles are similar if their corresponding angles are equal, and the corresponding sides are proportional.

Theorem:
If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.

      

Given: ΔABC ~  ΔPQR       

To prove:    ar(ABC) ar(PQR)   = ( AB PQ ) 2 = ( BC QR ) 2 = ( CA RP ) 2

Construction: AM ⊥ BC and PN ⊥ QR.

Proof:

In ΔABC, ar(ABC) = ½ x BC x AM

In ΔPQR, ar(PQR) = ½ x QR x PN

Taking the ratios of the areas , we get

ar(ABC) ar(PQR)   = 1 2 x BC x AM 1 2 x QR x PN      =   BC x AM QR x PN  

ar(ABC) ar(PQR)   =  BC QR   x  AM PN   ........ (1)

In ΔAMB and ΔPNQ,

∠B = ∠Q              (Hence ΔABC ~  ΔPQR)

∠AMB = ∠PNQ     (Each is of 90 o)

ΔAMB ~ ΔPNQ      (By AA Criterion of similarity)

AM PN   =  AB PQ          (Corresponding sides are proportional) ..... (2)

AB PQ   =  BC QR   =  CA RP   (Hence ΔABC ~  ΔPQR)

ar(ABC) ar(PQR)   = BC QR x AM PN = AB PQ   x AM PN (From equation (3))

= AB PQ   x AB PQ    .............. From Equation (2)


ar(ABC) ar(PQR)   = ( AB PQ  ) 2

Similarly, We can prove ar(ABC) ar(PQR)  = ( BC QR ) 2 = ( CA RP   ) 2

ar(ABC)/ar(PQR) = ( AB PQ  ) 2 = ( BC QR ) = ( CA RP     ) 2
  
If one of the angle is 90 degrees then it is a right angled triangle.

Theorem:
If a perpendicular drawn form vertex of the right angle of a right triangle to the hypotenuse, then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.
(or)
In ΔABC, ∠B = 90° BD ⊥ AC then ΔABC ~ ΔADB ~ ΔBDC .



Given: In ΔABC, ∠B = 90 o, BD ⊥ AC

To prove: ΔABC ~ ΔADB ~ ΔBDC

Proof:
In ΔADB and ΔABC, ∠A = ∠A.

 ∠ADB = ∠ABC = 90 o

∴  
ΔADB ~ ΔABC (By AA criterion of similarity) ........... (1)

ΔBDC ~ ΔABC ......... (2)

From (1) and (2), ΔADB ~ ΔBDC

⇒ ΔABC ~ ΔADB ~ ΔBDC

Theorem:
In a right angle triangle, the square of the hypotenuse is equal to sum of the squares of other two sides, this is known as Pythagoras theorem.
(or)
In ΔPQR, ∠PQR = 90° then PR 2 = PQ 2 + QR 2 .




Given: In ΔPQR, ∠PQR = 90°.

To proove: PR 2 = PQ 2 + QR 2.

Construction: QS ⊥ PR.

Proof:

In ΔPQR and ΔPSQ, ∠QPR = ∠SPQ

∠PQR = ∠PSQ = 90

ΔPQR ~ ΔPSQ (AA similarity criterion for two triangles)

  PQ PS    =  PR PQ   

⇒ PQ x PQ = PR x PS

PQ 2 = PR x PS ...... (1) 

In ΔPQR and ΔQSR, ∠QRP = ∠QRS

∠QSR = ∠PQR = 90 o

ΔPQR ~ ΔQSR (AA Similarity criterion)

QR SR    =  PR QR     

QR 2 = PR x SR ................ (2)

Adding equation (1) and equation (2) , we get

PQ 2 + QR 2 = PR x PS + PR x SR
                   
                  = PR(PS + SR)

∴ PQ 2 + QR 2 = PR 2
 
Theorem:
The converse of Pythagoras theorem states that in a triangle if the square of one side is equal to sum of the squares of other two sides then the angle opposite to the first side is a right angle.
(or)
In ΔABC, AC 2 = AB 2 + BC 2 then ∠B = 90°.




Given: In ΔABC, AC 2 = AB 2 + BC 2

To Prove: ∠B = 90 o

Construction: Draw ΔPQR such that ∠Q = 90 o, PQ = AB and QR = BC

Proof: PR 2 = PQ 2 + QR 2 (By Pythagoras Theorem)

PR 2 = AB 2 + BC 2 (Hence PQ = AB and QR = BC) ......... (1)

AC 2 = AB 2 + BC .........(2)

AC 2 = PR 2 ⇒ AC = PR ..............(3)

In ΔABC and ΔPQR , AB = PQ, BC = QR, AC = PR

∴ ΔABC ≅ ΔPQR (SSS congruence axiom)

⇒ ∠B = ∠Q

But ∠Q = 90 o

Hence ∠B = 90 o

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