Question

the length of a diameter of a circle is 20cm. if the distance of a chord from the centre of the circle is 8cm, find the length of the chord.​

Answer 1

Given,

• Diameter of the circle = 20 cm
• Radius of the circle = 20/2 = 10 cm
• distance of chord from the centre = 8 cm

To find,

• Length of the chord

In the figure (refer to the attachment)

Using pythagoras theorem in ΔOAM

→ AM² = OA² - OM²

→ AM² = 10² - 8² = 100 - 64

→ AM² = 36

→ AM = 6 cm

Using the theorem: A perpendicular drawn from the center of the circle to a chord bisects the chord. i.e, (AM = BM) so,

→ AB = 2 AM = 2 × 6 = 12 cm

Therefore,

Length of the chord will be 12 cm.

Answer 2

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 = $\rm{\dfrac{Diameter}{2}\:=\:\dfrac{20}{2}}$ = 10 cm

• OC = 8 cm

✯ ∆AOC is a right angle triangle.

Sᴏ,

Using Pythagoras theorem,

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

• Hypotenuse = OA

• Height = OC

• Base = AC = $\bf{\dfrac{AB}{2}}$

$:\implies\:\:\bf{(OA)^2\:=\:(OC)^2\:+\:(AC)^2\:}$

$:\implies\:\:\bf{(10)^2\:=\:(8)^2\:+\:(AC)^2\:}$

$:\implies\:\:\bf{100\:=\:64\:+\:(AC)^2\:}$

$:\implies\:\:\bf{(AC)^2\:=\:100\:-\:64}$

$:\implies\:\:\bf{(AC)^2\:=\:36}$

$:\implies\:\:\bf{AC\:=\:\sqrt{36}}$

$:\implies\:\:\bf\pink{AC\:=\:6\:cm}$

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

➣ Length of the chord = AB = 2 × AC

➛ Length of the chord = 2 × 6

➛ Length of the chord = 12 cm

$\\ \Large\bf\purple{Therefore,}$

The length of the chord is 12 cm.