Pythagoras Theorem

Pythagoras Theorem

Theorem 1: In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Given: A right-angled triangle ABC in which B = ∠90º.

To Prove: (Hypotenuse)² = (Base)² + (Perpendicular)².

i.e., AC² = AB² + BC²

Construction: From B draw BD ⊥ AC.



Proof: In triangle ADB and ABC, we have

∠ADB = ∠ABC         [Each equal to 90º]

and, ∠A = ∠A           [Common]

So, by AA-similarity criterion, we have

∆ADB ~ ∆ABC

(Rightarrow frac{AD}{AB}=frac{AB}{AC})   [∵ In similar triangles corresponding sides are proportional]

⇒ AB² = AD × AC                                ….(i)

In triangles BDC and ABC, we have

∠CDB = ∠ABC             [Each equal to 90º]

and, ∠C = ∠C                 [Common]

So, by AA-similarity criterion, we have

∆BDC ~ ∆ABC

(Rightarrow frac{DC}{BC}=frac{BC}{AC})         [∵ In similar triangles corresponding sides are proportional]

⇒ BC² = AC × DC                              ….(ii)

Adding equation (i) and (ii), we get

AB² + BC² = AD × AC + AC × DC

⇒ AB² + BC² = AC (AD + DC)

⇒ AB² + BC² = AC × AC

⇒ AC² = AB² + BC²

Hence, AC² = AB² + BC²

The converse of the above theorem is also true as proved below.

[youtube https://www.youtube.com/watch?v=JDmglbduKvE?feature=oembed]

Read also:

• Pythagorean Theorem in Algebra
• Pythagorean Theorem

Theorem 2: (Converse of Pythagoras Theorem).

In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to the side is a right angle.

Given: A triangle ABC such that AC² = AB² + BC²



Construction: Construct a triangle DEF such that DE = AB, EF = BC and ∠E = 90º,

Proof: In order to prove that B = ∠90º, it is sufficient to show that ∆ABC ~ ∆DEF.

For this we proceed as follows :

Since  ∆DEF is a right angled triangle with right angle at E. Therefore, by Pythagoras theorem, we have

DF² = DE² + EF²

⇒ DF² = AB² + BC²         [∵  DE = AB and EF = BC         (By construction)]

⇒ DF² = AC²                        [∵  AB² + BC² = AC² (Given)]

⇒ DF = AC                              ….(i)

Thus, in  ∆ABC and  ∆DEF, we have

AB = DE, BC = EF           [By construction]

and, AC = DF                   [From equation (i)]

∴ ∆ABC ~ ∆DEF

⇒ ∠B = ∠E = 90º

Hence, ∆ABC is a right triangle right angled at B.

Pythagoras Theorem With Examples

Example 1:    Side of a triangle is given, determine it is a right triangle.

(2a – 1) cm, (2sqrt { 2a } ) cm,  and (2a + 1) cm

Sol.    Let p = (2a – 1) cm, q = (2sqrt { 2a } ) cm and r = (2a + 1) cm.

Then, (p² + q²) = (2a – 1)² cm² + (2 )² cm²

= {(4a² + 1– 4a) + 8a}cm²

= (4a² + 4a + 1)cm²

= (2a + 1)² cm² = r².

(p² + q²) = r².

Hence, the given triangle is right angled.

Example 2:    A man goes 10 m due east and then 24 m due north. Find the distance from the starting point.

Sol.    Let the initial position of the man be O and his final position be B. Since the man goes   10 m due east and then 24 m due north. Therefore, ∆AOB is a right triangle right-angled at A such that OA = 10 m and AB = 24 m.



By Phythagoras theorem, we have

OB² = OA² + AB²

⇒ OB² = 10² + 24² = 100 + 576 = 676

⇒ OB = (sqrt { 676 } ) = 26 m

Hence, the man is at a distance of 26 m from the starting point.

