Question

Inthe given figre CE AND DE are equal chords of a circle with center O .if angle AOB= 90 FIND ar [triangle CED]:ar [triangle AOB]

Answer 1

Here OA=OB=Radius ⇒∆AOBisanisoscelestriangle. ⇒∠OAB=∠ABO=x ⇒∠AOB+∠OAB+∠ABO=180 ⇒90+x+x=180 ⇒2x=90 ⇒x=45 ⇒∠OAB=∠ABO=45 Also CE=DE(Given) And∠CED=90 (Angleinasemi−circle.) Hence,in∆ECD ⇒∠ECD=∠EDC=45 CD=2r EO=r In∆AOF AO=r sin45= OF r ⇒OF=rsin45 ⇒OF= r 2 √ cos45= AF r ⇒AF=rsin45 ⇒AF= r 2 √ AsperpendicularfromOwillbisect AB,hence AB=2. r 2 √ = 2 √ r Ar(∆CED) Ar(∆AOB) = 1 2 ×CD×EO 1 2 ×AB×FO = CD×EO AB×FO = 2r.r r 2 √ . 2 √ r ⇒ Ar(∆CED) Ar(∆AOB) =2(Answer)