This is of circle please answer in written as soon as possible
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This is very simple question.
See, u just have to remember the theorem that WHENEVER A PERPENDICULAR IS DRAWN TO THE CHORD FROM THE CENTRE, IT WILL DIVIDE THAT CHORD INTO TWO EQUAL HALVES.
Here CP and PD are chords and OA and OB are the perpendiculars respectively.
Now I can solve the question yourself without any problem...
See, u just have to remember the theorem that WHENEVER A PERPENDICULAR IS DRAWN TO THE CHORD FROM THE CENTRE, IT WILL DIVIDE THAT CHORD INTO TWO EQUAL HALVES.
Here CP and PD are chords and OA and OB are the perpendiculars respectively.
Now I can solve the question yourself without any problem...
shreya20041:
Butn there is nothing written that chords are equal
OK...Let me do it...
Given: Two circles O and O' intersect at P. ------ CD ll OO' ------ OA and O'B are the perpendiculars of CP and PD respectively. ----------To Prove: CD = 2 OO"---------PROOF: OA is perpendicular to CP so, CA = AP ( reason told earlier) Now AP+AP = CP ---> 2AP = = CP.......(this is equation i) Now same with another circle.......O'B I asked perpendicular to PD so, PB = BE (reason told earlier) so, PB+PB = PD that is 2PB = PD (this is equation ii) .............Now by usin
using both the equations we will get....... CD = CP + PD........OR CD = 2AP + 2PB.........that is CD = 2 (AP+PB).....and we can see that AP + PB = OO'.........So, by substituting the value of AP + PB we will get that CD = 2 (OO').......Hence Proved!!!!
Sorry, if you face any difficulty during reading these complicated comments.....but here I am typing for 29 minutes continuously....... : )
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