In the figure given above, $A D$ is a straight line, $O P$ perpendicular to $A D$ and O is the centre of both circles. If $O A = 20 c m, O B = 15 c m$ and $O P = 12 c m$ . what is $A B$ equal to?

7 c m

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8 c m

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10 c m

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12 c m

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Solution

The correct option is A $7 c m$
$B P^{2} = O B^{2} - O P^{2} = 15^{2} - 12^{2} = 81$
$∴ B P = 9 c m$
$A P^{2} = O A^{2} - O P^{2} = 20^{2} - 12^{2} = 256$
$∴ A P = 16 c m$
$\Rightarrow A B = A P - B P = (16 - 9) c m = 7 c m$

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In the figure given above, AD is a straight line, OP perpendicular to AD and O is the centre of both circles. If OA=20 cm,OB=15 cm and OP=12 cm. what is AB equal to?

The correct option is A 7 cmBP2=OB2−OP2=152−122=81∴BP=9 cmAP2=OA2−OP2=202−122=256∴AP=16 cm⇒AB=AP−BP=(16−9)cm=7 cm

In the figure given above, $A D$ is a straight line, $O P$ perpendicular to $A D$ and O is the centre of both circles. If $O A = 20 c m, O B = 15 c m$ and $O P = 12 c m$ . what is $A B$ equal to?

The correct option is A $7 c m$
$B P^{2} = O B^{2} - O P^{2} = 15^{2} - 12^{2} = 81$
$∴ B P = 9 c m$
$A P^{2} = O A^{2} - O P^{2} = 20^{2} - 12^{2} = 256$
$∴ A P = 16 c m$
$\Rightarrow A B = A P - B P = (16 - 9) c m = 7 c m$