Question

Calculate the length of a chord which is at a distance 6 cm from the centre of a circle of diameter 20 cm.​

Answer 1

EXPLANATION.

• GIVEN

Length of a chord which is at a distance = 6cm

From center of circle diameter = 20 cm

CALCULATE THE LENGTH OF CHORD.

According to the question,

Let AB is a chord From the center of circle

OM is perpendicular to AB

Diameter = 20 cm

Radius = Diameter / 2

=> Radius = 20/2

=> Radius = 10 cm

Therefore

=> OA = 10 cm and OM = 6 cm

By Pythagoras Theorm

=> h² = p² + b²

=> (AO) ² = (OM)² + ( AM)²

=> (10)² = (6)² + AM²

=> 100 = 36 + AM²

=> 64 = AM²

=> 8 = AM

Therefore,

=> AB = Am + MB

=> AB = 2AM

=> AB = 2 X 8

=> AB = 16 cm

Length of chord = 16 cm

Note = For diagram see image

attachment.

Answer 2

AnswEr:

⋆ Reference of image is shown in diagram

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

$\normalsize\;\;\bullet\;\sf Diameter\;of\;the\;circle\;:\; \bf{20\;cm.}\\\\\normalsize\;\;\bullet\;\sf Lenght\;of\;center\;to\;chord\; (OM)\;\bf{6\;cm.}$

$\rule{150}{2}$

$\underline{\bigstar\:\boldsymbol{As\;per\;given\; Question\;:}}$

$\normalsize\;\star\;\sf Diameter\;of\;the\;circle\;:\; \bf{20\;cm.}\\\\\normalsize\;\therefore\;\sf Radius\;of\;the\;circle\;:\; \bf{10\;cm}$

$\bf \underline{\bigstar\;Now,\;using\; Pythagoras\; theorem\;:}\\\\ \dashrightarrow\sf (Hypotenuse)^2 = (Perpendicular)^2 + (Base)^2 \\\\ \dashrightarrow\sf (OA)^2 = (OM)^2 + (AM)^2\\\\ \dashrightarrow\sf (10)^2 = (6)^2 + (AM)^2\\\\ \dashrightarrow\sf 100 = 36 = (AM)^2\\\\ \dashrightarrow\sf 100 - 36 = (AM)^2\\\\ \dashrightarrow\sf 64 = (AM)^2\\\\ \small\sf \;\; \underline{Taking\;sqrt.\;both\;sides\;:}\\\\ \dashrightarrow\sf \sqrt{64} = \sqrt{(AM)^2}\\\\ \dashrightarrow{\underline{\boxed{\sf{\pink{AM = 8\;cm}}}}}\;\bigstar$

$\rule{150}{2}$

$\bf \underline{\bigstar\; Now,\;we\;have\;to\;find\;chord\;of\;circle\;:}\\\\ \dashrightarrow\sf AB = AM + MB\\\\ \dashrightarrow\sf AB = AM + AM\qquad\bigg\lgroup\bf AM = MB \bigg\rgroup\\\\ \dashrightarrow\sf AB = 8\;cm + 8\;cm\\\\ \dashrightarrow{\underline{\boxed{\sf{\purple{AM = 16\;cm}}}}}\;\bigstar\\\\ \sf\dag\;\underline{Hence,\:the\; length\;of\;chord\;is\; \bf{16\;cm}}$