Question

Give this answer please ​

Answer 1

Answer:

Given−Oisthecentreofacircle,inscribedinaquadrilateral

ABCD.

Theradiusofthecircleis10cm.

∠BAD=90

o

.

AB,BC,CD&ADtouchtheinscribedcircleatP,Q,R&S

respectively.

CR=27cmandBC=38cm.

Tofindout−

AB=x=?

Solution−

WejoinOS&OP.ThenOS&OPareradiiofthe

inscribedcirclethroughthepointsofcontactofthetangents

AS&APrespectively.

∴OS=OP=10cm.

Again∠OSA=90

o

=∠OPAsincetheradiusthroughthepointof

contactofatangenttoacircleisperpendiculartothetangent.

NowinOSAP,∠OSA=90

o

=∠OPAandOS=OP=10cm.

∴OSAPisasquareofside10cm.

SoAP=OS=10cm.

AgainCR=CQ=27cmandBQ=BPsincethelengthsofthetangents,

fromapointtoacircle,areequal.

∴BQ=BC−CQ=(38−27)cm=11cm.

AlsoBP=BQ=11cm.

∴AB=x=AP+BP=(10+11)cm=21cm.

Ans−OptionD.

Answer 2

Answer:

thanks for freepoints........