Question

AB and CD are chords of a circle with centre P. Ir AB = 16 cm, CD = 30 cm and
the distance of AB from centre P is 15 cm, find the distance of CD from centre p.​

Answer 1

Answer:

8cm

MARK AS BRAINLIEST

Step-by-step explanation:

try yourself.

Construction: Draw the radius from Centre P to A and C.

Now PA = PC  (radii of same circle)

Also a perpendicular drawn from the centre of a circle on a chord, bisects it.

if E is the point on AB, then AE = BE = 16/2 = 8cm

In triangle AEP

angle AEP is 90

AE = 8cm

PE = 15cm

By Pythagoras theorem,

AP squared = PE squared + AE squared

= 8^2 + 15^2

= 64 + 225

= 289

thus AP = 17cm

Now, let F be the point the perpendicular makes with the chord CD.

therefore angle CFP is 90 degrees

CF = 30/2 = 15

again use the same Pythagoras theorem,

CP squared = FP squared + CF squared

289 = FP squared + 225

FP = sqrt 289 - 225

= sqrt 64

=  8cm

THUS DISTANCE FROM CD TO P IS 8CM

Answer 2

Answer :- 8 cm

Explaination:-