Question

1.In a circle of radius 6cm a chord is drawn 3cm from the center

a) Calculate the angle subtended by the chord at the centre of the circle b)Find the length of the minor and the perimeter of the minor segment formed by a chord and minor​

Answer 1

a) The angle subtended by the chord at the centre of the circle is 120°.

b) The length of the minor arc is 12.57 cm (2 d.p.).

The perimeter of the minor segment formed by a chord and minor​ arc is 22.96 cm (2 d.p.).

Explanation:

The chord of a circle is the base of an isosceles triangle, where its congruent sides are the radii of the circle.

The height of the isosceles triangle is the perpendicular bisector of the chord from the central angle.

The angle subtended by the chord at the centre of the circle is twice the angle formed by the radius and the perpendicular bisector (marked θ on the attached diagram).

To find the measure of angle θ, use the cosine trigonometric ratio.

`\begin{aligned}\implies \cos \theta&=\sf (adjacent\;side)/(hypotenuse)\\\\\cos \theta &=\sf (3)/(6)\\\\\cos \theta &=\sf (1)/(2)\\\\\theta &=\arccos\left(\sf (1)/(2)\right)\\\\\theta &=60^(\circ)\end{aligned}`

As the angle subtended by the chord at the centre of the circle is twice angle θ:

`\implies 2\theta=2 \cdot 60^(\circ)=120^(\circ)`

Therefore, the angle subtended by the chord at the centre of the circle is 120°.

A minor arc is less than 180° and is equal to the central angle.

Therefore, as the angle subtended by the chord at the centre of the circle is 120°, the minor arc is the part of the circumference between the two endpoints of the chord (marked in red on the attached diagram).

To find the length of the minor arc, use the arc length formula.

`\boxed{\begin{minipage}{6.4 cm}\underline{Arc length}\\\\Arc length $= (\pi r\theta)/(180^(\circ))$\\\\where:\\ \phantom{ww}$\bullet$ $r$ is the radius. \\ \phantom{ww}$\bullet$ $\theta$ is the angle measured in degrees.\\\end{minipage}}`

Therefore:

`\begin{aligned}\implies \textsf{Arc length}&= (\pi \cdot 6 \cdot 120^(\circ))/(180^(\circ))\\\\&= 4 \pi\\\\&= 12.57\; \sf cm\;(2\;d.p.)\end{aligned}`

Therefore, the length of the minor arc is 12.57 cm (2 d.p.).

To calculate the perimeter of the minor segment formed by the chord and minor​ arc, we need to calculate the length of the chord.

`\boxed{\begin{minipage}{10.2 cm}\underline{Chord length}\\\\Chord length $=2r\sin\left((\theta)/(2)\right)$\\\\where:\\ \phantom{ww}$\bullet$ $r$ is the radius. \\ \phantom{ww}$\bullet$ $\theta$ is the angle (in degrees) subtended at the center by the chord.\\\end{minipage}}`

Therefore:

`\begin{aligned}\implies \textsf{Chord length}&= 2 \cdot 6 \cdot \sin \left((120^(\circ))/(2)\right)\\\\&= 12 \cdot \sin \left(60^(\circ)\right)\\\\&= 12 \cdot (√(3))/(2)\\\\&= 6√(3)\; \sf cm\end{aligned}`

Therefore, the perimeter of the minor segment formed by the chord and minor​ arc is:

`\begin{aligned}\implies \textsf{Perimeter}&=\textsf{Minor arc length}+\textsf{Chord length}\\&=4 \pi + 6 √(3)\\&=22.96\; \sf cm\;(2\;d.p.)\end{aligned}`