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Pythagoras Theorem

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


Given    

ABC is a triangle in which ABC = 90°

 


To Prove   
AC2 = AB2 + BC2


Construction    
Draw BD
AC.

Proof    
In
ΔADB and ΔABC
A = A                          [Common angle]
ADB = ABC                      [each 90o]
ΔADB ~ ΔABC                   [AA criteria]

 

Example : AB2 = AD ´ AC                                 (i)
Similarly, BC2 = CD
´ AC                                  (ii)
Adding Equations (i) and (ii) we get,
AB2 + BC2 = AD
´ AC + CD ´ AC
               = AC(AD+CD)
               = AC
´ AC = AC2
       AC2 = AB2 + BC2

 

Theorem (Converse of the Pythagoras Theorem)

 
 
If in a triangle the square of a side is equal to the sum of the squares of other two sides, then it will be a right-triangle.
 

 


Given
In the
ΔABC, AC2 = AB2+BC2

To Prove 

D
ABC is a right angled triangle at B.

Construction   
Construct a right triangle PQR, right angled at Q, such that PQ = AB and QR = BC.

Proof  
In the
ΔPQR, Q=90°        
PR2 = PQ2 + QR2
 PR2 = AB2 + BC2                    [by construction] (i)

But   AB2 + BC2 = AC2                               (given) (ii)

Comparing Equations (i) and (ii) we get,

PR2 = AC2 PR = AC

Therefore, by SSS congruence rule we get,

ΔABC ≅ΔPQR  
B = Q
But 
Q = 90°      
ABC = 90°

ΔABC is a right triangle at B.  
 

Example:
The side of some triangles are given below. Determine which of them form right triangles.

(i)  6 cm, 6 cm, 6 cm;
(ii)  7 cm, 8 cm, 9 cm;
(iii) 8 cm, 15 cm, 17 cm;
(iv) 7 cm, 24 cm, 25 cm.


Solution
A triangle will be right- angled-triangle if the square of its longest side is equal to the sum of the squares of the other two sides.

(i)
  a = 6 cm, b = 6 cm, c = 6 cm
     a2 = 36, b2 = 36 and c2 = 72
Therefore,   c2 = a2 + b2
                 
(72 = 36 + 36)
Hence, it is a right triangle.

(ii)    
a = 7 cm, b = 8 cm, c = 9 cm
      a2 + 49, b2 = 64, c2 = 81
Here,             81
¹ 49+64
                   c2 ¹ a2+b2
Hence, it is not a right
triangle

(iii)  Here a = 8 cm, b = 15 and c =17 cm.
     a2 = 64, b2 = 225, c2 = 289
Here, 289 = 225 + 64 = 289
       
  c2 = a2+b2
Hence, it is a right
triangle.

 

(iv) Here 

a = 7 cm, b = 24 and c =25 cm.  

        a2 = 49, b2 = 576, c2 = 625
Here,  625 = 49 + 576 = 625
         
c2 = a2 + b2

Hence, it is a right triangle.  


Example
A ladder reaches a window which is 12 m above the ground on one side of the street. Keeping its foot at the same point the ladder is turned to the other side of the street to reach a window 9 m high. Find the width of the street if the length of the ladder is 15 m.

 

 
Solution
Let AE and CE represent the ladder and AB is the height of one window and CD is the height of the second window.   

AB = 12 m, CD = 9 m and AE = CE =15 m
Width of the street = BD= BE + DE
                      BE2 = AE2 - AB2 = 152 - 122

                   BE2 = 81
                   BE = 9m
                   
 DE2 = CE2 - CD2=152 - 92
                  DE2 = 144 
⇒                     DE = 12 m

Hence,the width of the street= BE + DE = 12 + 9 = 21 m.


Example
Two poles of heights 6 m and 11 m stand vertically on a plane ground. If the distance between their base is 12 m, find the distance between their tops.

