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In the diagram above, line JX is parallel to line KY. If the area of t

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In the diagram above, line JX is parallel to line KY. If the area of t  [#permalink]

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New post 10 Feb 2017, 08:48
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In the diagram above, line JX is parallel to line KY. If the area of triangle KYZ is 18 square feet, what is the length of side JX?

(1) Side JZ measures \(18\sqrt{2}\)

(2) Triangle KYZ is isosceles

Attachment:
TriangleJXZ.png
TriangleJXZ.png [ 9.05 KiB | Viewed 1242 times ]

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Re: In the diagram above, line JX is parallel to line KY. If the area of t  [#permalink]

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New post 10 Feb 2017, 09:29
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Bunuel wrote:
Image
(NOTE: Not drawn to scale)

In the diagram above, line JX is parallel to line KY. If the area of triangle KYZ is 18 square feet, what is the length of side JX?

(1) Side JZ measures \(18\sqrt{2}\)

(2) Triangle KYZ is isosceles

Attachment:
TriangleJXZ.png



Both triangle are AA similar
Since both have 90 deg. angle and two lines parallel
(1) no info of any sides of triangle KYZ
thus we cannot find any sides

(2) since KYZ is isosceles means its other two sides are 6 units each ( given area =18 or side^2/2=18)
but no info of bigger traingle JXZ sides

Combining we get one side of bigger JXZ triangle and we know two sides of KYZ=6units
thus all sides will be known (bigger triangle is also isosceles)
thus we can calculate the required data

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Re: In the diagram above, line JX is parallel to line KY. If the area of t  [#permalink]

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New post 11 Feb 2017, 10:05
its a question based on ratios,, since the area of smaller traingle is given,,we can find the ratios of the sides of the triangle and hence find the area of individual triangles and thereby sides...
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In the diagram above, line JX is parallel to line KY. If the area of t  [#permalink]

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New post 17 Feb 2017, 16:33
Bunuel wrote:
Image
(NOTE: Not drawn to scale)

In the diagram above, line JX is parallel to line KY. If the area of triangle KYZ is 18 square feet, what is the length of side JX?

(1) Side JZ measures \(18\sqrt{2}\)

(2) Triangle KYZ is isosceles

Attachment:
TriangleJXZ.png


Official solution from Veritas Prep.

C. Before you even get to the statements, you should recognize that the two triangles in the diagram are similar. Because the sides JX and KY are parallel, and each forms a 90 degree angle with the base, you know that all angles will be the same in both triangles, meaning that if you can find the relationship between the small and large triangles, you can extrapolate their sides and areas.

Statement 1 is not sufficient, as you don't know enough about any one side of the small triangle to use the similar triangles logic at this point. And statement 2 is not sufficient, as with that information you don't know any lengths of the large triangle.

But taken together, the statements are sufficient. Knowing that the small triangle is isosceles with an area of 18 tells you that \(\frac{1}{2}*a^2=18\), meaning that the short sides are each 6 feet in length, and that the hypotenuse measures \(6\sqrt{2}\). And since the hypotenuse of the large triangle is \(18\sqrt{2}\), you can find the relationship that the sides of the large triangle are three times the sides of the smaller triangle, making the length of side \(JX = 18\).
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Re: In the diagram above, line JX is parallel to line KY. If the area of t  [#permalink]

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Re: In the diagram above, line JX is parallel to line KY. If the area of t &nbs [#permalink] 08 Apr 2018, 06:21
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