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STATEMENT OF PYTHAGORAS THEOREM: FREE DRAWING OF IT

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   STATEMENT OF PYTHAGORAS THEOREM        

INTRODUCTION

A Greek philosopher of sixth century B.C.  named Pythagoras discovered a very important and useful property of right-angled triangles, named Pythagoras property. In a right-angled triangle, the sides have special names. 

The side opposite to the right angle is called the hypotenuse. (Pythagoras Theorem )

The other two sides are perpendicular and base (legs of the right-angled triangle)

In the adjoining triangle, ABC , <C = 90⁰, so, AB is its hypotenuse and BC and  CA are two legs. (Pythagoras theorem)

PYTHAGORAS THEOREM: 

In a right triangle, the square of the hypotenuse is equal to the sum of the square of the other two sides  (Pythagoras theorem)

Given, ABC is a right angled at A  i.e.<A  = 90⁰, so that BC is its hypotenuse. 

To prove,  BC2 = CA2 + AB2 i.e.  (Hypotenuse)2 =  BASE2 + PERPENDICULAR2   

Where,  BC =  hypotenuse (a);  AB = base (c) ; AC = perpendicular (b) (Pythagoras theorem)

CONSTRUCTION:  AB is extended to a point D such that BD = CA = b. At D DE is drawn on AD as perpendicular and DE  = AB =c  (Pythagoras theorem)

 PYTHAGORAS THEOREM PROOF: 

In ∆ ABC and ∆ DEB,

CA = BD (by construction)

AB = DE (by construction)

< A = <D ( each = 90⁰)

So, ∆ABC  ≡ ∆DBE (SAS congruency)

  • BC = BE (c.p.c.t.)
  • BC = BE = a
  • <ACB = <DBE (c.p.c.t.)

In ∆ABC, <A + <ABC + <ACB = 180⁰ 

  • 90⁰ + <ACB + <ABC = 180⁰
  • <ABC + <ACB = 90⁰ 

Since, sum of angles at a point on one side of a straight line is 180⁰  

Therefore, <ABC + < CBE + <DBE = 180⁰ sides

  • ( <ABC + <DBE) + <CBE = 180⁰ 
  • 90⁰ + <CBE = 180⁰ 
  • <CBE = 90⁰ 

So, <CBE is a right-angled triangle at B  

Now, <A + <D = 90⁰ + 90⁰ = 180⁰ 

Now, AC ‖ DE (Pythagoras theorem)

So, CADE is a trapezium (sum of co-int angles = 180⁰)  

So, from the figure, 

Area of trapezium CADE = Area of ∆CAB + Area of ∆BDE + Area of ∆CBE

  • ½ (CA + ED) X AD = ½ CA X AB + ½ BD X DE + ½ CB X EB  (area of a trapezium = ½ (sum of ‖ sides X height and area of a triangle = ½ base X area)
  • So,
  • (b +c )(c +b) = bc + bc + a x a {(AD = AB +BD) =  c + b) 
  • (b +c)2 = 2bc + a2        
  • b 2 + c2 + 2bc = 2bc + a2       
  • b2 + c2 = a2   
  • CA2 + AB2 = BC2
  • SO, BC2 = CA2 + AB2 

PROBLEMS RELATED TO PYTHOGOREAN THEOREM: 

Qs1: (Pythagorean theorem) For going to a city A, there is route via city C such that AC Ʇ. CB, AC = 2x km, and CB = (2x +7) km. It is proposed to construct a 26 km highway which directly connects the two cities A  and B. Find how much distance will be saved in reaching city B from city A after the construction of highway. (Pythagorean theorem)

Ans1  (Pythagorean theorem): The cities A,B and C are marked in the following figure:

In ∆ABC, <C = 90

 we get

AC2 + BC2 =   AB2  

  • (2x)2 + {2(x + 7)}2 = 262
  • 4{x2 + x2 + 14x + 49} = 676 
  • 2x2 + 14x + 49 = 169
  • 2x2 + 14x -120 = 0
  • x2 + 7x -12 = 0  
  • x2 + 12x – 5x – 60 = 0 
  • x( x + 12) -5 (x +12) = 0 
  • (x + 12) (x – 5) = 0 
  • x = -12 or 5 (x value cannot be negative), so, x = 5 

therefore , AC = 2x km = 2 X 5 = 10 km 

BC = 2(X +7) = 2(5 +7) =24 KM.

