Question

Find the length of a chord which is at a distance of
5 cm from the centre of a circle of radius 10 cm.​

Answer 1

Step-by-step explanation:

hey friend your answer is above

Answer 2

$\huge\sf\pink{Answer}$

☞ Length of chord = 17.32 cm

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$\huge\sf\blue{Given}$

✭ Radius of the circle (OA) = 10 cm

✭ OC = 5 cm

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$\huge\sf\gray{To \:Find}$

◈ Length of chord?

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$\huge\sf\purple{Steps}$

$\large\sf \star \ Diagram \ \star$

$\setlength{\unitlength}{1mm}\begin{picture}(50,55)\thicklines\qbezier(25.000,10.000)(33.284,10.000)(39.142,15.858)\qbezier(39.142,15.858)(45.000,21.716)(45.000,30.000)\qbezier(45.000,30.000)(45.000,38.284)(39.142,44.142)\qbezier(39.142,44.142)(33.284,50.000)(25.000,50.000)\qbezier(25.000,50.000)(16.716,50.000)(10.858,44.142)\qbezier(10.858,44.142)( 5.000,38.284)( 5.000,30.000)\qbezier( 5.000,30.000)( 5.000,21.716)(10.858,15.858)\qbezier(10.858,15.858)(16.716,10.000)(25.000,10.000)\put(10,17){\line(5,0){30}}\put(25,30){\circle*{1}}\put(7,25){\sf\large{10 cm}}\put(25,17){\line(0,2){13}}\qbezier(10,17)(20,26)(25,30)\put(26,22){\sf 5 cm}\put(26,30){\sf O}\put(6,16){\sf A}\put(23,13){\sf C}\put(42,16){\sf B}\end{picture}$

Let O be the centre of the circle whose radius is 10 cm

Let AB be a chord of the circle. Then,

$\sf{\dashrightarrow AC = CB \:and \: OC \perp AB }$

Applying Pythagoras theorem in ∆OCA

$\underline{\boxed{\sf(Hypotenuse)^2 = (Base)^2 + (Perpendicular)^2}}$

$\bullet\underline{\textsf{As Per The Question}}$

Substituting the given values,

$\sf{\dashrightarrow (OA)^2 = (AC)^2 + (OC)^2 }$

$\sf{\dashrightarrow (10)^2 = (AC)^2 + (5)^2}$

$\sf{\dashrightarrow 100 = AC^2 + 25 }$

$\sf{\dashrightarrow 100 - 25 = AC^2}$

$\sf{\dashrightarrow \sqrt{75} = AC}$

$\red{\sf{\dashrightarrow 8.66 \: cm = AC }}$

$\sf{\dashrightarrow AB = AC + CB }$

$\sf{\dashrightarrow AB = 8.66 + 8.66}$

$\sf\orange{\dashrightarrow AB = 17.32 \: cm }$

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