Question

solve it plzzzzzz..., ​

Answer 1

Figure of the question :-

$\mathtt{\bf{\underline{\red{Given:-}}}}$

A circle C (0, r).

A tangent l at point A touches circle at point A.

$\mathtt{\bf{\underline{\green{To\:Proof:-}}}}$

OA ⊥ l

$\mathtt{\bf{\underline{\purple{Construction:-}}}}$

Take a point B on the tangent l & Join OB.

$\mathtt{\bf{\underline{\orange{Proof:-}}}}$

We know that, if the the line segment joins the radius to the tangent l . So, the perpendicular will become shortest as compared to l.

OA = OC (Radii of the same circle)

According to the figure,

OB = OC + BC.

∴ OB > OC

➝ OB > OA

➝ OA < OB

We know that any arbitrary point B on the tangent l.

So , we can say that OA is shorter line segment as compared to other line segments joining O to any point on l.

Hence, the tangent at any point of circle is perpendicular to the radius.

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Answer 2

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \large \sf \pink{Given :-}$

A circle C ( O, r ) and a tangent l at point A.

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$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \large\sf \blue{To \: prove :- }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: OA \perp l$

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$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \large \sf \orange{Cønstructïøn:- }$

Take a point B, other than A,on the tangent l.

Join OB . Suppose OB meets the circle in C.

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$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \huge \sf \purple{Prøøf :-}$

We know that, among all line segment joining the point O to a point on l, the perpendicular is shortest to l.

$\sf{OA = OC ( Radius \: of \: the \: same \: circle ) }$

$\sf{now, OB = OC + BC }$

$\therefore{OB > OC}$

$\longrightarrow{OB > OA}$

$\longrightarrow{OA < OB}$

B is an arbitrary point on the tangent l. Thus, OA is shorter than any other line segment joining O to any point on l .

$\sf{Here,{ OA \perp l }}$

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