Question

Radius of circle is 10 cm. There are two chords of length 16 cm each what will be the distance of these chords from the center of the circle

Answer 1

in cirle draw perpendicular bisectors of chords through center's
that is oml_to ab and on l_ to cd
than join ao and oc
ao=oc=10cm (radius of circle)
now in triangle aom and triangle onc
Om and on will be equal since chords are equal therefore they will be at equal distance
now in triangle aom
hypotenuse =10cm
base =8cm
so by Pythagoras theoram we can find perpendicular

${h}^{2} = {b}^{2} + {p}^{2}$
${p}^{2} = {h }^{2} - {b}^{2}$
=10×10-8×8
=100-64
=36
$p = \sqrt{36 } \\ p = 6$
distance of chords from centre is 6cm
hope this will help you