Question

In each of the following figures. O is the centre of the circle. Find the values of x and y.
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Answer 1

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Finding value of x : -

OAB is a right angled triangle.

So , by using pythagoras theoram

OB ² = AO ² + AB²

x² = 30² + 16²

x² = 900 + 256

x² = 1156

√x² = √1156

x = 34

┄───➤ $\sf{x \ = \ 34 \ cm}$

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Finding value of y : -

AB = AC [ Tangents to B. ]

So , AC = 16

┄───➤ $\sf{AC \ = \ y \ = \ 16 \ cm}$

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Answer 2

❥︎Given :-

• A circle with centre 0, radius 30 cm.
• AB and BC are tangents from external point and AB = 16 cm.

❥︎To Find :-

• The values of x and y

❥︎Concept Used :-

⟶Pythagoras Theorem

$\begin{gathered}{\boxed{\bf{\pink{(Hypotenuse)^2 = (Base)^2 + (Perpendicular)^2}}}}\end{gathered}$

⟶Length of tangents drawn from external point are equal.

❥︎Solution :-

⟶Step :- 1. To find y

Since, AB & BC are tangents drawn to a circle from external point.

$\bf\implies \:AB = BC$

$\bf\implies \:y = 16 \: cm$

⟶ Step : - 2. To find y

❥︎ Now, In right triangle OAB

Using Pythagoras Theorem

$\bf \:⟶ {AB}^{2} + {OA}^{2} = {OB}^{2}$

$\bf\implies \: {16}^{2} + {30}^{2} = {x}^{2}$

$\bf\implies \: 256 + 900 = {x}^{2}$

$\bf\implies \: {x}^{2} = 1156$

$\bf\implies \: {x}^{2} = {34}^{2}$

$\bf\implies \:x = 34 \: cm$

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