Length of a tangent segment drawn from a point which is at a distance 12.5 cm from the centre of a circle is 12 cm, find the diameter of the circle.
25 cm
24 cm
7 cm
14 cm
Chapter 3 – Circle – Text Book Solution
Problem Set 3 | Q 1.04 | Page 83
Four alternative answers for the following question is given. Choose the correct alternative.
Length of a tangent segment drawn from a point which is at a distance 12.5 cm from the centre of a circle is 12 cm, find the diameter of the circle.
- 25 cm
- 24 cm
- 7 cm
- 14 cm
solution
Let O be the centre of the circle and AB be the tangent segment drawn from an external point A touching the circle at B.
The tangent at any point of a circle is perpendicular to the radius through the point of contact.
∴ ∠ABO = 90º
In right ∆ABO,
[{OA}^2 = {AB}^2 + {OB}^2 ]
[ Rightarrow OB = sqrt{{OA}^2 – {AB}^2}]
[ Rightarrow OB = sqrt{left( 12 . 5 right)^2 – left( 12 right)^2}]
[ Rightarrow OB = sqrt{156 . 25 – 144}]
[ Rightarrow OB = sqrt{12 . 25} = 3 . 5 cm]
Radius of the circle = OB = 3.5 cm
∴ Diameter of the circle = 2 × Radius of the circle = 2 × 3.5 = 7 cm
Hence, the correct answer is 7 cm .
Explanation:-
We have a circle with centre O, and AB is a tangent segment drawn from an external point A touching the circle at B.
Since AB is tangent to the circle at point B, the tangent at any point of a circle is perpendicular to the radius through the point of contact. Therefore, we have ∠ABO = 90º.
In right ∆ABO, using the Pythagorean theorem, we have:
[{OA}^2 = {AB}^2 + {OB}^2]
Substituting the values given in the problem, we get:
[ Rightarrow OB = sqrt{{OA}^2 – {AB}^2}]
[ Rightarrow OB = sqrt{left( 12 . 5 right)^2 – left( 12 right)^2}]
[ Rightarrow OB = sqrt{156 . 25 – 144}]
[ Rightarrow OB = sqrt{12 . 25} = 3 . 5 cm]
Therefore, the radius of the circle is OB = 3.5 cm.
The diameter of the circle is twice the radius, so diameter of the circle = 2 × Radius of the circle = 2 × 3.5 = 7 cm.
Hence, the correct answer is 7 cm.
Chapter 3 – Circle – Text Book Solution
Problem Set 3 | Q 1.04 | Page 83
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