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

$\large\bf\underline{Given:-}$

• chord is at a distance of 6cm from centre of circle
• Diameter of circle = 20cm.

$\large\bf\underline {To \: find:-}$

• Length of Chord

$\huge\bf\underline{Solution:-}$

Diameter of chord = 20cm

So, Radius = 20/2 = 10cm

Diagram:-

$\setlength{\unitlength}{1cm} \begin{picture}(6,6) \put(3,3){\circle{2}} \put(3,3){\circle*{0.1}} \put(3,3.1){\sf C}\put(2.5,3.5){\line(1, -1){1}} \put(2.2, 3.5){\sf D} \put(3.7,2.3){\sf B}\put(2.5, 2.5){\line(1,0){1}} \put(2.1,2.3){\sf A} \put(3,3){\line(0,-1){0.5}} \put(2.5, 2.7){\tiny\textsf{6 cm}} \put(3.2, 2.8){\tiny\textsf{10 cm}}\put(3, 2.6){\line(1,0){0.1}} \put(3.1, 2.6){\line(0,-1){0.1}}\end{picture}$

Let :-

DB be the diameter and BC is the Radius .

• CE is perpendicular to AB then, EB = ½×AB

By Pythagoras theorem:-

Perpendicular ² + Base² = hypotenuse ²

In right angled triangle EBC

➝ 6² + EB² = 10²

➝ 36 + EB² = 100

➝ EB² = 100 - 36

➝ EB² = 64

➝ EB = √64

➝ EB = 8cm

• AB = 2EB
• AB = 2 × 8
• AB = 16

So, the length of the chord is 16cm.

Answer 2

Answer:

✩ DIAGRAM :

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

⠀⠀⠀$\rule{160}{1}$

☢ Given :

1. Diameter = 20 cm

⇴ So , Radius ( OA ) = 10 cm

2. Length of Center to Chord ( OM ) = 6 cm

⠀

$\underline{\bigstar\:\textsf{By Pythagoras theorem, In \Delta OAM :}}$

$:\implies\sf (Hypotenuse)^2=(Perpendicular)^2+(Base)^2\\\\\\:\implies\sf (OA)^2=(OM)^2+(AM)^2\\\\\\:\implies\sf (10)^2=(6)^2+(AM)^2\\\\\\:\implies\sf (10)^2 - (6)^2 = (AM)^2\\\\\\:\implies\sf (10 - 6)(10 + 6) = (AM)^2\\\\\\:\implies\sf 4 \times 16 = (AM)^2\\\\\\:\implies\sf 64 = (AM)^2\\\\\\:\implies\sf \sqrt{64} = AM\\\\\\:\implies\sf AM = 8 \:cm$

$\rule{180}{1.8}$

$\underline{\bigstar\:\textsf{Length of Chord, AB :}}$

$\dashrightarrow\sf\:\:AB=AM+MB\\\\\\\dashrightarrow\sf\:\:AB=AM+AM\qquad[\because\:AM=MB]\\\\\\\dashrightarrow\sf\:\:AB=8\:cm+8\:cm\\\\\\\dashrightarrow\:\:\underline{\boxed{\sf AB=16\:cm}}$

⠀

$\therefore\:\underline{\textsf{Hence, the length of Chord is \textbf{16 cm}}}.$