Question

In a circle of radius 25cm, length of
the chard 14cm find the distance
of the chord do from the centre​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Distance\:between\:chord\:from\:the\:centre=24\:cm}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given:}} \\ \tt: {\implies Radius \: of \: sphere = 25 \: cm} \\ \\ \tt: {\implies Length \: of \: chord = 14 \: cm }\\ \\ \red{\underline \bold{To \: Find:}} \\ \tt:{ \implies Distance \: between \:Centre\: and \: chord = ?}$

• According to given question :

$\tt \circ \: Let \: OA \:be \: radius = 25 \: cm \\ \\ \tt \circ \: AB = 10 \: cm \\ \\ \tt \circ \: MB = 7 \: cm \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies {OA}^{2} = {OM}^{2} + {MB}^{2} \\ \\ \tt: \implies {25}^{2} = {OM}^{2} + {7}^{2} \\ \\ \tt: \implies 625 = {OM}^{2} + 49 \\ \\ \tt: \implies 625 - 49= {OM}^{2} \\ \\ \tt: \implies 576= {OM}^{2} \\ \\ \tt: \implies OM= \sqrt{576} \\ \\ \green{\tt: \implies OM = 24\: cm} \\ \\ \green{\tt \therefore length \: of\: OM \: is \: 24 \: cm}$

Answer 2

★ GiveN :

Radius of circle (AO) = 25 cm

Length of the chord (AC) = 14 cm

$\rule{200}{1}$

★ To FinD :

We have to find the distance of the chord from the centre.

$\rule{200}{1}$

★ SolutioN :

We know that,

Perpendicular from centre bisects the chord equally

So the Length of chord (AB) = 7 cm

Now, we will apply Pythagorean theorm

$\Large{\implies{\boxed{\boxed{\sf{(H)^2 = (P)^2 + (B)^2}}}}}$

Where,

• H is OA (25 cm)
• P is OB
• B is AB (7 cm)

★ Putting Values ★

$\sf{\dashrightarrow (25)^2 = (OB)^2 + (7)^2} \\ \\ \sf{\dashrightarrow (OB)^2 = (25)^2 - (7)^2} \\ \\ \sf{\dashrightarrow (OB)^2 = 625 - 49} \\ \\ \sf{\dashrightarrow (OB)^2 = 576} \\ \\ \sf{\dashrightarrow OB = \sqrt{576}} \\ \\ \sf{\dashrightarrow OB = (\sqrt{24})^2} \\ \\ \sf{\dashrightarrow OB = 24} \\ \\ \Large{\implies{\boxed{\boxed{\sf{OB = 24 \: cm}}}}}$

$\therefore$ Distance of the chord from the centre of the circle is 24 cm.