Question

In figure point B point Q and point P are the points of contact of the tangents of the circle seg AP || seg BR ,AP =12.5 , BR=8 find the radius ​

Answer 1

Answer:

radius of circle=11.52cm (approx)

Step-by-step explanation:

by construction join OQ

now by theorem

AP=PQ=12.5cm

BR=RQ=8cm

now in triangle POQ, angleQ=90

so

OPsquare=PQsquare+OQsquare

OQsquare=OPsquare - 156.25 equ 1

similarly in triangle OQR,

OQsquare=ORsquare - 64 equ 2

now,

PR=PQ+QR=12.5+8=20.5 cm

in Triangle POR,angle POR=90 by a theorem

so OPsqaure+ORsquare=420.25

also

ORsquare=Opsquare - 156.25 by equatting equ 1 & 2

so OPsquare+OPsquare - 156.25=420.25

OP=17cm (approx)

now

OQsquare

=289 - 156.25 (in triangle POQ)

=132.75

now radius

=OQ=11.52 cm