avni38
AB=CD=16cm chord AB=chord CD By theorem, Congruent chords are equidistant from the centre. therefore OM=ON...1 ON is perpendicular to chord CD CN=half of CD(perpendicular from the centre to chord bisects the chord.) CN=half 10 therefore CN =8cm Radius=OC=10cm In right angled triangle ONC OC sq.=ON sq.+CN sq. 10sq.=ON sq. + 8sq. 100=ON sq.+64 ON sq.=100-64 ON sq. =36cm therefore ON=6cm therefore distance of these chords from the centre is 6cm
it's have a construction ... draw OM perpendicular to AB and ON perpendicular to CD
2. construction.. Draw IM perpendicular to AB and ON perpendicular to CD Seg OM is perpendicular to AB Seg ON perpendicular to seg CD chord AB = chord CD. hence OM =ON = 5cm(congruent chords are equidistant from centre) radius=OA =13cm In right angled triangle OMN OA sq.=ON sq.+AM sq. 13 sq.=5 sq. +AM sq. 169= 25 + AM sq. AM sq. = 169-25 Am sq. = 144 AM = 12cm Now, AM =half of AB (perpendicular drawn from centre to chord bisects the chord) therefore half of AB = 12 cm.
3. Construction.. Draw seg CM and seg CM Proof...In triangle CPM and triangle CPN seg CM = seg CM (radius of same circle ) seg CP= seg CP (common side) seg PM = seg PN..(given ) therefore triangle CPN= triangle CPM.(SSS test) therefore angle CPM = angle CPN ( C a C t) hence ray PC bisects angle MPN. Hence proved.
Answers & Comments
chord AB=chord CD
By theorem,
Congruent chords are equidistant from the centre.
therefore OM=ON...1
ON is perpendicular to chord CD
CN=half of CD(perpendicular from the centre to chord bisects the chord.)
CN=half 10
therefore CN =8cm
Radius=OC=10cm
In right angled triangle ONC
OC sq.=ON sq.+CN sq.
10sq.=ON sq. + 8sq.
100=ON sq.+64
ON sq.=100-64
ON sq. =36cm
therefore ON=6cm
therefore distance of these chords from the centre is 6cm
it's have a construction ... draw OM perpendicular to AB and ON perpendicular to CD
2. construction.. Draw IM perpendicular to AB and ON perpendicular to CD
Seg OM is perpendicular to AB
Seg ON perpendicular to seg CD
chord AB = chord CD.
hence OM =ON = 5cm(congruent chords are equidistant from centre)
radius=OA =13cm
In right angled triangle OMN
OA sq.=ON sq.+AM sq.
13 sq.=5 sq. +AM sq.
169= 25 + AM sq.
AM sq. = 169-25
Am sq. = 144
AM = 12cm
Now,
AM =half of AB (perpendicular drawn from centre to chord bisects the chord)
therefore half of AB = 12 cm.
3. Construction.. Draw seg CM and seg CM
Proof...In triangle CPM and triangle CPN
seg CM = seg CM (radius of same circle )
seg CP= seg CP (common side)
seg PM = seg PN..(given )
therefore triangle CPN= triangle CPM.(SSS test)
therefore angle CPM = angle CPN ( C a C t)
hence ray PC bisects angle MPN.
Hence proved.
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