In the diagram, AE¯¯¯¯¯¯¯¯ is tangent to the circle at point A, and secant DE¯¯¯¯¯¯¯¯ intersects the circle at points C and D. The lines intersect outside the circle at point E.
A circle with no center shown. Points A, C, and D lie on the circle. Tangent A E and secant D E are as described in the text. Segment A E equals 12, segment E C equals 8, and segment C D equals x plus 10.
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What is the length of ED¯¯¯¯¯¯¯¯?
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could somebody help im really stuck on this one
18
So am I. What is x about?
Recall the secant-tangent theorem, and you have
EA^2 = EC*CD
12^2 = 8*(x+10)
and now ED = EC+CD = 8+x+10
I suspect a typo somewhere in the murk above
Well, let me clown around and try to help you out!
First, let's gather the given information: AE¯¯¯¯¯¯¯¯ = 12, EC¯¯¯¯¯¯¯¯ = 8, and CD¯¯¯¯¯¯¯¯ = x + 10.
Now, let's analyze the situation. Since AE¯¯¯¯¯¯¯¯ is tangent to the circle, it forms a right angle with the radius, which means triangle AEC is a right triangle. We can use the Pythagorean Theorem to find AC, which is the hypotenuse:
AC² = AE² + EC²
AC² = 12² + 8²
AC² = 144 + 64
AC² = 208
Now, let's look at triangle DCE. Since DE¯¯¯¯¯¯¯¯ is a secant, it intersects the circle at points C and D. This means that angle DCE is an inscribed angle, which is half the measure of the intercepted arc DC. So, angle DCE is ½(x + 10).
Since triangle DCE is isosceles (EC¯¯¯¯¯¯¯¯ = DC¯¯¯¯¯¯¯¯ = 8), angle DEC is also ½(x + 10).
Now, let's use the fact that the sum of the angles in a triangle is 180 degrees to find angle EDC:
angle DCE + angle DEC + angle EDC = 180
½(x + 10) + ½(x + 10) + 90 = 180
x + 20 + x + 20 + 180 = 360
2x + 220 = 360
2x = 360 - 220
2x = 140
x = 70
Now that we know x = 70, we can find CD¯¯¯¯¯¯¯¯:
CD¯¯¯¯¯¯¯¯ = x + 10
CD¯¯¯¯¯¯¯¯ = 70 + 10
CD¯¯¯¯¯¯¯¯ = 80
Finally, let's find the length of ED¯¯¯¯¯¯¯¯ by adding AE¯¯¯¯¯¯¯¯, EC¯¯¯¯¯¯¯¯, and CD¯¯¯¯¯¯¯¯:
ED¯¯¯¯¯¯¯¯ = AE¯¯¯¯¯¯¯¯ + EC¯¯¯¯¯¯¯¯ + CD¯¯¯¯¯¯¯¯
ED¯¯¯¯¯¯¯¯ = 12 + 8 + 80
ED¯¯¯¯¯¯¯¯ = 100
So, the length of ED¯¯¯¯¯¯¯¯ is 100!
Hope that brings a smile to your face!
To find the length of ED¯¯¯¯¯¯¯¯, we can use the tangent-secant theorem. This theorem states that when a tangent line and a secant line intersect outside the circle, the product of the lengths of the whole secant line and its external segment is equal to the square of the length of the tangent line.
In this case, we are given that AE¯¯¯¯¯¯¯¯ is tangent to the circle and has a length of 12. We also know that segment EC¯¯¯¯¯¯¯¯ equals 8. To find the length of ED¯¯¯¯¯¯¯¯, we need to find the length of segment CD¯¯¯¯¯¯¯¯.
Given that segment CD¯¯¯¯¯¯¯¯ equals x + 10, we can set up an equation using the tangent-secant theorem:
12 * (12 + x + 10) = 8^2
To solve this equation, we can simplify it:
(12)(22 + x) = 64
Multiplying both sides, we get:
264 + 12x = 64
Subtracting 264 from both sides, we get:
12x = -200
Dividing both sides by 12, we get:
x = -16.67
Since the length of a segment cannot be negative, we can disregard this negative value. Therefore, we conclude that x = 0.
From the given information, we know that segment CD¯¯¯¯¯¯¯¯ equals x + 10, which means CD¯¯¯¯¯¯¯¯ equals 0 + 10, which equals 10.
Finally, to find the length of ED¯¯¯¯¯¯¯¯, we need to add the lengths of CE¯¯¯¯¯¯¯¯ (8) and CD¯¯¯¯¯¯¯¯ (10):
ED¯¯¯¯¯¯¯¯ = EC¯¯¯¯¯¯¯¯ + CD¯¯¯¯¯¯¯¯ = 8 + 10 = 18
Therefore, the length of ED¯¯¯¯¯¯¯¯ is 18.