## STATEMENT OF PYTHAGORAS THEOREM: FREE DRAWING OF IT

** 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**, BC ^{2} = CA^{2} + AB^{2} **i.e. (

**Hypotenuse)**

^{2}= BASE^{2}+ PERPENDICULAR^{2 }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)**a^{2}= 2bc +^{2 }**b**^{2}+ c^{2}+ 2bc = 2bc + a^{2 }**b**^{2}+ c^{2}= a^{2 }**CA**^{2}+ AB^{2}= BC^{2}**SO, BC**^{2}= CA^{2}+ AB^{2}

**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****⁰**

** we get**

**AC ^{2} + BC^{2} = AB^{2}**

**(2x)**^{2}+ {2(x + 7)}^{2}= 26^{2}**4{x**^{2}+ x^{2}+ 14x + 49} = 676**2x**^{2}+ 14x + 49 = 169**2x**^{2}+ 14x -120 = 0**x**^{2}+ 7x -12 = 0**x**^{2}+ 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 = 90**⁰ and D is md point of BC, prove that AC*^{2} = AD^{2}+ 3CD^{2.}

^{2}= AD

^{2}+ 3CD

^{2.}

**Ans2 ****As d is the mid-point of BC, **

**So, BD =CD and BC = 2CD**

**In ****∆ABD, <ABD = 90****⁰, **

**SO, AD ^{2} = AB^{2} + BD^{2} **

**AB ^{2}= AD^{2 } – BD^{2} **

**In ****∆ABC <B = 90****⁰,**

**So, AC ^{2} = ( AD^{2}– BD^{2}) + BC^{2 } **

**AD ^{2} – CD^{2 }+ (2CD)^{2 } {As BD = CD, BC = 2CD} ** =

**AC**

^{2}** We can say AD ^{2}– CD^{2} + 4CD^{2} **=

**AC**

^{2}**So, AC**^{2} = AD^{2} + 3CD^{2} PROVED.

^{2}= AD

^{2}+ 3CD

^{2}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, AB ^{2} = AD^{2 }+ BD^{2} **

**(2a) ^{2} = AD^{2} + a^{2} => AD^{2} = 3a^{2} => 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****⁰**

**So, OA = ½ AC**

**OB = ½ BD **

**In ****∆AOB, <AOB = 90****⁰ **

**So, OA ^{2} + OB^{2} = AB^{2}**

**AB ^{2} = ( ½ AC)^{2} + ( ½ BD )^{2}**

**4AB ^{2} = AC^{2} + BD^{2} **

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

**So, AB**^{2} + BC^{2} + CD^{2} + DA^{2} = AC^{2} + BD^{2}

^{2}+ BC

^{2}+ CD

^{2}+ DA

^{2}= AC

^{2}+ BD

^{2}

** **

Qs5 ** ****In the figure, AD ****Ʇ**** BC , if d divides BC in the ratio 1:3 , prove that 2 AC ^{2} = 2AB^{2} + BC^{2}**

**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****⁰**** **

**So, AC ^{2} = AD^{2} + DC^{2} = AD^{2} + (3BD)^{2} = AD^{2} + 9BD^{2} **

**In ****∆****ABD, <D = 90****⁰**** ****So, AB ^{2} = AD^{2} + BD^{2} **

**AD**^{2} = AB^{2} – BD^{2}

^{2}= AB

^{2}– BD

^{2}

**So, together we get,****AC**^{2}=(AB^{2}– BD^{2}) + 9BD^{2}**AB**^{2}+ 8 BD^{2}= AB^{2}+ 8 x ( ¼ BC)^{2}**½ BC**^{2}+ AB^{2}**2AC**^{2}= 2AB^{2}+*BC*^{2}*Qs6**In the given figure, <B of**∆**ABC is obtuse and AD**Ʇ**BC (Produced), prove that AC*^{2}= AB^{2}+ BC^{2}+ 2BC X BD

* ***Ans6** **In ****∆****ABD, < ADB = 90****⁰****, **

**AB**^{2} = AD^{2} + DB^{2}

^{2}= AD

^{2}+ DB

^{2}

**In ****∆****ADC, <ADC = 90****⁰**

**So, AD ^{2}+ DC^{2} = AC^{2} **

**AC**^{2}= AD^{2}+ (DB + BC)^{2}=> AD^{2}+ DB^{2}+ BC^{2}+ 2BC X BD**AB**^{2}+ BC^{2}+ 2BC X BD = AC^{2}

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