In a circle with centre P,the radius is 13 , | vedclass

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(A) Let $P$ be the center of the circle with radius $r = 13 \, cm$. Let $M$ and $N$ be the midpoints of chords $AB$ and $CD$ respectively. Since the p

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$10$

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(A) Let $P$ be the center of the circle with radius $r = 13 \, cm$. Let $M$ and $N$ be the midpoints of chords $AB$ and $CD$ respectively. Since the perpendicular from the center to a chord bisects the chord,$AM = MB = 12 \, cm$ and $CN = ND$. In $	riangle PMA$,by the Pythagorean theorem,$PM^2 + AM^2 = PA^2$,so $PM^2 + 12^2 = 13^2$,which gives $PM^2 = 169 - 144 = 25$,so $PM = 5 \, cm$. There are two cases: chords are on the same side or opposite sides of the center. Case $1$: Chords are on opposite sides. The distance $MN = PM + PN = 17 \, cm$. Since $PM = 5 \, cm$,$PN = 17 - 5 = 12 \, cm$. In $	riangle PNC$,$PN^2 + CN^2 = PC^2$,so $12^2 + CN^2 = 13^2$,which gives $CN^2 = 169 - 144 = 25$,so $CN = 5 \, cm$. Thus,$CD = 2 	imes CN = 10 \, cm$. Case $2$: Chords are on the same side. The distance $MN = |PM - PN| = 17 \, cm$. Since $PM = 5 \, cm$,$PN$ would be $22 \, cm$ or $-12 \, cm$,which is impossible as the radius is $13 \, cm$. Thus,$CD = 10 \, cm$.

In a circle with centre $P$,the radius is $13 \, cm$. The distance between two parallel chords $AB$ and $CD$ is $17 \, cm$. If $AB = 24 \, cm$,find the length of $CD$. (in $, cm$)

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