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]
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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)
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