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Quadrilateral RSTU shown above is a site plan for a parking lot in
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14 Dec 2012, 07:53
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Quadrilateral RSTU shown above is a site plan for a parking lot in which side RU is parallel to side ST and RU is longer than ST. What is the area of the parking lot? (1) RU = 80 meters (2) TU = \(20\sqrt{10}\) meters Attachment:
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Quadrilateral RSTU shown above is a site plan for a parking lot in
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14 Dec 2012, 08:02



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15 Jul 2016, 04:26
Bunuel wrote: Quadrilateral RSTU shown above is a site plan for a parking lot in which side RU is parallel to side ST and RU is longer than ST. What is the area of the parking lot?Given figure is a trapezoid, thus its area is SW*(ST+RU)/2=60*(45+RU)/2. So, all we need to know to answer the question is the length of RU. (1) RU = 80 meters. Sufficient. (2) TU = \(20\sqrt{10}\) meters. Draw altitude from vertex T to RU as shown below: Attachment: The attachment trapezoid.png is no longer available Since TX=SW=60 and TU = \(20\sqrt{10}\), then we can find XU. Therefore we can find RU=RW+WX+XU. Sufficient. Answer: D. Hi Bunuel, I am referring to a similar question in Sackmann's Challenge sets (as attached). At first it seemed a simple and a straightforward option D, but the solution (in spoiler) provided in the set says we do not know about the symmetry and hence we cannot use pythagoras theorem to deduce the base of the triangle. I am not sure if this reasoning is correct. Please advise. Statement (1) is insufficient, but it does help. Knowing ST, combined with the given length, allows us to use the pythagorean theorem to
and the length of SW. However, that does not give us the entire length of RS unless we knew the
figure was symmetrical, which we dont. Thanks.
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Re: Quadrilateral RSTU shown above is a site plan for a parking lot in
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15 Jul 2016, 05:37
aditi2013 wrote: Bunuel wrote: Quadrilateral RSTU shown above is a site plan for a parking lot in which side RU is parallel to side ST and RU is longer than ST. What is the area of the parking lot?Given figure is a trapezoid, thus its area is SW*(ST+RU)/2=60*(45+RU)/2. So, all we need to know to answer the question is the length of RU. (1) RU = 80 meters. Sufficient. (2) TU = \(20\sqrt{10}\) meters. Draw altitude from vertex T to RU as shown below: Attachment: trapezoid.png Since TX=SW=60 and TU = \(20\sqrt{10}\), then we can find XU. Therefore we can find RU=RW+WX+XU. Sufficient. Answer: D. Hi Bunuel, I am referring to a similar question in Sackmann's Challenge sets (as attached). At first it seemed a simple and a straightforward option D, but the solution (in spoiler) provided in the set says we do not know about the symmetry and hence we cannot use pythagoras theorem to deduce the base of the triangle. I am not sure if this reasoning is correct. Please advise. Statement (1) is insufficient, but it does help. Knowing ST, combined with the given length, allows us to use the pythagorean theorem to
and the length of SW. However, that does not give us the entire length of RS unless we knew the
figure was symmetrical, which we dont. Thanks. Hi! aditi2013, If you notice in question from Sackmann's Challenge sets, length of RO (if UO is perpendicular drawn from U) is not given. Because of which we wouldn't be able to calculate the length of one base. Whereas, in question by Bunuel, length of RW is given for us to calculate length of base. Hope I am able to answer your query.
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Re: Quadrilateral RSTU shown above is a site plan for a parking lot in
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15 Jul 2016, 07:37
Walkabout wrote: Attachment: lot.png Quadrilateral RSTU shown above is a site plan for a parking lot in which side RU is parallel to side ST and RU is longer than ST. What is the area of the parking lot? (1) RU = 80 meters (2) TU = \(20\sqrt{10}\) meters Area of Quadrilateral = \(\frac{1}{2}\)*(SW)*(ST+RU) We have SW & ST  We need RU to solve for area. Statement 1) gives RU. SufficientStatement 2) gives TU  Perpendicular from RU to T should be the same length as SW. Using Pythagoras therom we can find RU = RW+ST+Ans of Pythagoras theroem. Sufficient.Ans D)
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Re: Quadrilateral RSTU shown above is a site plan for a parking lot in
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15 Jul 2016, 12:12
Divyadisha wrote: Hi! aditi2013, If you notice in question from Sackmann's Challenge sets, length of RO (if UO is perpendicular drawn from U) is not given. Because of which we wouldn't be able to calculate the length of one base. Whereas, in question by Bunuel, length of RW is given for us to calculate length of base. Hope I am able to answer your query.[/quote] Hi Divyadisha, My query is about statement 1 , which gives us length of ST (we already have height). However, the official solution does not use pythagoras theorem to determine the length of the base, as it mentions that we are not sure about the symmetry. I am unable to understand that part. Please refer to spoiler in my post and see if it can help to resolve my query. Regards Aditi



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Re: Quadrilateral RSTU shown above is a site plan for a parking lot in
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21 Apr 2018, 03:42
Bunuel wrote: Quadrilateral RSTU shown above is a site plan for a parking lot in which side RU is parallel to side ST and RU is longer than ST. What is the area of the parking lot?Given figure is a trapezoid, thus its area is SW*(ST+RU)/2=60*(45+RU)/2. So, all we need to know to answer the question is the length of RU. (1) RU = 80 meters. Sufficient. (2) TU = \(20\sqrt{10}\) meters. Draw altitude from vertex T to RU as shown below: Since TX=SW=60 and TU = \(20\sqrt{10}\), then we can find XU. Therefore we can find RU=RW+WX+XU. Sufficient. Answer: D. Hi pushpitkc, hope you are having fantastic gmat weekend this DS question and answers are clear, just want to solve to find XU Since TX=SW=60 and TU = \(20\sqrt{10}\), then we can find XU. so we we use pythegorean theorem \(a^2+b^2 =c^2\) let XU be \(a\) \(a^2+60 =20\sqrt{10}\) \(a^2=20\sqrt{10}  60\) can you please explain what am i doing here wrong ?



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Re: Quadrilateral RSTU shown above is a site plan for a parking lot in
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21 Apr 2018, 05:47
dave13 wrote: Bunuel wrote: Quadrilateral RSTU shown above is a site plan for a parking lot in which side RU is parallel to side ST and RU is longer than ST. What is the area of the parking lot?Given figure is a trapezoid, thus its area is SW*(ST+RU)/2=60*(45+RU)/2. So, all we need to know to answer the question is the length of RU. (1) RU = 80 meters. Sufficient. (2) TU = \(20\sqrt{10}\) meters. Draw altitude from vertex T to RU as shown below: Since TX=SW=60 and TU = \(20\sqrt{10}\), then we can find XU. Therefore we can find RU=RW+WX+XU. Sufficient. Answer: D. Hi pushpitkc, hope you are having fantastic gmat weekend this DS question and answers are clear, just want to solve to find XU Since TX=SW=60 and TU = \(20\sqrt{10}\), then we can find XU. so we we use pythegorean theorem \(a^2+b^2 =c^2\) let XU be \(a\) \(a^2+60 =20\sqrt{10}\) \(a^2=20\sqrt{10}  60\) can you please explain what am i doing here wrong ? Hi dave13The mistake you have made is that \(b^2 = 60^2 = 3600\) and \(c^2 = (20\sqrt{10})^2 = 4000\) Here, \(a^2 + 3600 = 4000\) > \(a^2 = 4000  3600 = 400\) > \(a = \sqrt{400} = 20\) Hope that helps you!
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Re: Quadrilateral RSTU shown above is a site plan for a parking lot in
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