 


Pole    CD = 6 m
Pole    AB = 11 m


Solution
Distance between the two poles = BC = 12 m

Draw    DE
|| CB,
Hence,  BCDE is a rectangle.
...  CD = BE = 6 m and AE = AB
BE = 11 6 = 5 m
Now in
ΔAED, E = 90°
AD2  = DE2 + AE2
           = BC2 + AE2
           = 122 + 52 =169

   AD = 13 m   
 

Example
A ladder 25 m long, reaches a window of a building 20 m above the ground.  Determine the distance of the foot of the ladder from the building.

Solution
Ref to fig. AC represents the ladder
and A is the window. In ΔABC, B = 90°
AC2 = AB2 + BC2
 


       252 = 202 + BC2
   BC = 15 m
Hence distance of the foot of ladder from the wall = 15 m.

 

Example
In the obtuse ΔABC (Obtuse angled at B), if AD CD (produced).  Prove that; AC2 = AB2 + BC2 + 2 BC ´ BD.
 


Solution
In  
ΔADC, D = 90°
 AC2 = AD2 + DC2
In 
ΔADB, D = 90°
       AB2 = AD2 + BD2
From Equation (i) and (ii) we get,
AC2 = AB2 - BD2 + DC2
AC2 = AB2 - BD2 + (DB+BC)2
      
= AB2 - BD2 + BD2 + BC2 + 2 (BD
´ BC)
AC2 = AB2 + BC2 + 2 (BD
´ BC)

 

Example
In the figure F is the mid-point of BC.  AB = 9 cm, DE = 7 cm and BC = 24 cm.
Prove that
AFE = 90°

Solution
In
ΔABF, B = 90°


 CF = BF = BC,

 

 BC = () ´ 24 = 12 cm

AF2 = AB2 + BF2   [Pythagoras Theorem]
      = 92 + 122 = 225

AF = 15 cm

Again, ABCD is a rectangle,

  AB = CD = 9 cm
Now, In
ΔECF, C=90°
CF = 12 cm and CE = CD + DE = 9 + 7 =16 cm.
Now,   EF2 = CF2 + CE2  

[Pythagoras Theorem]
               = 122+162 = 400

        EF = 20 cm
Join AE

Again in,  ΔADE,             
AD = BC = 24 cm.
DE = 7 cm and
ADE=90°
AE2 = AD2+DE2      

[Pythagoras Theorem]
          =  242+72 = 625
Now in 
ΔAFE, AE2 = 625, EF2 = 400
And   AF2= 225
...    AE2 = EF2+AF2
625 = 400 + 225

...AFE = 90°     [converse of Pythagoras theorem]  


Example
In a right angled triangle whose sides a and b and hypotenuse c, the altitude drawn on the hypotenuse is x.  Prove that.




Solution
Let ABC a right triangle, such that,
ACB = 90° and AC = b, BC = a and AB = c and CP AB and CP = x.
 


To Prove  

 

Proof   
In 
ΔABC, C = 90°
AB2 = AC2 + BC2                  [

Pythagoras theorem.]
 
     c2 = b2 + a2

Now in
  ΔABC and ΔACP                   
A =A                                             [Common]
ACB =APC                                  [each of 90o]

ΔABC ~ ΔACP                                [AA Corollary]     


  

x =

L.H.S =

        = = R.H.S.  


Example
ABC is an isosceles triangles, in which AB = AC and BD AC.  Prove that BC2 = 2AC ´ CD, if A is acute.

Solution
In the figure AB = AC and BD AC and A is acute in ΔABC.
 



In  
ΔBCD, D = 90°     
..
BC2 = BD2+CD2                                             (i)
In
ΔABD, AB2 = AD2 + BD2                               (ii)

From Equation (i) and (ii) we get,
BC2 - AB2 = CD2 - AD2
or  BC2 = AB2 + CD2 - AD2
            
= AB2 + (AC - AD)2 - AD2
            
= AB2 + AC2 - 2AC
´AD
           = AC2+AC2 - 2AC
´AD
           = 2AC2 - 2AC
´AD
          
= 2AC(AC - AD)
           = 2AC
´ CD

 

Example
Prove that, three times the square of any side of an equilateral triangle is equal to four times the square of its altitude.

 

Solution
 

 

 

Given   
An equilateral triangle ABC, in which AD
BC.
     
