Question

find the length of the chord. diameter 20cm distance 8cm​

Answer 1

Answer:

6 cm

Chord=x

Diameter=2r

R=10

Distance =10

By pythagoras theorem

${10}^{2} = x {}^{2} + 8 {}^{2}$

$x {}^{2} { = 10}^{2} - 8 {}^{2}$

$x {}^{2} = 100 - 64$

$x {}^{2} = 36$

$x = 6$

Chord is 6cm

Answer 2

Answer:

Gɪᴠᴇɴ :

Diameter of a circle is 20 cm.

Distance of a chord from the centre of the circle is 8 cm.

Tᴏ Fɪɴᴅ :

The length of the chord.

Cᴀʟᴄᴜʟᴀᴛɪᴏɴ :

As sʜᴏᴡɴ ɪɴ ᴛʜᴇ ғɪɢᴜʀᴇ,

O is the centre of the circle.

OA is the radius.

AB be the chord.

OC be the distance from the centre of the circle to the chord.

Aᴄᴄᴏʀᴅɪɴɢ ᴛᴏ ᴛʜᴇ ǫᴜᴇsᴛɪᴏɴ,

OA = \rm{\dfrac{Diameter}{2}\:=\:\dfrac{20}{2}}

2

Diameter

=

2

20

= 10 cm

OC = 8 cm

✯ ∆AOC is a right angle triangle.

Sᴏ,

Using Pythagoras theorem,

\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{(Hypotenuse)^2\:=\:(Height)^2\:+\:(Base)^2\:}}}}}} \\ \end{gathered}

★

(Hypotenuse)

2

=(Height)

2

+(Base)

2

Hypotenuse = OA

Height = OC

Base = AC = \begin{gathered}\bf{\dfrac{AB}{2}} \\ \end{gathered}

2

AB

\begin{gathered}:\implies\:\:\bf{(OA)^2\:=\:(OC)^2\:+\:(AC)^2\:} \\ \end{gathered}

:⟹(OA)

2

=(OC)

2

+(AC)

2

\begin{gathered}:\implies\:\:\bf{(10)^2\:=\:(8)^2\:+\:(AC)^2\:} \\ \end{gathered}

:⟹(10)

2

=(8)

2

+(AC)

2

\begin{gathered}:\implies\:\:\bf{100\:=\:64\:+\:(AC)^2\:} \\ \end{gathered}

:⟹100=64+(AC)

2

\begin{gathered}:\implies\:\:\bf{(AC)^2\:=\:100\:-\:64} \\ \end{gathered}

:⟹(AC)

2

=100−64

\begin{gathered}:\implies\:\:\bf{(AC)^2\:=\:36} \\ \end{gathered}

:⟹(AC)

2

=36

\begin{gathered}:\implies\:\:\bf{AC\:=\:\sqrt{36}} \\ \end{gathered}

:⟹AC=

36

\begin{gathered}:\implies\:\:\bf\pink{AC\:=\:6\:cm} \\ \end{gathered}

:⟹AC=6cm

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

➣ Length of the chord = AB = 2 × AC

➛ Length of the chord = 2 × 6

➛ Length of the chord = 12 cm

\begin{gathered} \\ \Large\bf\purple{Therefore,} \\ \end{gathered}

Therefore,

The length of the chord is 12 cm.

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