Answer to Question #126885 in Geometry for jungkookiee

Question

A chord of length 12 cm is drawn in a circle of radius 9 cm. Calculate the perpendicular distance from the center of the circle to the chord.

Answer 1

Consider a circle with center $O$ and radius $OA=9\ cm.$ Let $AB$ be a chord of length $12\ cm.$

We see that $OA$ and $OB$ are two radii of the circle: $OA=OB=9 \ cm.$

We have the equilateral triangle $\Delta AOB: OA=OB.$

$AD$ is the height of the triangle $AOB.$ Then $AD$ is the perpendicular bisector and

$$AD=DB={1\over 2}AB$$

Consider the right triangle $OAD$

The Pythagorean Theorem

$$OA^2=OD^2+AD^2$$ $$OD=\sqrt{OA^2-AD^2}$$

$OA=9 \ cm, AD=\dfrac{1}{2}(12\ cm)=6\ cm$

$$OD=\sqrt{(9\ cm)^2-(6\ cm)^2}=3\sqrt{5}\ cm$$

The perpendicular distance from the center of the circle to the chord is $3\sqrt{5}\ cm.$