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 = 90⁰

 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+ 3CD^2. 

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 = 90⁰

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 Ʇ 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 = 90⁰ 

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

In ∆ABD, <D = 90⁰  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 ꞱBC (Produced), prove that  AC² = AB² + BC²+ 2BC X  BD  

 Ans6 In ∆ABD, < ADB = 90⁰,  

AB² = AD² + DB²  

In ∆ADC, <ADC = 90⁰

So,  AD²+ DC² = AC² 

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