Question

$\large\bold{\underline{\underline{Question:-}}}$

The Radius of a Circle is 13cm and length of one of its Chords is 24cm. Find the distance of the chord from the Centre.

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Answer 1

ʜᴇʏ,

Gɪᴠᴇɴ:-

• ʀᴀᴅɪᴜs ᴏғ ᴛʜᴇ ᴄɪʀᴄʟᴇ ɪs 13 ᴄᴍ.
• ʟᴇɴɢᴛʜ ᴏғ ᴏɴᴇ ᴄʜᴏʀᴅ ɪs 24 ᴄᴍ.

ᴛᴏ ғɪɴᴅ:-

• ᴅɪsᴛᴀɴᴄᴇ ᴏғ ᴛʜᴇ ᴄʜᴏʀᴅ ғʀᴏᴍ ᴛʜᴇ ᴄᴇɴᴛʀᴇ.

sᴏʟᴜᴛɪᴏɴ:-

ᴀʙ ɪs ᴀ ᴄʜᴏʀᴅ.

ᴏᴍ ᴘᴀʀᴇʟʟᴇʟ ᴛᴏ ᴀʙ.

ɪɴ ᴛʀɪᴀɴɢʟᴇ ᴀᴏᴍ, ᴀɴɢʟᴇ ᴏᴍᴀ = 90

ᴛʜᴇʀᴇғᴏʀᴇ,

ᴏA^2 = ᴏᴍ^2 + ᴍᴀ^2

ᴍᴀ ^2 = ᴏᴀ^2 - ᴏᴍ^2

= 13^2 - 5^2

= 169 - 25

Mᴀ = ʀᴏᴏᴛ 144

Mᴀ = 12

ᴄʜᴏʀᴅ ᴀʙ = 2 (Mᴀ)

= 2(12)

= 24

Answer 2

$\huge{\underline{\bf{\red{Solution}}}}$

GIVEN:

• A circle with centre O
• Radius AO = 13 cm
• Chord AB = 24 cm

TO FIND :

The perpendicular distance of the chord from O. Let it be called OM.

OM is perpendicular to the chord AB.

Perpendicular from the centre O to a chord bisects the chord.

So AM = AB/2 = 12 cm

In right triangle AMO , Using Pythagoras’s Theorem,

AO² = AM² + OM²

=> 13² = 12² + OM²

=> 169 = 144 + OM²

=> OM² = 169 - 144 = 25

=> OM = √25 = 5 cm

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