Question

A tangent AB at a point A of a circle of radius 6 cm meets a line through the centre O at a point B . If OB = 10 cm , find the length of AB

Answer 1

OA is radius i.e, 6cm

Line through O meets tangent from A at B

Angle OAB =90° [AB is tangent]

By Pythagoras theorem:-

OB = 10cm (given)

OA^2 + AB^2 = OB^2

Answer 2

Answer:

The length of AB is 8cm.

Explanation:

Given:

• A tangent AB at a point A of a circle of radius 6 cm meets a line through the center O at a point B.
• OB = 10 cm

To find:

the length of AB.

Step 1

AB exists as a tangent

OA be the radius of the circle.

OA = 6cm

The line through O meets tangent from A at B

Angle OAB = 90° [AB is tangent]

Step 2

By using Pythagoras theorem, we get

OB = 10cm (given)

$OA^2 + AB^2 = OB^2$

$\angle O A B=90^{\circ}$

OA = 6 cm = radius

OB = 10 cm

Step 3

In $\triangle A B O$.

$\angle A=90^{\circ}$

By using Pythagoras theorem, we get

$A O^{2}+A B^{2}=O B^{2}$

$6^{2}+A B^{2}=10^{2}$

$A B^{2}=100-36$

$A B=\sqrt{64}$

AB = 8cm

Therefore, the length of AB is 8cm.

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