Question

Calculate the length of a chord which is at a distance 12cm from the centre of a circle of radius
13cm. ​

Answer 1

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$\odot \: \: \: \: \: \: \frak{ \color{gold}{Length \: of \: the \: Chord = 10cm}}$

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Step-by-step explanation:

Given :

• The distance from the centre of the circle to the chord is 12 cm
• The radius of the circle is 13 cm

To Find :

• the length of the chord

Solution :

$\underline{ \bf{ \dag \: as \: we \: know}}$

$\\ \boxed{ \pink{ \sf{Pythagoras \: Theorem \implies AC² = AB² + BC²}}}$

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$\\ \dashrightarrow \sf OA² = AC² + CO²$

$\\ \dashrightarrow \sf{13}^{2} = AC² + {12 }^{2}$

$\\ \dashrightarrow \sf169 = AC² + 144$

$\\ \dashrightarrow \sf \: AC² = 169 - 144$

$\\ \sf \dashrightarrow \: AC² = 25$

$\\ \sf \dashrightarrow \: AC = \sqrt{25 }$

$\\ \sf \dashrightarrow \: AC = 5$

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C is the mid point of AB

$\therefore \bf AC = \frac{1}{2} AB$

$\\ \to \tt \: 5 = \frac{1}{2} AB$

$\\ \tt \to \: AB = 5 \times 2$

$\\ \to \large{ \underline{ \boxed{ \purple{ \frak{10cm}}}} \star}$

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$\underline{ \text{Therefore, the Length of the chord = 10cm}}$

Answer 2

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\odot \: \: \: \: \: \: \frak{ \color{gold}{Length \: of \: the \: Chord = 10cm}}⊙LengthoftheChord=10cm

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Step-by-step explanation:

Given :

The distance from the centre of the circle to the chord is 12 cm

The radius of the circle is 13 cm

To Find :

the length of the chord

Solution :

\underline{ \bf{ \dag \: as \: we \: know}}

†asweknow

\begin{gathered} \\ \boxed{ \pink{ \sf{Pythagoras \: Theorem \implies AC² = AB² + BC²}}}\end{gathered}

PythagorasTheorem⟹AC²=AB²+BC²

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\begin{gathered} \\ \dashrightarrow \sf OA² = AC² + CO² \\ \\ \end{gathered}

⇢OA²=AC²+CO²

\begin{gathered} \\ \dashrightarrow \sf{13}^{2} = AC² + {12 }^{2} \\ \\ \end{gathered}

⇢13

2

=AC²+12

2

\begin{gathered} \\ \dashrightarrow \sf169 = AC² + 144 \\ \\ \end{gathered}

⇢169=AC²+144

\begin{gathered} \\ \dashrightarrow \sf \: AC² = 169 - 144 \\ \\ \end{gathered}

⇢AC²=169−144

\begin{gathered} \\ \sf \dashrightarrow \: AC² = 25 \\ \\ \end{gathered}

⇢AC²=25

\begin{gathered} \\ \sf \dashrightarrow \: AC = \sqrt{25 } \\ \\ \end{gathered}

⇢AC=

25

\begin{gathered} \\ \sf \dashrightarrow \: AC = 5 \\ \\ \end{gathered}

⇢AC=5

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C is the mid point of AB

\therefore \bf AC = \frac{1}{2} AB∴AC=

2

1

AB

\begin{gathered} \\ \to \tt \: 5 = \frac{1}{2} AB \\ \end{gathered}

→5=

2

1

AB

\begin{gathered} \\ \tt \to \: AB = 5 \times 2 \\ \end{gathered}

→AB=5×2

\begin{gathered} \\ \to \large{ \underline{ \boxed{ \purple{ \frak{10cm}}}} \star} \\ \end{gathered}

→

10cm

⋆

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\underline{ \text{Therefore, the Length of the chord = 10cm}}

Therefore, the Length of the chord = 10cm