Question

answer my questions please I need to know

Answer 1

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.


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