In the given figure, AD is a diameter of a circle and AB is a chord. If AD=34cm,AB=30 cm. Then, which of the following is the distance of AB from the center of the circle?

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A
$17 cm$
B
$15 cm$
C
$4 cm$
D
$8 cm$

Solution:

Consider the given figure,

From the figure,
Diameter,

AD = 34

cm.
Chord,

AB = 30

cm.

∴

Radius,

AO = \frac{1}{2} AD

AO = \frac{1}{2} \times 34

AO = 17

cm.

∵ ON ⊥ AB

∴ ON

bisects

AB

N

and

∠ ANO = 90^{0}

.
So,

AN = \frac{1}{2} AB

AN = \frac{1}{2} \times 30

AN = 15

cm
Now,

AO = 17

AN = 15

∠ ANO = 90^{0}

∴ ΔAON

is a right triangle with hypotenuse

AO .

By Pythagoras theorem,

{AO}^{2} = {ON}^{2} + {AN}^{2}

{AO}^{2} - {AN}^{2} = {ON}^{2}

\Rightarrow {ON}^{2} = {(17}^{2} {- 15}^{2}) {cm}^{2}

\Rightarrow {ON}^{2} = (289 - 225) {cm}^{2}

\Rightarrow {ON}^{2} = 64 {cm}^{2}

\Rightarrow ON = 8

cm
Hence,

AB

is at a distance of

8

cm from the center

O

.
Therefore, option 4 is correct.

In the given figure, $AD$ is a diameter of a circle and $AB$ is a chord. If $AD = 34 cm, AB = 30 cm$ . Then, which of the following is the distance of $AB$ from the center of the circle?

Solution:

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