Question

14. In the figure, quadrilateral ABCD is circumscribing a circle with centre O an
AD 1 AB. If radius of the incircle is 10cm, then find the value of x?
[3]
27 cm
38 am
P
x cm
15. In the given figure, PA and PB are tangents to a circle with centre O drawn
from an external point P. QR is a tangent at point S. Prove that PQ+ QS =P​

Answer 1

Step-by-step explanation:

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.