Example 3:    Two towers of heights 10 m and 30 m stand on a plane ground. If the distance between their feet is 15 m, find the distance between their tops.

Sol.



By Phythagoras theorem, we have

AC² = CE² + AE²

⇒ AC² = 15² + 20² = 225 + 400 = 625

⇒ AC = (sqrt { 625 } ) = 25 m.

Example 4:     In Fig., ∆ABC is an obtuse triangle, obtuse angled at B. If AD ⊥ CB, prove that

AC² = AB² + BC² + 2BC × BD

Sol.    Given: An obtuse triangle ABC, obtuse-angled at B and AD is perpendicular to CB produced.

To Prove: AC² = AB² + BC² + 2BC × BD

Proof: Since ∆ADB is a right triangle right angled at D. Therefore, by Pythagoras theorem, we have AB² = AD² + DB²              ….(i)



Again ∆ADC is a right triangle right angled at D.

Therefore, by Phythagoras theorem, we have

AC² = AD² + DC²

⇒ AC² = AD² + (DB + BC)²


⇒ AC² = AD² + DB² + BC² + 2BC • BD

⇒ AC² = AB² + BC² + 2BC • BD            [Using (i)]

Hence, AC² = AB² + BC² + 2BC • BD

Example 5:     In figure, ∠B of ∆ABC is an acute angle and AD ⊥ BC, prove that

AC² = AB² + BC² – 2BC × BD

Sol.    Given: A ∆ABC in which ∠B is an acute angle and AD ⊥ BC.

To Prove: AC² = AB² + BC² – 2BC × BD.

Proof: Since ∆ADB is a right triangle right-angled at D. So, by Pythagoras theorem, we have

AB² = AD² + BD²          ….(i)

Again ∆ADC is a right triangle right angled at D.



So, by Pythagoras theorem, we have

AC² = AD² + DC²

⇒ AC² = AD² + (BC – BD)²

⇒ AC² = AD² + (BC² + BD² – 2BC • BD)

⇒ AC² = (AD² + BD²) + BC² – 2BC • BD

⇒ AC² = AB² + BC² – 2BC • BD            [Using (i)]

Hence, AC² = AB² + BC² – 2BC • BD

Example 6:     If ABC is an equilateral triangle of side a, prove that its altitude = (frac { sqrt { 3 }  }{ 2 } a).

Sol.    ∆ABD is an equilateral triangle.

We are given that AB = BC = CA = a.

AD is the altitude, i.e., AD ⊥ BC.

Now, in right angled triangles ABD and ACD, we have

AB = AC                  (Given)

and AD = AD         (Common side)

∆ABD ≅ ∆ACD     (By RHS congruence)

⇒ BD = CD ⇒ BD = DC = (frac { 1 }{ 2 }BC) = (frac { a }{ 2 })



From right triangle ABD.

AB² = AD² + BD²

(Rightarrow =A+ )

(Rightarrow A=-frac}{4}=frac{3}{4})

(Rightarrow AD=frac{sqrt{3}}{2}a )

Example 7:     ABC is a right-angled triangle, right-angled at A. A circle is inscribed in it. The lengths of the two sides containing the right angle are 5 cm and 12 cm. Find the radius of the circle.

Sol.    Given that ∆ABC is right angled at A.

AC = 5 cm and AB = 12 cm

BC² = AC² + AB² = 25 + 144 = 169

BC = 13 cm

Join OA, OB, OC



Let the radius of the inscribed circle be r

Area of ∆ABC = Area of ∆OAB + Area of ∆OBC + Area of ∆OCA

⇒ 1/2 × AB × AC

(=frac{1}{2}left( 12text{ }times text{ }r right)text{ }+frac{1}{2}left( 13text{ }times text{ }r right)text{ }+frac{1}{2}left( 5text{ }times text{ }r right))

⇒  12 × 5 = r × {12 + 13 + 5}

⇒  60 = r × 30 ⇒   r = 2 cm

Example 7:     ABCD is a rhombus. Prove that

AB² + BC² + CD² + DA² = AC² + BD²

Sol.    Let the diagonals AC and BD of rhombus ABCD intersect at O.

