AB and CD are two chords of a circle with cen | vedclass

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Question

(A) Let $r$ be the radius of the circle. Draw a perpendicular from $P$ to chord $AB$ at point $M$ and to chord $CD$ at point $N$. $PM = 15 \, cm$ and

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(A) Let $r$ be the radius of the circle. Draw a perpendicular from $P$ to chord $AB$ at point $M$ and to chord $CD$ at point $N$. $PM = 15 \, cm$ and $PN = 8 \, cm$. Since the perpendicular from the centre bisects the chord,$AM = MB = AB / 2 = 16 / 2 = 8 \, cm$. In right-angled triangle $	riangle PAM$,by Pythagoras theorem: $r^2 = PM^2 + AM^2 = 15^2 + 8^2 = 225 + 64 = 289$. So,$r = \sqrt{289} = 17 \, cm$. Now,in right-angled triangle $	riangle PNC$,by Pythagoras theorem: $r^2 = PN^2 + CN^2$. $17^2 = 8^2 + CN^2 \implies 289 = 64 + CN^2$. $CN^2 = 289 - 64 = 225$. $CN = \sqrt{225} = 15 \, cm$. Since $N$ is the midpoint of $CD$,$CD = 2 	imes CN = 2 	imes 15 = 30 \, cm$.

$AB$ and $CD$ are two chords of a circle with centre $P$. The distance of chord $AB$ from the centre $P$ is $15 \, cm$ and the distance of chord $CD$ from the centre $P$ is $8 \, cm$. If $AB = 16 \, cm$,then find the length of chord $CD$. (in $, cm$)

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