Question

Of any two chords of a circle show that the one which is nearer to the centre is larger.

Answer 1

Consider above the circle, Consider two chords $AB,CD$.

at distance $a,b$ respectively $a>b$.

let $AB=2l_1$

$CD=2l_2$

from the figure,

$a^2+l_1^2=r^2---\left(1\right)$

$a^2=r^2-l_1^2----\left(2\right)$

$b^2+l_2^2=r^2$

$b^2=r^2-l_2^2-----\left(3\right)$

$a^2>b^2$

$\therefore r^2-l_1^2>r^2-l_2^2$

$\therefore l_2^2>l_1^2$

$\Rightarrow l_2>l_1$

$\Rightarrow 2l_2>2l_1$

$\Rightarrow CD>AB$

$\Rightarrow$ chord $AB$ is farther.
and chord $CD$ the nearer chord is longer. $[\therefore b<a]$