# Question is attached . Plz answer the question

2
by Sanalkumar433

2016-02-13T21:24:13+05:30
Construction: Radii OA, OB, OC and OD.

In triangle OAQ,
=> Angle OQA = 90
=> OQ = 6cm (distance between AB and O)
=> AQ = AB/2 = 16/2 = 8cm
(Theorem: If a line is drawn perpendicular to a chord on the same circle, it bisects the chord)
Therefore, by Pythagoras theorem,
=> AO = (OQ^2 + AQ^2)^1/2
= (6^2+8^2)^1/2
=> (100)^1/2
=> 10cm

In triangle OPC,
=> OC = 10cm
=> CP = CD/2 = 12/2 = 6cm (same theorem as earlier)
Therefore, by Pythagoras theorem,
=> OC^2 = OP^2+CP^2
=> OP^2 = OC^2 - CP^2
=> OP = (100-36)^1/2 = (64)^1/2 = 8cm

Which is also perpendicular distance of CD from centre.
great
Thanks!
mst welcme
• Brainly User
2016-02-13T21:33:04+05:30
Join OD...now we know that perpendicular drawn from the centre to the chord bisects the chord..so CD=1/2PD=6cm..so IN Triangle OPD...let PQ be x..so by using Pythagoras theorem... we have..ODsq.=POsq.+PDsq...=>ODsq.=(x+6)^2 +6^2=x^2+36+12x+36=x^2+72+12x...let this be equation (1)...now..join OB..so in Triangle..OBQ..same way using Pythagoras theorem.. we have...OBsq.=8^2+6^2=64+36=100..so..OB=10cm..now...radius=10cm..=OB=OD..so putting the value of OB in equation(1)..so..100=x^2+72+12x=>0=x^2+12x-28=>0=x^2+14x-2x-28=>(x+14)(x-2)so..x=-14 and x=2..so...PO=2+6=8cm