Question

AD is diameter of a circle, O being the centre and
AB is a chord. Let the centre of AB be denoted by
M, then find OM

options are
8cm
5cm
7cm
6cm​

Answer 1

Answer:

OX is the perpendicular distance from centre O to the chord AB

∴AX=XB

Now,

AX+XB=AB

⇒2AX=AB

⇒2AX=30

⇒AX=

2

30

⇒AX=15

Again,

AO+OD=AD

⇒2AO=AD

⇒2AO=34

⇒AO=17

△AOX is a right angled triangle

∴AO

2

=OX

2

+AX

2

⇒17

2

=OX

2

+15

2

⇒OX

2

=17

2

−15

2

⇒OX

2

=

64

⇒OX=8cm

right answer is 8 cm

Answer 2

Given:

AD is the diameter of a circle, O being the circle.

AB is the chord of the circle with center M.

To Find:

OM

Solution:

It is given that AD = 34cm and AB = 30cm

AOM is a right-angled triangle.

Now,

AO = OD = 34/2 = 17cm [at equal distance from the center of the circle]

OP is perpendicular to AB, and we know that when a perpendicular is drawn to the chord from it bisects the chord.

Then according to the statement AM = 30/2

                                                        AM = 15cm

So, the measure of AM = 15cm

Now, AOM is a right-angled triangle so by using Pythagoras theorem

⇒ h² = b² + p²

⇒ OM² + MA² = OA²

Removing the squares by using roots

⇒ OM = √(OA²- MA²

Substituting the values

⇒ OM = √ 17² - 15²

⇒ OM = √2²

⇒ OM = 8cm

Therefore, the measure of OM is 8cm.