In a circle with centre P,the length of chord | vedclass

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Question

(A) Let $P$ be the centre of the circle and $AB$ be the chord of length $40 \, cm$. Let $M$ be the midpoint of $AB$. The perpendicular from the centre

Vedclass · Accepted Answer

$58$

Vedclass · Answer

(A) Let $P$ be the centre of the circle and $AB$ be the chord of length $40 \, cm$. Let $M$ be the midpoint of $AB$. The perpendicular from the centre to a chord bisects the chord. Therefore,$AM = MB = \frac{40}{2} = 20 \, cm$. Given the distance from the centre to the chord is $PM = 21 \, cm$. In the right-angled triangle $	riangle PMA$,by the Pythagorean theorem: $PA^2 = PM^2 + AM^2$ $PA^2 = 21^2 + 20^2$ $PA^2 = 441 + 400 = 841$ $PA = \sqrt{841} = 29 \, cm$. Here,$PA$ is the radius $(r)$ of the circle. Diameter $(d)$ = $2 	imes r = 2 	imes 29 = 58 \, cm$.

In a circle with centre $P$,the length of chord $AB$ is $40 \, cm$. $AB$ lies at a distance of $21 \, cm$ from the centre $P$. Find the diameter of the circle. (in $, cm$)

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