Question

In a circle of radius 5 cm, a chord is of length 6 cm. Find the distance of the chord from
the centre of the circle.​

Answer 1

It should be 4 centimeters

Answer 2

${\underline{\underline{\frak{AnswEr\;:}}}}$

$\setlength{\unitlength}{1.2mm}\begin{picture}(50,55)\thicklines\qbezier(25.000,10.000)(33.284,10.000)(39.142,15.858)\qbezier(39.142,15.858)(45.000,21.716)(45.000,30.000)\qbezier(45.000,30.000)(45.000,38.284)(39.142,44.142)\qbezier(39.142,44.142)(33.284,50.000)(25.000,50.000)\qbezier(25.000,50.000)(16.716,50.000)(10.858,44.142)\qbezier(10.858,44.142)( 5.000,38.284)( 5.000,30.000)\qbezier( 5.000,30.000)( 5.000,21.716)(10.858,15.858)\qbezier(10.858,15.858)(16.716,10.000)(25.000,10.000)\put(25,30){\line(5, - 4){16}}\put(25,30){\circle*{1}}\put(24,32){\sf\large{O}}\put(25,30){\line(0, - 5){13}}\put(6,14){\sf\large{A}}\put(25,30){\line(- 5, - 4){16}}\put(10,17){\line(5, 0){30}}\put(42,14){\sf\large{B}}\put(24,14){\sf\large{C}}\end{picture}$

⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Here,

• Length of chord, AB = 6 cm

• Perpendicular line from the centre of the circle to the chord = OC

• Radius of Circle = OA = OB = 5 cm

⠀⠀

THEOREM: A perpendicular dropped from the centre of the circle to a chord bisect it.

Therefore, both the halves of the chord are equal.

⠀⠀

Here, OB bisects AB

So, AC = BC = $\sf \dfrac{AB}{2} = \cancel{ \dfrac{6}{2}}$ = 3

${\underline{\sf{\bigstar\;In\; \triangle\;OCB}}}$

Using Pythagoras Theorem,

$\star\;{\boxed{\sf{\purple{H^2 = B^2 + P^2}}}}$

$\qquad:\implies\sf OB^2 = OC^2 + BC^2$

$\qquad\: \: \quad:\implies\sf 5^2 = OC^2 + 3^2$

$\qquad \: \: \quad:\implies\sf 25 = OC^2 + 9$

$\qquad\: \: \quad:\implies\sf 25 - 9 = OC^2$

$\qquad\qquad \: \: \: \: :\implies\sf 16 = OC^2$

$\qquad\qquad:\implies\sf OC = \sqrt{16}$

$\qquad\quad\quad\:\:\::\implies{\boxed{\frak{\pink{OC = 4}}}}\;\bigstar$

$\therefore\;{\underline{\sf{Hence,\;The\;distance\;of\;chord\;from\;the\;centre\;of\;the\;circle\;is\; \bf{4\;cm}.}}}$