To Prove   
3 AB2 = 4 AD2

Proof   
Let AB = BC = CA = a
In
ΔABD and ΔACD,
AB = AC, AD = AD

and   ADB = ADC     (each 90o)
...          ΔABD≅ΔACD

...   BD = CD = (CPCTE)

Now in,   ΔABD, D = 90°
...           AB2 = BD2 + AD2
or   AB2 =
or 3 AB2 = 4 AD2 


Example
In an equilateral triangle ABC, this side BC is trisected at D.  Prove that, 9AD2 = 7 AB2.
 


Solution
 Here ABC is a equilateral triangle, AB = BC = AC and BD =() BC.

Construction   
Draw AE
BC.

Proof   

ΔABE ≅ΔACE
BE = EC =
Now in ΔABE,
AB2 = BE2+AE2                                  (i)
Also, AD2 = AE2+DE2                       (ii)
 AB2 - AD2 = BE2 - DE2

= BE2 - (BE - BD)2          
= 
= 

  AB2 - AD2 = 
or   7AB2 = 9AD2  


Example
In any triangle, the sum of the squares of any two sides is equal to the sum of twice the square of the median, bisecting the third side and twice the square of, half of the third side.

Solution

Given  
In
ΔABC, AD is the median.

 
 
To Prove   
AB2 + AC2 = 2 AD2 + 2

Construction     
Draw AE
BC

Proof   
In
ΔAEC, E = 90°                      
...   AC2 = AE2 + EC2       [Pythagoras’ Theorem] (i)

In  
ΔAED, E = 90°
...     AD2 = AE2+ED2                               (ii)

Similarly,  AB2 = AE2 + BE2                       
(iii)
From Equation (i) and (ii) we get,

AC2 =(AD2 - ED2) + (ED + CD)2
      
= AD2 - ED2 + ED2 + CD2 + 2(ED
´CD)
      = AD2 + CD2 + 2(ED
´CD)
     
= AD2+       
     
= AD2+                           (iv)

Similarly we can prove;  
AB2 = AD2 +                       (v)

Adding Equation (iv) and (v) we get,
 
AB2+AC2 = 2AD2+

AB2+AC2 = 2AD2+ 2    

 

Example
In a rectangle ABCD, there is a point O, which is joined to each of the vertices A, B, C, D.  Prove that, OA2+OC2 = OB2+OD2

Solution
Given  

ABCD is a rectangle and O is a point in the interior of ABCD.

 


To Prove 

OA2 + OC2 = OB2 + OD2

Construction 

Draw EF
|| DC through ‘O’

Proof 

In rectangle ABCD, AB = DC, and AD = BC.

As  EF
|| DC || AB
ABFE and CDEF are also rectangles.

Hence,  AE = BF and DE = CF
In right-angled-triangle BOF,
BO2 = BF2 + OF2      (i) 
In right-angled triangle DEO,
OD2 = OE2 + ED2     (ii) 


Adding Equation (i) and (ii) we get
BO2 + OD2 = BF2 + OF2 + OE2 + ED2
                = (BF2 + OE2) + (OF2 + ED2)
                = (AE2 + OE2) + (OF + CF2)                

= AO2 + OC2


Example

In the figure S and T trisect QR, prove that, 
8
PT2 = 3 PR2 + 5 PS2
 


Solution
In 
ΔPQR, Q = 90°
PR2 = PQ2 + QR2                                (i)
Similarly,  PT2 = PQ2 + QT2                 (ii)
and   PS2 = PQ2 + QS2                         (iii)

From Equation (i),

3PR2
= 3PQ2 + 3QR2      (iv)
From Eq
uation (iii)
5PS2
= 5PQ2 + 5QS2     (v)

Adding Equations (iv) and (v) we get,

3PR2 + 5PS2 = 8PQ2 + 3QR2 + 5QS2

   
                 = 8PQ2 + 3

                     = 8PQ2 + 8QT2

    
                     = 8(PQ2+QT2)

   
                 = 8 PT2  

 

Example
P and Q are the mid-points of the sides AC and BC respectively of a triangle ABC, right-angled at C. Prove that,

