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Relation between distance between 
the chord and the centre of the circle, 
length of the chord and radius
d l/2 
OB= radius = r 
OC = distance between the centre and 
chord= d 
AB = length of the chord =l 
BC = AB/2 =l/2 
O 
C B 
l
OBC Right angled triangle 
we can Apply Pythagoras 
theorem 
O 
d 
l/2 
O 
d 
C 
l/2 
C B 
If we do not know 
By Pythagoras theorem 
Where r is the hypotenuse of OBC 
B 
r 
풉풚풑풐풕풆풏풐풔풆ퟐ = 풂풍풕풊풕풖풅풆ퟐ + 풃풂풔풆ퟐ 
풉풚풑풐풕풆풏풐풔풆 = 풂풍풕풊풕풖풅풆ퟐ + 풃풂풔풆ퟐ 
OB= 푩푪ퟐ + 푶푪ퟐ 
= dퟐ + l/2 ퟐ r
O 
d 
l/2 
O 
d 
C 
l/2 
C B 
If we do not know d 
By Pythagoras theorem 
Where d is the altitude of OBC 
풉풚풑풐풕풆풏풐풔풆ퟐ = 풂풍풕풊풕풖풅풆ퟐ + 풃풂풔풆ퟐ 
풂풍풕풊풕풖풅풆ퟐ + 풃풂풔풆ퟐ = 풉풚풑풐풕풆풏풐풔풆ퟐ 
B 
풂풍풕풊풕풖풅풆ퟐ = 풉풚풑풐풕풆풏풐풔풆ퟐ − 풃풂풔풆ퟐ 
풂풍풕풊풕풖풅풆 = 풉풚풑풐풕풆풏풐풔풆ퟐ − 풃풂풔풆ퟐ 
푶푪 = 푶푩ퟐ − 푩푪ퟐ 
= rퟐ − l/2 ퟐ d
O 
d 
l/2 
O 
d 
C 
l/2 
C B 
If we do not know 
l/2 
By Pythagoras theorem 
Where l/2 is the base of OBC 
풉풚풑풐풕풆풏풐풔풆ퟐ = 풂풍풕풊풕풖풅풆ퟐ + 풃풂풔풆ퟐ 
B 
풃풂풔풆ퟐ = 풉풚풑풐풕풆풏풐풔풆ퟐ − 풂풍풕풊풕풖풅풆ퟐ 
풃풂풔풆 = 풉풚풑풐풕풆풏풐풔풆ퟐ − 풂풍풕풊풕풖풅풆ퟐ 
푩푪 = 푶푩ퟐ − 푶푪ퟐ 
= rퟐ − dퟐ 
l/2
rퟐ − l/2 ퟐ = 
= rퟐ − dퟐ 
O 
d 
C 
l/2 
B
REVIEW AND APPLICATION 
8CM 
P Q 
A B 
16CM 
AB is the diameter 
PQ is a chord 
AB PQ 
Area of ABPQ ?
8CM 
16CM 
P 
Trapezium 
a 
h 
b 
Area=h/2(a+b) 
a=16cm 
b=8cm 
h = ?
P 8CM 
Q 
o 
A B 
16CM 
To find ‘h’ 
P 
4 cm T 
O 
Where TO is the 
height of the 
trapezium also 
the distance 
from the centre 
to chord. 
= rퟐ − l/2 ퟐ d 
풅 = rퟐ − l/2 ퟐ 
= 8ퟐ − (8/2) ퟐ 
= ퟔퟒ − ퟏퟔퟐ 
= ퟒퟖ 
= ퟐ ퟑ cm
8CM 
16CM 
P 
Trapezium 
a=8 cm 
h=2 ퟑ 
b = 16cm 
Area=h/2(a+b) 
= ퟐ ퟑ /2(8+16) 
= ퟑ (ퟐퟒ) 
=24 ퟑ cmퟐ a=16cm 
b=8cm 
h = ퟐ ퟑ
Thank you 
Submitted by , 
priya

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Circles 1

  • 1. Relation between distance between the chord and the centre of the circle, length of the chord and radius
  • 2. d l/2 OB= radius = r OC = distance between the centre and chord= d AB = length of the chord =l BC = AB/2 =l/2 O C B l
  • 3. OBC Right angled triangle we can Apply Pythagoras theorem O d l/2 O d C l/2 C B If we do not know By Pythagoras theorem Where r is the hypotenuse of OBC B r 풉풚풑풐풕풆풏풐풔풆ퟐ = 풂풍풕풊풕풖풅풆ퟐ + 풃풂풔풆ퟐ 풉풚풑풐풕풆풏풐풔풆 = 풂풍풕풊풕풖풅풆ퟐ + 풃풂풔풆ퟐ OB= 푩푪ퟐ + 푶푪ퟐ = dퟐ + l/2 ퟐ r
  • 4. O d l/2 O d C l/2 C B If we do not know d By Pythagoras theorem Where d is the altitude of OBC 풉풚풑풐풕풆풏풐풔풆ퟐ = 풂풍풕풊풕풖풅풆ퟐ + 풃풂풔풆ퟐ 풂풍풕풊풕풖풅풆ퟐ + 풃풂풔풆ퟐ = 풉풚풑풐풕풆풏풐풔풆ퟐ B 풂풍풕풊풕풖풅풆ퟐ = 풉풚풑풐풕풆풏풐풔풆ퟐ − 풃풂풔풆ퟐ 풂풍풕풊풕풖풅풆 = 풉풚풑풐풕풆풏풐풔풆ퟐ − 풃풂풔풆ퟐ 푶푪 = 푶푩ퟐ − 푩푪ퟐ = rퟐ − l/2 ퟐ d
  • 5. O d l/2 O d C l/2 C B If we do not know l/2 By Pythagoras theorem Where l/2 is the base of OBC 풉풚풑풐풕풆풏풐풔풆ퟐ = 풂풍풕풊풕풖풅풆ퟐ + 풃풂풔풆ퟐ B 풃풂풔풆ퟐ = 풉풚풑풐풕풆풏풐풔풆ퟐ − 풂풍풕풊풕풖풅풆ퟐ 풃풂풔풆 = 풉풚풑풐풕풆풏풐풔풆ퟐ − 풂풍풕풊풕풖풅풆ퟐ 푩푪 = 푶푩ퟐ − 푶푪ퟐ = rퟐ − dퟐ l/2
  • 6. rퟐ − l/2 ퟐ = = rퟐ − dퟐ O d C l/2 B
  • 7. REVIEW AND APPLICATION 8CM P Q A B 16CM AB is the diameter PQ is a chord AB PQ Area of ABPQ ?
  • 8. 8CM 16CM P Trapezium a h b Area=h/2(a+b) a=16cm b=8cm h = ?
  • 9. P 8CM Q o A B 16CM To find ‘h’ P 4 cm T O Where TO is the height of the trapezium also the distance from the centre to chord. = rퟐ − l/2 ퟐ d 풅 = rퟐ − l/2 ퟐ = 8ퟐ − (8/2) ퟐ = ퟔퟒ − ퟏퟔퟐ = ퟒퟖ = ퟐ ퟑ cm
  • 10. 8CM 16CM P Trapezium a=8 cm h=2 ퟑ b = 16cm Area=h/2(a+b) = ퟐ ퟑ /2(8+16) = ퟑ (ퟐퟒ) =24 ퟑ cmퟐ a=16cm b=8cm h = ퟐ ퟑ
  • 11. Thank you Submitted by , priya