Question

Radius of a circle with Centre O is 26 cm find the length of the chord if the chord is at a distance of 10 cm from the centre.​

Answer 1

From the rough diagram attached above,

O is the centre of the circle.

Radius = OA = OB = 26 cm

Let AB is a chord .

Draw OP perpendicular to AB .

OP = 10 cm

In right ∆OAP ,

<OPA = 90° , ( construction )

By Pythagoras Theorem :

OA² = AP² + OP²

=> 26² = AP² + 10²

=> 676 = AP² + 100

=> 676 - 100 = AP²

=> 576 = AP²

=> AP = √576

=> AP = 24 cm

$Length \: of \:the \: Chord = AB \\= AP + PB \\= AP + AP$

$\blue { The \: perpendicular \:from \:the \: centre }$

$\blue { of \:a \:circle \:to \:a \:chord \: bisects }$

$\blue { the \:chord . }$

$= 2AP \\= 2 \times 24 \\= 48\: cm$

Therefore.,

$\red { Length \: of \:the \: Chord } \green {= 48\: cm}$

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Answer 2

Answer:

From the rough diagram attached above,

O is the centre of the circle.

Radius = OA = OB = 26 cm

Let AB is a chord .

Draw OP perpendicular to AB .

OP = 10 cm

In right ∆OAP ,

<OPA = 90° , ( construction )

By Pythagoras Theorem :

OA² = AP² + OP²

=> 26² = AP² + 10²

=> 676 = AP² + 100

=> 676 - 100 = AP²

=> 576 = AP²

=> AP = √576

=> AP = 24 cm

Therefore.,

Step-by-step explanation: