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O is the center of the circle. If line segment DC has length

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O is the center of the circle. If line segment DC has length [#permalink] New post 09 Aug 2019, 10:09
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O is the center of the circle. If line segment DC has length \(\sqrt{8}\), and side AB has length \(\sqrt{30}\), what is the length of x?

A) \(\sqrt{10}\)

B) \(2\sqrt{3}\)

C) \(4\)

D) \(3\sqrt{2}\)

E) \(4\sqrt{2}\)
[Reveal] Spoiler: OA

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Re: O is the center of the circle. If line segment DC has length [#permalink] New post 09 Aug 2019, 10:48
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GreenlightTestPrep wrote:
Image
O is the center of the circle. If line segment DC has length \(\sqrt{8}\), and side AB has length \(\sqrt{30}\), what is the length of x?

A) \(\sqrt{10}\)

B) \(2\sqrt{3}\)

C) \(4\)

D) \(3\sqrt{2}\)

E) \(4\sqrt{2}\)


First recognize that, because ∠ABC is an inscribed angle "holding" (i.e., containing) the diameter, we know that ∠ABC = 90°
Image

At this point, it's crucial that we recognize that there are two similar triangles (one red and one blue) "hiding" in this diagram.
Image

ASIDE: We know that ∠ABC = 90° and ∠BDC = 90°, and we know that the angles denoted by stars are both equal (since they are actually the same angle in both cases).
If two triangles have two angles in common, their 3rd angles must be equal as well.
In other words, ∠BAC = ∠DBC (I have denoted both sides with hearts)

Since the red triangle and the blue triangle have the same 3 angles, they are similar triangles

KEY CONCEPT: With any two similar triangles, the ratios of corresponding sides will be equal.

If we remove all of the extra information . . .
Image
. . . we can see that:
AB and BD are corresponding sides (they are both between the right angle and the angle denoted by a heart)
BC and DC are corresponding sides (they are both between the right angle and the angle denoted by a star)

Before we can apply our nice property for similar triangles, let's let y = the length of side BC to get:
Image
At this point, we can create an equation using corresponding sides.
We can write: AB/BD = BC/DC
Plug in our values to get: √30/x = y/√8
Cross multiply to get: xy = √240
Divide both sides by x to get: y = (√240)/x
Of course if y = (√240)/x , then side BC in the blue triangle must also be (√240)/x

We can now add this information to our diagrams to get:
Image
ALMOST DONE!

Notice that we can apply the Pythagorean Theorem to the 3 sides of the BLUE triangle to get: x² + (√8)² = [(√240)/x]²
Expand both sides: x² + 8 = 240/x²
Multiply both sides by x² to get: x⁴ + 8x² = 240
Rewrite as: x⁴ + 8x² - 240 = 0
Factor: (x² + 20)(x² - 12) = 0

So, EITHER x² + 20 = 0 OR x² - 12 = 0
Since it's impossible for x² + 20 to equal 0, it must be the case that x² - 12 = 0

If x² - 12 = 0 then x² = 12, which means x = √12 or x = -√12
Since x cannot be negative, we know that x = √12
Simplify to get: x = √12 = 2√3

Answer: B


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Re: O is the center of the circle. If line segment DC has length [#permalink] New post 09 Aug 2019, 10:56
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Amazing.

Post for FB or Instagram account for the next week: Geometry week :)
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Re: O is the center of the circle. If line segment DC has length   [#permalink] 09 Aug 2019, 10:56
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