6.1 The Right Triangle  6.2 The Theorem of Pythagoras  6.3 Special Right Triangles

1. Chapter 6: Theorem of
Pythagoras And The Right
Triangle
Subtitle

2. Chapter 6: Theorem of Pythagoras And
The Right Triangle
•6.1 The Right Triangle
•6.2 The Theorem of Pythagoras
•6.3 Special Right Triangles

3. 6.1 The Right Triangle
• DABC is a right triangle, hence m ∠ABC = 900. Therefore m ∠A
and m ∠C are complementary ( figure 6.1).
figure 6.1

4. 6.1 The Right Triangle
• Now seg.BD is a perpendicular onto seg.AC (figure 6.2).
figure 6.2

5. 6.1 The Right Triangle
• Seg.BD divides DABC into two right triangles DBDC and DADB (
figure 6.2). It can be easily proven that these two triangles are
similar to the parent DABC and therefore similar to each other.
figure 6.2

6. 6.1 The Right Triangle
• Proof:
• Consider DABC and DBDC
• ∠ ABC ≅ ∠ BDC right angles and
• ∠ BCA ≅ ∠ DCB same angle
• by AA test D ABC ~ D BDC  (1)
• Similarly consider D ABC and D ADB.
• ∠ ABC ≅ ∠ADB right angle
• ∠ CAB ≅ ∠ DAB same angle
• by AA test DABC ~ DADB  (2)
figure 6.2

7. 6.1 The Right Triangle
• from (1) and (2)
• DABC ~ DADB ~ DBDC.
• Since DABC ~ DBDC
• and DABC ~ DADB
figure 6.2
In (A), (seg.BC) is repeated and in
(B), (seg.AB) is repeated at the means. This is
referred to as the geometric mean.

8. 6.1 The Right Triangle
• The two proportions (A) and (B) obtained by the
similarity of DADB and DBDC with the original
triangle are stated as a theorem as follows:
• If an altitude seg.BD is drawn to the hypotenuse
seg.AC of a right triangle DABC then each leg ,
• i.e. seg.AB and seg.BC is the geometric mean
between the hypotenuse and seg.DA and seg.DC
respectively ( refer figure 6.2).
• The similarity of DBDC and DADB gives the
proportion.
figure 6.2

9. 6.1 The Right Triangle
• Example 1
• Find the geometric mean between :
• a) 2 and 18
• b) 4 and 16
• c) 9 and 25
Solution: a)

10. 6.1 The Right Triangle
• Example 2
• Find x
• Solution :
• The square of the altitude to the hypotenuse is equal to the product of the
segments cut on the hypotenuse.
x2 = (12 )( 3)
x2 = 36
x = 6.

11. 6.1 The Right Triangle
• Example 3
• Find y

12. 6.2 The Theorem of Pythagoras
• DABC is a right triangle.
• (seg.AB) = c
• (seg.BC) = a
• (seg.CA) = b
• CD is perpendicular to AB such that
Figure 6.3

13. 6.2 The Theorem of Pythagoras
• DABC ~ DCBD
• or (BC)2 = (AB)(CD)
• a2 = (c)(x) = cx  (1)
• DABC ~ DACD
Figure 6.3
• or (AC)2 = (AB)(AD)
• b2 = (c )( y) = cy  (2)
• Therefore, from (1) and (2)
• a2 + b2 = cx + cy
• = c ( x + y )
• = c (c)
• = c2
• a2 + b2 = c2
The square of the hypotenuse is equal to the sum of the squares of the legs.

14. 6.2 The Theorem of Pythagoras
• Converse of Pythagoras Theorem : In a triangle if the square of the
longest side is equal to the sum of the squares of the remaining
two sides then the longest side is the hypotenuse and the angle
opposite to it, is a right angle.
• Since m∠Q = 900 ∠B is a right angle and DABC is a right triangle.
Therefore if the square of the length of the longest side is equal to
the sum of the squares of the other two sides the triangles is a
right triangle.

15. 6.2 The Theorem of Pythagoras
• Example 1
• The length of a rectangle is 4 ft. and
the breadth is 3ft. What is the length
of its diagonal?
• Solution:
• ABCD is a rectangle such that seg.AB = 4,
seg.BC= 3. m ∠ABC is 900.
• DABC is a right triangle.
(AC) 2 = (AB)2 + (BC)2
= (4)2 + (3)2
= 16 + 9
= 25
(AC) = 5 ft.
The length of the diagonal is 5 ft.

16. 6.2 The Theorem of Pythagoras
• Example 2
• A man drives south along a straight road for 17 miles. Then turns
west at right angles and drives for 24 miles where he turns north
and continue driving for 10 miles before coming to a halt. What is
the straight distance from his starting point to his terminal point ?

17. 6.2 The Theorem of Pythagoras
Example 3
D ABC is a right triangle. m  ACB = 900 , seg.AQ bisects seg.BC at Q.
Prove that 4(AQ)2 = 4( AC)2 + BC2

18. 6.3 Special Right Triangles
• The 300 - 600 - 900 triangle : If the angles of a triangle are 300 ,
600 and 900 then the side opposite to 300 is half the hypotenuse
and the side opposite to 600 is
3
2
times the hypotenuse.
Figure 6.5

19. The 300 - 600 - 900 triangle
• To prove that AB = 1/2(BC) and AC=
3
2
(BC)
• but (seg. AB) = 1/2(DB)  2
In D ABC  A = 900  B = 600 and  C = 300
 D DCB is an equilateral triangle
(seg.DC) = (seg.BC) = (seg.DB)  (1)
Therefore AB = ½(BC)

20. The 300 - 600 - 900 triangle
• In right triangle ABC
• (seg.AB)2 + (seg.AC)2 = (seg.BC)2 Pythagoras
• ½ (seg.BC)2 + (seg.AC)2 = (seg.BC)2
• (seg.AC)2 =(seg.BC)2 - 1/2(seg.BC)2
• (seg.AC)2 =3/4(seg.BC)2
• (seg.AC) =
3
4
(seg.BC)2
Therefore AC=
3
2
(𝐵𝐶)

21. The 450 - 450 - 900 triangle
• If the angles of a triangle are 450 - 450 - 900 then the
perpendicular sides are
2
2
times the hypotenuse.
• To prove that AB = BC =
2
2
AC
• By Pythagoras theorem
• (seg.AB)2 + (seg.BC)2 = (seg.AC)2
• (seg.AB) = (seg.BC) . DABC is isosceles.
• (seg.AB)2 + (seg.BC)2 = 2(seg.AB)2 = (seg.AC)2
In D ABC  A = 450 ,  B = 900 and  C = 450.
Figure 6.6
(AB)2=1/2(AC)2
AB =
2
2
AC 2
and
BC =
2
2
AC 2

22. Exercises:
• Example 1
LMNO is a parallelogram such that m  LON = 300 and (seg.LO) = 12 cm.
If seg.MP is the perpendicular distance between seg.LM and seg.ON ,
find (seg.MP).

23. Exercises:
Since m  LON = m  MNP = 300
D MNP is a 300 - 600 - 900 triangle
(MP) = ½(MN)
(MN) = (LO) = 12cm
MP = 1/2 (12cm)
MP = 6cm.

24. Exercises:
• Example 2
D PQR is an acute triangle seg.PS is perpendicular to seg.QR and seg.PT
bisects QR.
Prove that (seg.PR)2 + (seg.PQ)2 = (seg.PT)2 + (seg.QT)2

25.  end 
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