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In circle O, chords AB and CD intersect at E, AE = 3 inches, BE = 8 inches, and CE is 2 inches longer than DE. What is the length of DE, expressed in inches?

User Kayne
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1 Answer

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Okay, let's break this down step-by-step:

* Circle O has chords AB and CD that intersect at E

* AE = 3 inches

* BE = 8 inches

* CE is 2 inches longer than DE

So we know:

AE = 3 inches

BE = 8 inches

CE = DE + 2 inches

To find DE, we can use the Pythagorean theorem for any right triangle:

a^2 + b^2 = c^2

In this case:

3^2 + 8^2 = c^2

9 + 64 = 73

c = 8 inches

So the hypotenuse AE forms a right triangle with leg BE of length 8 inches.

Now we have the length of one leg (BE = 8 inches) and the hypotenuse (AE = 8 inches).

We can use the relation between leg, hypotenuse and sine to calculate the other leg (DE):

sin(A) = DE / 8

DE = 8 * sin(A)

Without knowing the angle A, we can make an estimate:

sin(A) ≈ A (for small A)

So DE ≈ 8A inches

Now we know:

CE = DE + 2 inches

And DE ≈ 8A inches

So: 8A + 2 = DE

=> DE = 10A

And since A is small, we can say: DE ≈ 10 * A inches

To summary, if AE = 3 inches and BE = 8 inches,

then DE ≈ 10 * A inches

DE is approximately 10 times some small angle A.

Does this make sense? Let me know if you have any other questions!

User Ccjmne
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