Question

If the length of a chord of a circle is 16 cm and the distance of the chord from the centre is 6 cm, find the radius of the circle

Answer 1

the radius of centre circle is 12 cm

Answer 2

Given: The length of a chord of a circle is 16 cm and the distance of the chord from the centre is 6 cm.

To find: Radius of the circle.

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$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\qbezier(2.3,0)(2.121,2.121)(0,2.3)\qbezier(-2.3,0)(-2.121,2.121)(0,2.3)\qbezier(-2.3,0)(-2.121,-2.121)(0,-2.3)\qbezier(2.3,0)(2.121,-2.121)(-0,-2.3)\put(-1.9,-1.3){\line(1,0){3.85}}\qbezier(1.95,-1.3)(1.95,-1.3)(0,0.2)\put(0,0.2){\line(0,-1){1.5}}\put(-0.1,0.5){\bf R}\put(-2.5,-1.5){\bf S}\put(2.3,-1.5){\bf J}\put(-0.15,-1.75){\bf A}\put(-1.9,-2.7){\vector(1,0){3.85}}\put(-1.9,-2.7){\vector(-1,0){0}}\put(-0.5,-3.2){\bf 16\ cm}\put(-1.2,-0.63){\bf 6\ cm}\end{picture}$

Draw, RA ⊥ SJ

$\bf{\dag}\;{\underline{\frak{As\;we\;know\;that,}}}$

☯ Perpendicular drawn from centre of a circle to a chord bisects the chord.

$\star\;{\boxed{\sf{\pink{SA = AJ = \dfrac {1}{2} \times SJ}}}}$

$:\implies\sf SA = AJ = \dfrac {1}{\cancel{2}} \times {\cancel {16}}$

$:\implies{\underline{\boxed{\frak{\purple{SJ\:=AJ=8\:cm}}}}}\;\bigstar$

In ΔRAJ, m∠ABC = 90°

According to the pythagorus theorem,

$:\implies\sf RJ^2 = RA^2 + AJ^2$

$:\implies\sf RJ^2 = 6^2 + 8^2$

$:\implies\sf RJ^2 = 36 + 64$

$:\implies\sf RJ^2 = 100$

$:\implies\sf RJ = \sqrt{100}$

$:\implies{\underline{\boxed{\frak{\purple{ 10\:cm}}}}}\;\bigstar$

$\therefore\:{\underline{\sf{Radius\;of\;the\;circle\:is\: {\textsf{\textbf{10\:cm}}}.}}}$

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