Solution

(a)  
4AQ2 = 4AC2 + BC2
(b)   4(AQ2+BP2) = 5 AB
2



Given  
ACB is a right-angled Δ in which P is the mid-point of AC and Q is the mid-point of BC

To Prove   

4AQ2 = 4AC2 + BC2
and 4(AQ2 + BP2) = 5 AB2


Proof  

In
ΔABC, C = 90°             
AB2 = AC2 + BC2            (i)

In   
ΔACQ, C = 90°
AQ2 = AC2 + CQ2           (ii)

In
ΔPCB, PB2 = PC2 + BC2      (iii)

From Equation (ii) we get, 

AQ2 = AC2+      [BC = 2 CQ]


⇒ 4AQ2= 4AC2 + BC2      (iv)

Similarly, 4BP2 = 4BC2 + AC2  (v)

Adding equations (iv) and (v) we get,

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


4 (AQ2+BP2) = 4AB2+AB2=5AB2  

 

Example
In rhombus ABCD prove that, 
AB2 + BC2 + CD2 + DA2 = AC2 + BD2


Solution

The diagonals of a rhombus bisect each other at 90°.
 


...    OA = OC and OB = OD
and  
AOB = BOC = COD = DOA = 90°

Hence,  AB2 = OA2 + OB2             
  (i)
              BC2 = OB2 + OC2               
(ii)
              CD2 = OC2 + OD2              
(iii)
and        DA2 = OD2 + OA2               
(iv)

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

AB2 + BC2 +CD2 + DA2 = OA2 + OB2 + OB2 + OC2 + OC2 + OD2 + OD2 + OA2  
                                      
=2(OA2+OB2+OC2+OD2)
                                 =2(2OA2+2OB2)
                                  =(2OA)2+(2OB2)
                                  =AC2+BD2  

 

Example
From a point O in the interior of a ΔABC, perpendiculars OD, OE and OF are drawn to the sides BC, CA AB respectively. Prove that,

(i)       AF2 + BD2 + CE2 = OA2 + OB2 + OC2 - OD2 - OE2 - OF2
(ii)     AF2 +BD2 + CE2 = AE2 + CD2 + BF2

Solution
Given   
In
ΔABC, OD BC, OE AC and OF AB.
 
 
To Prove 
AF2 + BD2 + CE2 = OA2 + OB2 + OC2 - OF2 - OD2 - OE2


Construction 

Join point O to A, B and C.


Proof   

In the rightτριανγλεAFO, AFO = 90°       
 
...          
AO2 = AF2 + OF2

            AF2 = AO2 - OF2                (1)
Similarly,  BD2 = BO2 - OD2               (2)
and         CE2 = CO2 - OE2                 (3)

Adding Equations (1),(2), (3), we get,
AF2 + BD2 + CE2 = OA2 + BO2 + CO2 - OD2 - OE2 - OF2 (iii)

(ii)   AF2 + BD2 + CE2 = (AO2 - OE2) + (BO2 - OF2) + (CO2 - OD2)  

                                                

= AE2 + BF2 + CD2 

 

Example
The perpendicular AD, on the base BC of a ΔABC, intersects BC at D, so that BD = 3 CD.  Prove that, 2AB2 = 2AC2 + BC2
 


Solution
In the triangle ABC, AD
BC and BD = 3CD
BD =
In
ΔABD, D=90°
AB2 = BD2 + AD2
 
      2AB2 = 2BD2 + 2AD2
                 = 2BD2 + 2(AC2 - CD2)

                 = 2AC2 + 2(BD2 - CD2)

                 = AC2 + 2

=2AC2

=2 AC2 + BC2  


Example

In an equilateral triangle with side ‘a’, prove that,

(i)     the altitude is of length

(ii)    the area of the triangle is .

 

 

 

InΔABC,AB = BC = CA = a
Draw  AD BC
Α=B

AD is common 

...       ΔABD ΔACD

(RHS property)

...       BD = CD =

Now in  ΔABD, D = 90°

...      AB2 = AD2 + BD2

a2 = AD2 +

 ⇒ AD2 = a2

 Hence, Altitude = AD =

Area of the triangle = ´ base ´ height

                     =

                           = .  


 




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