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In the diagram above, triangle ABC is equilateral, figure SQRE is a square, and A is the midpoint of SQ. If the perimeter of triangle ABC is 6 inches, what is the length, in inches, of segment RY ?

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04 May 2015, 14:40

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find side of triangle: 6/3 = 2 inch per side find area of equilateral triangle: 2²* sqrt(3) / 4 = sqrt (3) find height: area of triangle = sqrt(3) = 1/2 * base * height -> solve for height: sqrt (3) = 1/2 * 2 * height -> height= sqrt (3) find third side: its a 90° triangle -> ratio is 1:sqrt(3):2 -> third site = 1 inch. find RB: (2 - sqrt(3)) / 2 find RY: use Thales' theorem -> 1 / sqrt(3) = RB / RY -> 1 / sqrt(3) = ((2 - sqrt(3)) / 2) / RY -> solve for RY -> RY = sqrt(3) - 1.5

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05 May 2015, 01:26

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Perimeter of equilateral \(\triangle\) = 6 Side of triangle = AB=BC=AC = 2

Side of Square = \(\sqrt{3}\)/2 * a = \(\sqrt{3}\)/2 * 2 = \(\sqrt{3}\)

Now \(\triangle\)YQA and \(\triangle\)YRB are similar triangles ; ----> As one angle \(90^{\circ}\), \(\angle\)y vertically opposite; all angles equal (AAA).

As \(\triangle\)YQA ~ \(\triangle\)YRB ; implies ratio of sides equal \(\frac{YR}{RB} = \frac{YQ}{AQ}\)

let YR = x , YQ = \(\sqrt{3}\) - x ; ----> ( as QR = \(\sqrt{3}\) )

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01 Jun 2015, 10:11

hsbinfy : Can you please explain how did you get QY=3/2.I am getting it as 1/2. Please explain how did you apply 30-6-90 to that triangle and how did you got tht value.

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01 Jun 2015, 11:16

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Shree9975 wrote:

hsbinfy : Can you please explain how did you get QY=3/2.I am getting it as 1/2. Please explain how did you apply 30-6-90 to that triangle and how did you got tht value.

Thanks

See the attachment.

For 30-60-90 triangle, sides are in the ratio 1 : √3 : 2. You just have to apply this rule to the triangles in the figure.

You will see that RY = √3 - 1.5

Answer C

Attachments

tria.jpg [ 35.17 KiB | Viewed 2727 times ]

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~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Kindly press Kudos if the explanation is clear. Thank you Ambarish

Equation 3 contains two unknowns: x and a. So, to find a unique value of x, we now know that we should try to find another relation involving x and/or a.

We get it by dropping a perpendicular from A on side BC.

In right triangle APB,

\(\frac{AP}{BP} = tan60 = \sqrt{3}\)

That is, \(\frac{2a}{1} = \sqrt{3}\)

That is, \(a = \frac{(sqrt3)}{2}\) . . . (4)

By solving (3) and (4), we get \(x = \sqrt{3} - 1.5\)

In the diagram above, triangle ABC is equilateral, figure SQRE is a square, and A is the midpoint of SQ. If the perimeter of triangle ABC is 6 inches, what is the length, in inches, of segment RY ?

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In the diagram above, triangle ABC is equilateral, figure SQRE is a square, and A is the midpoint of SQ. If the perimeter of triangle ABC is 6 inches, what is the length, in inches, of segment RY ?

Re: In the diagram above, triangle ABC is equilateral, figure SQRE is a sq [#permalink]

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20 Jun 2016, 12:43

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All of you are doing this problem incorrectly if you even begin to do the algebra or geometry.

If you know that the side of the triangle is 2 inches, then you know the length of RY has to be significantly less than that, so take a look at the answer choices.

A) 1/2 - way too big. B) 1.5 - way too big. C) .2 - close, keep this in mind D) .3 - close, but still too big. E) .85 - way too big.

The answer is between C and D, and because that segment looks a lot closer to 1/5 than to 1/3, go with answer C.

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21 Sep 2017, 01:04

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Here is my solution. Since, this one had a diagram and lot of rules, I posted an image. Hope this helps.

Attachments

21903586_10207991228218545_79220063_n.jpg [ 74.7 KiB | Viewed 348 times ]

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