Thus, the original distance between two cities A  and B = 10 km + 24 km = 34 km. 

New distance between A and B= 26 km 

Therefore, the distance saved = 34 km – 26 km = 8 km 

Qs2  In ∆ABC  <B = 90⁰ and D is md point of BC, prove that AC2 = AD2+ 3CD2. 

Ans2 As d is the mid-point of BC,

So, BD =CD and BC = 2CD

In ∆ABD, <ABD = 90⁰, 

SO, AD2 = AB2 + BD2 

AB2=  AD2  – BD2 

In ∆ABC <B = 90⁰,

So, AC2 = ( AD2– BD2) + BC2        

AD2 – CD2 + (2CD)2  {As BD = CD, BC = 2CD}  = AC2

 We can say AD2– CD2 + 4CD2  = AC2

So, AC2 = AD2 + 3CD2 PROVED.

 

Qs3 (Pythagoras theorem)  ∆ABC is an equilateral triangle of side 2a, find each of its altitude. 

Ans3 (Pythagorean theorem) ABC is an equilateral triangle with sides 2a, (given) 

A perpendicular AD is drawn on  BC (AD Ʇ BC) 

In ∆ABD  and  ∆ACD,

AB = AC (Given)

 

AD common 

By RHS rule of congruency

∆ABD≡ ∆ACD 

  •  DC =BD (c.p.c.t.)

BD = ½ BC = ½ X 2a = a 

In  ∆ABD, <ADB = 90⁰

So, AB2 = AD2 + BD2 

(2a)2 = AD2 + a2 =>  AD2 = 3a2 =>  AD = Ѵ3a 

In an equilateral triangle , all altitudes are equal

Hence, the length of each altitude = Ѵ3a. 

 

Qs4  Prove that the sum of the squares on the sides of a rhombus is equal to the sum of squares on its diagonal. (Pythagorean theorem)

Ans4 (Pythagorean theorem) Let ABCD be a rhombus whose diagonals AC and BD intersect at the point O. (Pythagorean theorem)

As the diagonals of a rhombus bisect each other at right angles then <AOB = 90

So, OA = ½ AC

OB = ½ BD 

In ∆AOB, <AOB = 90⁰ 

So,  OA2 + OB2 = AB2

AB2 = ( ½ AC)2 + ( ½ BD )2

4AB2 = AC2 + BD2  

But AB = BC = CD = DA ( as in a rhombus, sides are equal)

So, AB2 + BC2 + CD2 + DA2 = AC2 + BD2 

 

Qs5  In the figure, AD BC , if d divides BC in the ratio 1:3 , prove that 2 AC2 =   2AB2 + BC2

So, BD/DC = 1/3  => DC = 3BD  

BC = BD + DC = BD + 3 BD

 BD + DC = BC = BD + 3 BD = 4 BD

¼  BC = BD

In, ADC, <D = 90 

So, AC2 = AD2 + DC2 = AD2 + (3BD)2 = AD2 +  9BD2 

In ABD, <D = 90  So, AB2 = AD2 + BD2  

AD2 = AB2 – BD2 

  • So, together we get,
  • AC2 =(AB2 – BD2) + 9BD2
  • AB2 + 8 BD2 = AB2 + 8 x ( ¼ BC)2 
  •  ½ BC2 + AB2
  • 2AC2 = 2AB2 +BC2  
  • Qs6  In the given figure, <B of ABC is obtuse and AD BC (Produced), prove that  AC2 = AB2 + BC2+ 2BC X  BD  

 Ans6 In ABD, < ADB = 90,  

AB2 = AD2 + DB2  

In ADC, <ADC = 90

So,  AD2+ DC2 = AC2 

  • AC2 = AD2 + (DB + BC)2 => AD2 + DB2 + BC2 + 2BC X BD
  • AB2 + BC2 + 2BC X BD  = AC2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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