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Home » Prove that, of any two chords of a circle, the greater chord is nearer to the center.
Prove that, of any two chords of a circle, the greater chord is nearer to the center.
Class 10 | Exercise 18C | ICSE | Maths | Selina | Tangents and Intersecting Chords
Prove that, of any two chords of a circle, the greater chord is nearer to the center.

Selina Solutions Concise Class 10 Maths Chapter 18 ex. 18(C) - 1

Given:

    \[A\]

circle with center

    \[O\]

and radius

    \[r\]

    \[AB\text{ }and\text{ }CD\]

are two chords such that

    \[AB\text{ }>\text{ }CD\]

Also,

    \[OM\bot AB\text{ }and\text{ }ON\bot CD\]

Required to prove:

    \[OM\text{ }<\text{ }ON\]

Proof:

Join

    \[OA\text{ }and\text{ }OC\]

Then in right

    \[\vartriangle AOM\]

We have

    \[A{{O}^{2}}~=\text{ }A{{M}^{2}}~+\text{ }O{{M}^{2}}\]

    \[{{r}^{2}}~=\text{ }{{\left( {\scriptscriptstyle 1\!/\!{ }_2}AB \right)}^{2}}~+\text{ }O{{M}^{2}}\]

    \[{{r}^{2}}~=\text{ }{\scriptscriptstyle 1\!/\!{ }_4}\text{ }A{{B}^{2}}~+\text{ }O{{M}^{2}}~\ldots ..\text{ }\left( i \right)\]

Again, in right

    \[\vartriangle ONC\]

We have

    \[O{{C}^{2}}~=\text{ }N{{C}^{2}}~+\text{ }O{{N}^{2}}\]

    \[{{r}^{2}}~=\text{ }{{\left( {\scriptscriptstyle 1\!/\!{ }_2}CD \right)}^{2}}~+\text{ }O{{N}^{2}}\]

Or,

    \[{{r}^{2}}~=\text{ }{\scriptscriptstyle 1\!/\!{ }_4}\text{ }C{{D}^{2}}~+\text{ }O{{N}^{2}}~\ldots ..\text{ }\left( ii \right)\]

On equating (i) and (ii), we get

    \[{\scriptscriptstyle 1\!/\!{ }_4}\text{ }A{{B}^{2}}~+\text{ }O{{M}^{2}}~=\text{ }{\scriptscriptstyle 1\!/\!{ }_4}\text{ }C{{D}^{2}}~+\text{ }O{{N}^{2}}\]

But,

    \[AB\text{ }>\text{ }CD\]

[Given]

So,

    \[ON\]

will be greater than

    \[OM\]

to be equal on both sides.

Thus,

    \[OM\text{ }<\text{ }ON\]

Hence,

    \[AB\]

is nearer to the centre than

    \[CD\]

i

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