Question

the length of a chord at a distance of 52 cm from the centre of a circle is 78cm. what is length of the another chord at a distance of 60 cm from the centre of the same circle?​

Answer 1

Answer:

The chord is bisected by the perpendicular from the center. Let the perpendicular from the center O meets the chord AB at C .

In triangle AOC, <C=90 . AO = radius , AC=1/2 AB = 78/2 =39cm

OC =distance of the chord from the center = 52 cm

In ∆AOC,

AO^2= OC^2+AC^2

AO^2 = 52^2 +39^2

AO= 65cm.

Let the other chord be PQ which is 60 cm away from the center O.

Let OR = 60 .

In ∆POR ,

PO ^2= PR^2+OR^2

PR^2= PO^2-OR^2

=65^2–60^2 = 625

PR=25 cm.

Length of the chord PQ=2PR=50 cm.

Answer 2

Answer:-

The chord is bisected by the perpendicular from the center. Let the perpendicular from the center O meets the chord AB at C .

$\small \rm{\red{In \: triangle \: AOC, } }$

$\small \rm{ < C=90}$

$\small \rm{AO = radius}$

$\small \rm{AC= \frac{1}{2} }$

$\small{ \rm{AB = \frac{78}{2 }=39 \: cm}}$

$\small \rm{OC \: = \: distance \: of \: the \: chord \: from \: the \: center \:= 52 \: cm}$

$\small \rm{\red{In \: ∆AOC,}}$

$\small \rm{AO^2= OC^2+AC^2}$

$\small {\rm{AO^2 = 52^2 +39^2}}$

$\small \rm{\boxed{AO= 65 \: cm.}}$

$\small \rm{\underline{Let \: the \: other \: chord \: be \: PQ \: which \: is \: 60 \: cm\:away \: from \: the \: center \: O.}}$

$\small \rm{Let \: OR = 60 .}$

$\small\red{ \rm{In \: ∆POR ,}}$

$\small \rm{PO ^2= PR^2+OR^2}$

$\small \rm{PR^2= PO^2-OR^2}$

$\small \rm{=65^2–60^2 = 625}$

$\small \rm{\boxed{PR=25 \: cm.}}$

$\therefore \small\underline{ \underline{\blue{ \rm{Length \: of \: the \: chord \: PQ = 2 \: PR=50 \: cm.}}}}$