Question

in fig seg AB is the chord of circle with centre o seg pm perpendicular to chord AB If AM=5cm Find AB

Answer 1

Answer:

Radius of circle=PB=25 cm

AB=48 cm

We know that the perpendicular drawn from center of circle to the chord is always bisect the chord.

Therefore, AM=MB=\frac{1}{2}(48)=24 cm

2

1

(48)=24cm

In triangle PMB

PB^2=PM^2+MB^2PB

2

=PM

2

+MB

2

Using Pythagoras theorem

(hypotenuse)^2=(Base)^2+(perpendicular\;side)^2(hypotenuse)

2

=(Base)

2

+(perpendicularside)

2

Substitute the values

(25)^2=(24)^2+PM^2(25)

2

=(24)

2

+PM

2

625=576+PM^2625=576+PM

2

PM^2=625-576PM

2

=625−576

PM^2=49PM

2

=49

PM=\sqrt{49}=7 cmPM=

49

=7cm

Hence, PM=7 cm