8 junio, 2024

Chord (geometry): length, theorem and exercises

A rope, in plane geometry, is the line segment that joins two points on a curve. The line containing said segment is said to be a secant line to the curve. Often this is 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 the points A and B belong. The chord between A and B is the green segment. On the right is a circle and one of its chords, since it is possible to draw infinite numbers.

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 circle and measures twice the radius.

In the following figure are represented the radius, the diameter, a chord and also the arc of a circumference. Correctly identifying each one is important when solving problems.

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Chord length of a circle

We can calculate the length of the chord in a circle starting from figures 3a and 3b. Note that a triangle is always formed with two equal sides (isosceles): the 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 chord C to bisect the angle θ that exists between the two radii and whose vertex is the center O of the circle. This is a central angle -because its vertex is the center- and the bisector line is also a secant to the circumference.

Two right triangles are immediately 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 lengths of the segments that appear in one of the chords is equal to the product of the lengths of the segments that are defined in the other chord.

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

PA. BS = CP . P.S.

Solved exercises of strings

– Exercise 1

A circle has a chord of 48 cm, 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 fulfilled:

A = π.R2

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 R2, the Pythagorean theorem c2 = a2 + b2 is directly applied, since R is the hypotenuse of the triangle:

R2 = (7 cm)2 + (24 cm)2 = 625 cm2

So the required area is:

A = π. 625 cm2 = 1963.5 cm2

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

L = 2π. R.

Substituting values:

R = √625 cm2 = 25 cm

L = 2π. 25 cm = 157.1 cm.

– Exercise 2

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

x2 + y2 – 6x – 14y -111 = 0

It is known that the coordinates of the midpoint of the chord are P (17/2 ; 7/2).

Solution

The midpoint of the chord P does not belong to the circle, but the endpoints of the chord do. The problem can be solved using the previously stated string theorem, but first it is convenient to write the equation of the circle 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:

(xh)2 + (yk)2 = R2

To obtain it, it is necessary to complete squares:

(x2 – 6x) + (y2 – 14y) -111 = 0

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

(x2 – 6x+32-32) + (y2 – 14y+72-72) -111 = 0

And now, remembering the definition of remarkable product (ab)2 = a2 – 2ab + b2, we can write:

(x – 3)2 – 32 + (y – 7)2 – 72 – 111 = 0

= (x – 3)2 + (y – 7)2 = 111 + 32 + 72 → (x – 3)2 + (y – 7)2 = 169

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

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

The segments to use are the chords CD and AB, according to figure 6, both intersect at point P, therefore:

PC . PS = AP. bp

Now we are going to find the distance between the 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 dOP between two coordinate points (x1,y1) and (x2,y2) is:

dOP2 =OP2 = (x2 – x1)2 + (y2 – y1)2 = (3- 17/2)2 + (7- 7/2)2 = 121/4 + 49/4 = 170 /4

dOP = OP = √170 /2

With all the results obtained, plus the graph, we build 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 = BP

2.AP = length of the rope

Substituting in the string theorem:

PC . PS = AP. BP = [(13 +√170 /2) . (13 -√170 /2)] =AP2

[169 -170/4] =AP2

253/2 = AP2

AP = √(253/2)

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

Could the reader solve the problem in another way?

References

Baldor, A. 2004. Plane Geometry and Space with Trigonometry. Cultural Publications SA de CV Mexico. C-K12. Length of a Chord. Retrieved from: ck12.org. Escobar, J. The Circumference. Retrieved from: matematicas.udea.edu.co. Villena, M. Conics. Retrieved from: dspace.espol.edu.ec. Wikipedia. Chord (Geometry). Recovered from: es.wikipedia.org.

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