Circular badges are distributed in a school with a message of saving water. Each badge is designed by drawing two intersecting chords equidistant from the centre. Show that the segments of one chord are respectively equal to the segments of the other chord.
Which value is depicted through this question?
Explain with complete calculations & justifications.
Points : 20 ☺
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Answered by
3
It is given that chords are equidistant from the centre so, according to the theorem
the chords are equal
As both chords are in the same circle and of same length
that's why we can say that segment formed by both chords are equal
value depicted is caring about water
the chords are equal
As both chords are in the same circle and of same length
that's why we can say that segment formed by both chords are equal
value depicted is caring about water
GovindKrishnan:
Thanks for helping! ☺
Answered by
8
AB and CD are cords of a circle with center O . AB and CD intersect at P and AB = CD
TO PROVE :
(1) AP = PD
(2)PB = CP
CONSTRUCTION :
draw OM perpendicular to AB , ON perpendicular to CD . join OP.
PROOF:
AM = MB = 1/2 x AB
[ perpendicular from center insects the chord]
CN =ND = 1/2 x CD
[perpendicular from center insects the chord]
AM = ND -----------(1)
And
MB = CN-----------(2)
[ AB =CD]
In triangle OMP and triangles ONP
OM = ON
[ equal chords of a circle are equidistant from the center]
/_OMP = /_ONP [each 90°]
OP = OP[common]
therefore
OMP is congruent to ONP
[by RHS axiom]
MP =PN [cpct]--------------(3)
by adding (1) and (3) :-
AM + MP = ND + PN
AP = PD
by subtracting (3) from (1)
MB - MP =CN - PN
PB = CP
therefore
AP =PD
And
PB = CP
(proved)
_______________________
the value of saving water is depicted through this question
TO PROVE :
(1) AP = PD
(2)PB = CP
CONSTRUCTION :
draw OM perpendicular to AB , ON perpendicular to CD . join OP.
PROOF:
AM = MB = 1/2 x AB
[ perpendicular from center insects the chord]
CN =ND = 1/2 x CD
[perpendicular from center insects the chord]
AM = ND -----------(1)
And
MB = CN-----------(2)
[ AB =CD]
In triangle OMP and triangles ONP
OM = ON
[ equal chords of a circle are equidistant from the center]
/_OMP = /_ONP [each 90°]
OP = OP[common]
therefore
OMP is congruent to ONP
[by RHS axiom]
MP =PN [cpct]--------------(3)
by adding (1) and (3) :-
AM + MP = ND + PN
AP = PD
by subtracting (3) from (1)
MB - MP =CN - PN
PB = CP
therefore
AP =PD
And
PB = CP
(proved)
_______________________
the value of saving water is depicted through this question
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