Since the diagonals of a rhombus bisect each other at right angles.

∴ ∠AOB = ∠BOC = ∠COD = ∠DOA = 90º

and AO = CO, BO = OD.

Since ∆AOB is a right triangle right-angle at O.



∴ AB² = OA² + OB²

(Rightarrow A=+)         [∵ OA = OC and OB = OD]

⇒ 4AB² = AC² + BD²                      ….(i)

Similarly, we have

4BC² = AC² + BD²                    ….(ii)

4CD² = AC² + BD²                    ….(iii)

and, 4AD² = AC² + BD²           ….(iv)

Adding all these results, we get

4(AB² + BC² + AD²) = 4(AC² + BD²)

⇒ AB² + BC² + AD² + DA² = AC² + BD²

Example 8:     P and Q are the mid-points of the sides CA and CB respectively of a ∆ABC, right angled at C. Prove that:

(i) 4AQ² = 4AC² + BC²

(ii) 4BP² = 4BC² + AC²

(iii) (4AQ² + BP²) = 5AB²

Sol.    



(i)  Since ∆AQC is a right triangle right-angled at C.

∴ AQ² = AC² + QC²

⇒ 4AQ² = 4AC² + 4QC²         [Multiplying both sides by 4]

⇒ 4AQ² = 4AC² + (2QC)²

⇒ 4AQ² = 4AC² + BC² [∵  BC = 2QC]

(ii)  Since ∆BPC is a right triangle right-angled at C.

∴ BP² = BC² + CP²

⇒ 4BP² = 4BC² + 4CP²      [Multiplying both sides by 4]

⇒ 4BP² = 4BC² + (2CP)²

⇒ 4BP² = 4BC² + AC² [∵  AC = 2CP]

(iii)  From (i) and (ii), we have

4AQ² = 4AC² + BC² and, 4BC² = 4BC² + AC²

∴ 4AQ² + 4BP² = (4AC² + BC²) + (4BC² + AC²)

⇒ 4(AQ² + BP²) = 5 (AC² + BC²)

⇒ 4(AQ² + BP²) = 5 AB²

[In ∆ABC, we have AB² = AC² + BC²]

Example 9:     From a point O in the interior of a ∆ABC, perpendicular OD, OE and OF are drawn to the sides BC, CA and AB respectively. Prove that :

(i) AF² + BD² + CE² = OA² + OB² + OC² – OD² – OE² – OF²

(ii) AF² + BD² + CE² = AE² + CD² + BF2

Sol.     



Let O be a point in the interior of ∆ABC and let OD ⊥ BC, OE ⊥ CA and OF ⊥ AB.

(i)  In right triangles ∆OFA, ∆ODB and ∆OEC, we have

OA² = AF² + OF²

OB² = BD² + OD²

and, OC² = CE² + OE²

Adding all these results, we get

OA² + OB² + OC² = AF² + BD² + CE² + OF² + OD² + OE²

⇒ AF² + BD² + CE² = OA² + OB² + OC² – OD² – OE² – OF²

(ii)  In right triangles ∆ODB and ∆ODC, we have

OB² = OD² + BD²

and, OC² = OD² + CD²

OB² – OC² = (OD² + BD²) – (OD² + CD²)

⇒ OB² – OC² = BD² – CD²        ….(i)

Similarity, we have

OC² – OA² = CE² – AE²             ….(ii)

and, OA² – OB² = AF² – BF²            ….(iii)

Adding (i), (ii) and (iii), we get

(OB² – OC²) + (OC² – OA²) + (OA² – OB²)

= (BD² – CD²) + (CE² – AE²) + (AF² – BF²)

⇒ (BD² + CE² + AF²) – (AE² + CD² + BF²) = 0

⇒ AF² + BD² + CE² = AE² + CD² + BF²