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.
Pythagoras Theorem 1
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.

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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
Pythagoras Theorem 2
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, \(2\sqrt { 2a } \) cm,  and (2a + 1) cm
Sol.    Let p = (2a – 1) cm, q = \(2\sqrt { 2a } \) 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.
Pythagoras Theorem 3
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.
Pythagoras Theorem 4
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)
Pythagoras Theorem 5
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.
Pythagoras Theorem 6
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 = \(\frac { 1 }{ 2 }BC\) = \(\frac { a }{ 2 }\)
Pythagoras Theorem 7
From right triangle ABD.
AB2 = AD2 + BD2
\(\Rightarrow {{a}^{2}}=A{{D}^{2}}+{{\left( \frac{a}{2} \right)}^{2}} \)
\(\Rightarrow A{{D}^{2}}={{a}^{2}}-\frac{{{a}^{2}}}{4}=\frac{3}{4}{{a}^{2}}\)
\(\Rightarrow AD=\frac{\sqrt{3}}{2}a \)

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
Pythagoras Theorem 8
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
\(=\frac{1}{2}\left( 12\text{ }\times \text{ }r \right)\text{ }+\frac{1}{2}\left( 13\text{ }\times \text{ }r \right)\text{ }+\frac{1}{2}\left( 5\text{ }\times \text{ }r \right)\)
⇒  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.
Pythagoras Theorem 9
∴ AB2 = OA2 + OB2
\(\Rightarrow A{{B}^{2}}={{\left( \frac{1}{2}AC \right)}^{2}}+{{\left( \frac{1}{2}BD \right)}^{2}}\)         [∵ 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.    
Pythagoras Theorem 10
(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.     
Pythagoras Theorem 11
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

Example 10:     In a right triangle ABC right-angled at C, P and Q are the points on the sides CA and CB respectively, which divide these sides in the ratio 2 : 1. Prove that
(i) 9 AQ2 = 9 AC2 + 4 BC2
(ii) 9 BP2 = 9 BC2 + 4 AC2
(iii) 9 (AQ2 + BP2) = 13 AB2
Sol.    It is given that P divides CA in the ratio 2 : 1. Therefore,
\(CP=\frac { 2 }{ 3 }AC\)      ….(i)
Also, Q divides CB in the ratio 2 : 1.
∴ \(QC=\frac { 2 }{ 3 }BC\)      ….(ii)
Pythagoras Theorem 12
(i)  Applying pythagoras theorem in right-angled triangle ACQ, we have
AQ2 = QC2 + AC2
⇒ AQ2 = \(\frac { 4 }{ 9 }\) BC2 + AC2              [Using (ii)]
⇒ 9 AQ2 = 4 BC2 + 9 AC2            ….(iii)
(ii)  Applying pythagoras theorem in right triangle BCP, we have
BP2 = BC2 + CP2
⇒ BP2 = BC2 + AC2               [Using (i)]
⇒ 9 BP2 = 9 BC2 + 4 AC2             ….(iv)
(iii)  Adding (iii) and (iv), we get
9 (AQ2 + BP2) = 13 (BC2 + AC2)
⇒ 9 (AQ2 + BP2) = 13 AB2 [∵  BC2 = AC2 + AB2]

Example 11:     In a ∆ABC, AD ⊥ BC and AD2 = BC × CD. Prove ∆ABC is a right triangle.
Sol.    
Pythagoras Theorem 13
In right triangles ADB and ADC, we have
AB2 = AD2 + BD2             ….(i)
and, AC2 = AD2 + DC2           ….(ii)
Adding (i) and (ii), we get
AB2 + AC2 = 2 AD2 × BD2 + DC2
⇒ AB2 + AC2 = 2BD × CD + BD2 + DC2         [∵  AD2 = BD × CD (Given)]
⇒ AB2 + AC2 = (BD + CD)2 = BC2
Thus, in ∆ABC, we have
AB2 = AC2 + BC2
Hence, ∆ABC, is a right triangle right-angled at A.

Example 12:      The perpendicular AD on the base BC of a ∆ABC intersects BC at D so that DB = 3 CD. Prove that 2AB2 = 2AC2 + BC2.
Sol.    We have,
Pythagoras Theorem 14
DB = 3CD
BC = BD + DC
The perpendicular AD on the base BC of a ∆ABC intersects BC at D so that DB = 3 CD. Prove that
2AC2 + BC2.
We have,
DB = 3CD
∴ BC = BD + DC
⇒ BC = 3 CD + CD
⇒ BD = 4 CD ⇒  CD = \(\frac { 1 }{ 4 }\) BC
∴ CD = \(\frac { 1 }{ 4 }\) BC  and BD = 3CD = \(\frac { 1 }{ 4 }\) BC           ….(i)
Since ∆ABD is a right triangle right-angled at D.
∴ AB2 = AD2 + BD2           ….(ii)
Similarly, ∆ACD is a right triangle right angled at D.
∴ AC2 = AD2 + CD2         ….(iii)
Subtracting equation (iii) from equation (ii)  we get
AB2 – AC2 = BD2 – CD2
⇒ AB2 – AC2 = \({{\left( \frac{3}{4}BC \right)}^{2}}-{{\left( \frac{1}{4}BC \right)}^{2}}\)       \(\left[ From\ (i)\ CD=\frac{1}{4}BC,\ BD=\frac{3}{4}BC \right]\)
⇒ AB2 – AC2 = \(\frac { 9 }{ 16 }\) BC2 – \(\frac { 1 }{ 16 }\) BC2
⇒ AB2 – AC2 = \(\frac { 1 }{ 2 }\) BC2
⇒ 2(AB2 – AC2) = BC2
⇒ 2AB2 = 2AC2 + BC2.

Example 13:      ABC is a right triangle right-angled at C. Let BC = a, CA = b, AB = c and let p be the length of perpendicular from C on AB, prove that
(i) cp = ab
(ii) \(\frac{1}{{{p}^{2}}}=\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}\)
Sol.    (i)  Let CD ⊥ AB. Then, CD = p.
Pythagoras Theorem 14
∴ Area of ∆ABC = \(\frac { 1 }{ 2 }\) (Base × Height)
⇒  Area of ∆ABC = \(\frac { 1 }{ 2 }\) (AB × CD) =  \(\frac { 1 }{ 2 }\) cp
Also,
Area of ∆ABC = \(\frac { 1 }{ 2 }\) (BC × AC) = \(\frac { 1 }{ 2 }\) ab
∴ \(\frac { 1 }{ 2 }\) cp = \(\frac { 1 }{ 2 }\) ab
⇒  cp = ab
(ii)  Since ∆ABC is right triangle right-angled at C.
∴ AB2 = BC2 + AC2
⇒  c2 = a2 + b2
\(\Rightarrow {{\left( \frac{ab}{p} \right)}^{2}}={{a}^{2}}+{{b}^{2}} \)         \(\left[ \because \ cp\,=\,ab\ \ \therefore c=\frac{ab}{p} \right]\)
\(\Rightarrow \frac{{{a}^{2}}{{b}^{2}}}{{{p}^{2}}}={{a}^{2}}+{{b}^{2}} \)
\(\Rightarrow \frac{1}{{{p}^{2}}}=\frac{{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}{{b}^{2}}}\Rightarrow \frac{1}{{{p}^{2}}}=\frac{1}{{{b}^{2}}}+\frac{1}{{{a}^{2}}} \)
\(\Rightarrow \frac{1}{{{p}^{2}}}=\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}} \)

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