CHORD AND PERIMETER OF A SEGMENT OF A CIRCLE

Perimeter of Segment of a Circle and length of a chord. Have you ever found it difficult calculating the length of a chord? or perhaps finding the area or perimeter of a segment of a circle? In this article, we will discuss with practical examples, the length of a chord of a circle, the perimeter of a segment, and the area of a segment. Recall that in our previous article, we discussed the perimeter and area of a sector of a circle and the length of an arc.  Let us start with the segment of a circle after which we will take some few practical examples.

WHAT IS SEGMENT OF A CIRCLE FORMULA

The Segment of a circle – The segment of a circle is the region that is bounded by an arc and chord of a circle. There are two types of the segment, one is the minor segment and the other a major segment. The major segment of a circle takes the major part of the circumference of the circle. While the minor segmnet takes just a small part of the circumference. Also recall that an arc is a part of the circumference of a circle. And a chord is a line segment that joins any two points on the circumference of the circle. Many students wonders about how to solve questions on the perimeter of a segment of a circle. The next subheading will explain the steps.

WHAT IS THE PERIMETER OF A SEGMENT OF A CIRCLE

The perimeter of a segment of a circle is the sum of the length of the arc of the circle from which the segment is form and the length of the chord.  The shaded region in the diagram below is the region bounded by the arc and chord AB.

segment of a circle jpg

Before calculating the perimeter of a segment of a circle, we first need to know how to calculate the length of a chord. We have known that the length of an arc of a circle is

length of an arc jpg

and in our previous articles we have learnt how to calculate the length of an arc. Let us now derive the formula for calculating the length of a chord and see practical examples.

LENGTH OF A CHORD OF A CIRCLE

The chord is a line segment that joins any two points on a circle’s circumference. The line AB, in the diagram below is the length of the chord.

length of a chord of a circle

To calculate the length of the chord AB in the diagram above, the following steps are to be followed:

  1. Bisected angle <AOB (AOB is an isosceles triangle). OP which is perpendicular to AB. if angle AOB is θ, then <BOP = <AOP = \frac{\ominus }{2}
  2. Using Trigonometric ratio, consider triangle AOP=\frac{\ominus }{2}, Sin \frac{\ominus }{2}=\frac{opp}{hyp}
  3. Simplifying Sin \frac{\ominus }{2}=\frac{AP}{AO}.  Sin\frac{\ominus }{2}=\frac{AP}{r},  ∴ AP = rSin\frac{\ominus }{2}

But AP = PB (by the construction of the perpendicular bisector). Hence, AB which is the chord = AP + PB

=rSin\frac{\ominus }{2}+rSin\frac{\ominus }{2}

=2rSin\frac{\ominus }{2}

So the Chord AB=2rSin\frac{\ominus }{2} units. Where θ is the angle subtended by the chord at the center.

This is the formula for finding the length of any chord of a circle which subtends angle θ at the center of the circle. Let us consider some few practical examples on how to calculate the length of the chord of a circle. After which we can now calculate the perimeter of the segment of a circle.

Recommended: Logarithm of Numbers.

Recommended: Watch WAEC, GCE and NECO Past Questions Solutions. Click HERE to watch from our Past Questions Solutions Playlist (Archives). Hit the Subscribe button on any of the video for more Mathematics Tutorial Videos.
Example 1:  Find the length of the chord which subtends an angle 120^{0} at the centre of a circl of radius 14cm.

Firstly, let the length of the chord be PQ.

Given that r=14cm and θ = 120^{0}.

Length of the Chord PQ = 2rSin\frac{\ominus }{2}

Remember that \frac{\ominus }{2}=\frac{120^{0}}{2}=60

∴ Length of the chord, PQ = 2rSin60^{0}

PQ=2\times 14\times 0.8660

∴ PQ = 24.24cm

Example 2: In a circle radius rcm, a chord 16\sqrt{3}cm long is 10cm from the centre of the circle. Find correct to the nearest cm, the value of r. (WAEC)  a. 22cm  b. 17cm  c. 16cm  d. 15cm

Length of the chord = \sqrt[16]{3} cm, so half of it is \sqrt[8]{3}cm

So using the Pythagoras theorem to solve for r in the the right angle triangle

r^{2}=10^{2}+(\sqrt[8]{3})^{2}

r^{2}=100+(8^{2}\times \sqrt{3}\times \sqrt{3})

r^{2}=100+(64\times 3)

r^{2}=100+192

r^{2}=292

Square root both sides to clear the square

(\sqrt{r})^{2}=\sqrt{292}

∴ r=17.08 ≅ 17cm

Click HERE to watch a video solution of the above question with diagrams and detailed explanation. 

Example 3: A chord AB of a circle, radius 9.4cm is 12.8cm. Find the angle subtended by the chord at the centre.

Let θ be the angle subtended by the chord, r=9.4cm and the length of the chord, AB = 12.8cm

Recall the length of a chord = 2rsin\frac{\ominus }{2}

∴ Length of the chord, AB = 2rSin\frac{\ominus }{2}

12.8cm=2\times 9.4cm\times Sin\frac{\ominus }{2}

12.8cm=18.8cm\times Sin\frac{\ominus }{2}

So Sin\frac{\ominus }{2}=\frac{12.8cm}{18.8cm}

Sin\frac{\ominus }{2}=0.6805

\frac{\ominus }{2}=Sin^{-1}0.6805

Note: To find the arc Sin of a value using scientific calculator, press the value (0.6805) then 2nd function and then Sin. Using the four figure mathematical table, under Sine of Angles, look for 0.680….. , that will give you the 42^{0} write it down. Then check the differnce that you will add to that value to make up 0.6805. That will give you the decimal value.

So \frac{\ominus }{2}=42.88

∴ θ = 2\times 42.88

θ = 85.77^{0}

In the above examples, you have learnt how to carry out calculation on chord, when the length of the chord, the radius and the angle subtended by the chord at the centre is required to be calculated. Finally, let us now consider two examples on how to calculate the perimeter of a segment of a circle.

Example 1: Find the perimeter of the segment of a circle of radius 14cm, if the chord subtends angle 104^{0} at the centre.  {Take π = \frac{22}{7}}.
Example 2: AB is a chord of a circle with centre O and radius 4cm. <AOB = 120^{0}. Calculate the perimeter of the minor segment. {Take π = \frac{22}{7}}

From the question, we are given the following parameters: r=4cm, <AOB =120^{0}, π = \frac{22}{7}.

Perimeter of the Minor Segment of the circle = Length of arc + Length of chord

=\frac{120}{360}\times 2\times \frac{22}{7}\times 4cm

=\frac{1}{3}\times 8\times \frac{22}{7}

Length of arc AB = \frac{176}{21}= 8.38cm

The length of chord AB = 2rSin\frac{\ominus }{2}

=2\times 4Sin\frac{120}{2}

=8\times Sin60

=8\times \frac{\sqrt{3}}{2}

=\sqrt[4]{3}cm=6.92cm

∴ Perimeter of the Minor segment = 8.38cm+6.92cm

=15.30cm

In conclusion, solving questions on the chord of a circle and the segment of a circle is pretty simple, just follow the steps above, and the few examples already given above. If you want to practice more questions the chord and segments of a circle, download the worksheet. Click HERE to download and try solving them. Anticipate our next article on the area of a segment of a circle.

Thank you for reading this article “Chord and Perimeter of a Segment of a circle”. Hope it was educative. Click HERE to join our discussion forum. In addition, click the various links below to follow and subscribe to our various social media handles.

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October 15, 2021
MATH ARTICLES

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