Chord (geometry) length, theorem and exercises

A string, in plane geometry, it is the line segment that joins two points on a curve. The line that contains this segment is said to be a secant line to the curve. This is often a circle, but chords can certainly be drawn on many other curves, such as ellipses and parabolas..

In figure 1 on the left there is a curve, to which points A and B belong. The chord between A and B is the green segment. To the right is a circumference and one of its strings, since it is possible to draw infinite.

In the circumference its diameter is particularly interesting, which is also known as major chord. It is a chord that always contains the center of the circumference and measures twice the radius.

The following figure shows the radius, the diameter, a chord and also the arc of a circle. Correctly identifying each one is important when solving problems.

Article index

• 1 Chord length of a circumference
  • 1.1 String theorem 
• 2 Solved exercises of strings
  • 2.1 - Exercise 1
  • 2.2 - Exercise 2
• 3 References

Chord length of a circumference

We can calculate the length of the chord in a circle from Figures 3a and 3b. Note that a triangle is always formed with two equal sides (isosceles): segments OA and OB, which measure R, the radius of the circumference. The third side of the triangle is the segment AB, called C, which is precisely the length of the chord.

It is necessary to draw a line perpendicular to the chord C to bisect the angle θ that exists between the two radii and whose vertex is the center O of the circumference. This is a central angle -because its vertex is the center- and the bisector line is also a secant to the circumference.

Immediately two right triangles are formed, whose hypotenuse measures R. Since the bisector, and with it the diameter, divides the chord into two equal parts, it turns out that one of the legs is half of C, as indicated in figure 3b.

From the definition of the sine of an angle:

sin (θ / 2) = opposite leg / hypotenuse = (C / 2) / R

Therefore:

sin (θ / 2) = C / 2R

C = 2R sin (θ / 2)

String theorem 

The string theorem goes like this:

If any two chords of a circle intersect at a point, the product of the length of the segments that appear on one of the chords is equal to the product of the lengths of the segments that are defined on the other chord..

The following figure shows two chords of the same circumference: AB and CD, which intersect at point P. In the chord AB the segments AP and PB are defined, while in the chord CD CP and PD are defined. So, according to the theorem:

AP. PB = CP. P.S.

Solved exercises of strings

- Exercise 1

A circumference has a 48 cm chord, which is 7 cm from the center. Calculate the area of ​​the circle and the perimeter of the circumference.

Solution  

To calculate the area of ​​circle A, it is enough to know the radius of the circumference squared, since it is true:

A = π.R^two

Now, the figure that is formed with the data provided is a right triangle, whose legs are 7 and 24 cm respectively.

Therefore to find the value of R^two the Pythagorean theorem is applied directly c^two = a^two + b^two, since R is the hypotenuse of the triangle:

R^two = (7 cm)^two + (24 cm)^two = 625 cm^two

So the requested area is:

A = π. 625 cm^two = 1963.5 cm^two

Regarding the perimeter or length L of the circumference, it is calculated by:

L = 2π. R

Substituting values:

R = √625 cm^two = 25 cm

L = 2π. 25 cm = 157.1 cm.

- Exercise 2

Determine the length of the chord of a circle whose equation is:

x^two + Y^two - 6x - 14y -111 = 0

The coordinates of the midpoint of the chord are known to be P (17/2; 7/2).

Solution

The midpoint of the chord P not it belongs to the circumference, but the end points of the chord do. The problem can be solved using the string theorem previously enunciated, but first it is convenient to write the equation of the circumference in canonical form, to determine its radius R and its center O.

Step 1: obtain the canonical equation of the circumference

The canonical equation of the circle with center (h, k) is:

(x-h)^two + (y-k)^two = R^two

To obtain it, it is necessary to complete squares:

(x^two - 6x) + (and^two - 14y) -111 = 0

Note that 6x = 2. (3x) and 14y = 2. (7y), so that the previous expression is rewritten like this, remaining unchanged:

(x^two - 6x + 3^two-3^two) + (and^two - 14y + 7^two-7^two) -111 = 0

And now, remembering the definition of remarkable product (a-b)^two = a^two - 2ab + b^two It can be written:

(x - 3)^two - 3^two + (and - 7)^two - 7^two - 111 = 0

= (x - 3)^two + (and - 7)^two = 111 + 3^two + 7^two → (x - 3)^two + (and - 7)^two = 169

The circumference has center (3,7) and radius R = √169 = 13. The following figure shows the graph of the circumference and the chords that will be used in the theorem:

Step 2: determine the segments to use in the string theorem

The segments to be used are the strings CD and AB, according to figure 6, both are cut at point P, therefore:

CP. PD = AP. PB

Now we are going to find the distance between points O and P, since this will give us the length of the segment OP. If we add the radius to this length, we will have the segment CP.

The distance d_OP between two coordinate points (x₁,Y₁) and (x_two,Y_two) it is:

d_OP^two = OP^two = (x_two - x₁)^two + (Y_two - Y₁)^two = (3- 17/2)^two + (7/7/2)^two = 121/4 + 49/4 = 170/4

d_OP = OP = √170 / 2

With all the results obtained, plus the graph, we construct the following list of segments (see figure 6):

CO = 13 cm = R

OP = √170 / 2 cm

CP = OP + R = 13 + √170 / 2 cm

PD = OD - OP = 13 - √170 / 2 cm

AP = PB

2.AP = chord length

Substituting in the string theorem:

CP. PD = AP. PB = [(13 + √170 / 2). (13 -√170 / 2)] = AP^two

[169 -170/4] = AP^two

253/2 = AP^two

AP = √ (253/2)

The length of the chord is 2.AP = 2 (√253 / 2) = √506

Could the reader solve the problem in another way?

References

1. Baldor, A. 2004. Plane and space geometry with Trigonometry. Publicaciones Cultural S.A. de C.V. Mexico.
2. C-K12. Lenght of a Chord. Recovered from: ck12.org.
3. Escobar, J. The Circumference. Recovered from: matematicas.udea.edu.co.
4. Villena, M. Cónicas. Recovered from: dspace.espol.edu.ec.
5. Wikipedia. Rope (Geometry). Recovered from: es.wikipedia.org.