Question

The radius of a circle is 13 cm and the length of one of its chords is 10 cm. Find the distance of the chord from the centre.​

Answer 1

☞ See the attachment figure .

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$\sf\underbrace{\red{To\:Find:-}}$

• The distance of the chord from the center of the circle .

$\sf\underbrace{\red{SOLUTION:-}}$

☞ Let “AB” be a chord of a circle with center “O” and radius “13 c.m”, such that

• AB = 10 c.m

☞ From “O”, draw “OL” perpendicular to “AB” . Join “OA” .

☞ Since, the perpendicular distance from the center of the circle to a chord bisects the chord .

$\checkmark\:\rm{\therefore\:AL\:=\:LB\:=\:{\dfrac{1}{2}}\:AB\:=\:5cm\:}$

☞ Now, In Right angle triangle “OLA” , we have,

$\rm{(OA)^2\:=\:(OL)^2\:+\:(AL)^2\:}$

$\rm{\implies\:(13)^2\:=\:(OL)^2\:+\:(5)^2\:}$

$\rm{\implies\:(OL)^2\:=\:144\:}$

$\rm{\purple{\boxed{\implies\:OL\:=\:12cm\:}}}$

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♻️ Hence, the distance of the chord from the center of the circle is “12cm” .

Answer 2

$\huge\mathfrak\blue{Answer:}$

Given:

• We have been given a circle of radius 13 cm
• Length of one of its chord is 10 cm

To Find:

• We have to find the distance of chord from the center of the circle

Construction:

• Dropping a perpendicular from center O to chord AB which intersect AB at D
• Joining center O to A

Solution:

We have been given that

$\large\boxed{\sf{Radius \: of \: circle = 13 cm }}$

Let there be a chord AB in given circle

Such that Length of AB = 10 cm

Droping a perpendicular from center O of the circle to the chord AB which intersect AB at D

$\large\boxed{\sf{Distance \: of \: Chord = OD}}$

We know that , Perpendicular from the center to the chord divides the chord in two equal segments

$\boxed{\sf{AD = BD = \dfrac{10}{2} = 5 \: cm}}$

Joining line segment OA

$\large\boxed{\sf{OA = 13 cm}}$

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In ∆ OAD , Using Pythagoras Theorm

$\implies \sf{(OA)^2 = (AD)^2 + (DO)^2 }$

Substitung the values

$\implies \sf{(13)^2 = (5)^2 + (DO)^2 }$

$\implies \sf{169 = 25 + (DO)^2 }$

$\implies \sf{169- 25 = (DO)^2 }$

$\implies \sf{(DO)^2 = 144}$

Taking Square Root on Both Sides

$\implies \sf{DO = \sqrt{144}}$

$\implies \boxed{\sf{DO = 12 \: cm }}$

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$\huge\underline{\sf{\red{A}\orange{n}\green{s}\pink{w}\blue{e}\purple{r}}}$

$\boxed{\sf{Distance \: of \: Chord \: From \: Center = 12 \: cm}}$

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$\large\purple{\underline{\underline{\sf{Extra \: Information:}}}}$

• Circle is locus of all the points in a plane that maintains a constant distance from a fixed point
• Fixed point is know as Center
• Line Joining fix point to any point on the circle is known as Radius