Question

Answer me question no 18?

Answer 1

Answer:

11 cm

Step-by-step explanation:

AB, BC, CD and AD are tangents to the circle with centre O at Q, P,S and R.

Given that AB = 29 cm, AD = 23 cm, DS = 5 cm and ∠B = 90°.

The length of tangents drawn from an external point to a circle are equal.

∴ DS = DR = 5 cm.

AR = AD - DR

        = 23 - 5

        = 18 cm.

∴ AQ = AR = 18 cm.

QB = AB - AQ

     = 29 - 18

     = 11 cm .

∴ QB = BP = 11 cm.

Since, OQBP is a square.

∴ QB = BP = r = 11 cm.

Therefore, radius of circle = 11 cm.

Hope it helps!

Answer 2

proof- AD=23
AB=29
DS=5

AR=AQ -( TANGENT FROM
DS=DR EXTERNAL POINTS
BQ=BP OF A CIRCLE ARE EQUAL)

THEREFORE : AR=AD-DS
AR= 23-5= 18.
SO, AR=AQ=18 cm
BQ=AB-AR
BQ= 29-18=11 cm
SO, BQ=BP= 11 cm

ANGLE BPO AND ANGLE BPQ= 90 (TANGENT TO ANY POINT OF A CIRCLE IS PERPENDICULAR TO RADIUS)

SO NOW IN QUADRILATERAL BPOQ
ANGLE B+ ANGLE P + ANGLE O + ANGLEQ= 360
90+90+ ANGLE O + 90= 360
270+ ANGLE O = 360
ANGLE O= 360 - 270
ANGLE O = 90

THEREFORE : QUADRILATERAL BPOQ IS SQUARE
SO RADIUS = 11 cm

HENCE PROVE

I HOPE IT WILL HELP YOU....