Two secant segments are drawn to a circle from a point outside the circle. The external segment of the first secant segment is 8 centimeters and its internal segment is 6 centimeters. If the entire length of the second secant segment is 28 centimeters, what is the length of its external segment?
let the outer segment of the 2nd secant be x
then the inner segment is 28-x
8/6 = x/(28-x)
6x = 224 - 8x
14x = 224
x = 16
To find the length of the external segment of the second secant segment, we can use the Secant-Secant Theorem.
The Secant-Secant Theorem states that if two secant segments are drawn from the same point outside a circle, the product of the lengths of the external segment of one secant segment and the length of the entire other secant segment is equal to the product of the lengths of the internal segment of the first secant segment and the length of the entire second secant segment.
In this case, we have:
External segment of first secant segment: 8 cm
Internal segment of first secant segment: 6 cm
Length of second secant segment: 28 cm
Using the Secant-Secant Theorem, we can set up the following equation:
8 cm * (8 cm + x) = 6 cm * 28 cm
Where x represents the length of the external segment of the second secant segment.
Simplifying the equation, we get:
64 cm^2 + 8 cm * x = 168 cm^2
Rearranging the equation, we have:
8 cm * x = 168 cm^2 - 64 cm^2
8 cm * x = 104 cm^2
Dividing both sides of the equation by 8 cm, we find:
x = 13 cm
Therefore, the length of the external segment of the second secant segment is 13 centimeters.
To solve this problem, we can use a property of secant segments drawn to a circle:
When two secant segments are drawn to a circle from the same external point, the product of the lengths of the external segment of one secant and the whole secant will be equal to the product of the lengths of the external segment of the other secant and the whole secant.
Let's denote:
x = length of the external segment of the second secant
Given information:
External segment of the first secant = 8 cm
Internal segment of the first secant = 6 cm
Length of the second secant = 28 cm
Applying the property mentioned above, we can set up the equation:
8 * (8 + 6) = x * (x + 28)
Now, let's solve the equation:
64 = (x^2 + 28x)
Rearranging the equation:
x^2 + 28x - 64 = 0
To solve this quadratic equation, we can either factor it or use the quadratic formula. Let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Here, a = 1, b = 28, and c = -64. Plugging in these values:
x = (-28 ± √(28^2 - 4 * 1 * -64)) / (2 * 1)
Simplifying:
x = (-28 ± √(784 + 256)) / 2
x = (-28 ± √(1040)) / 2
x = (-28 ± 32.25) / 2
Simplifying further:
x = (-28 + 32.25) / 2 or x = (-28 - 32.25) / 2
x = 4.25 / 2 or x = -60.25 / 2
x = 2.125 or x = -30.125
Since length cannot be negative, the length of the external segment of the second secant is 2.125 centimeters.