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# Quadrilateral RSTU shown above is a site plan for a parking lot in

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Re: Quadrilateral RSTU shown above is a site plan for a parking lot in [#permalink]
Bunuel

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.

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 not give us the entire length of RS unless we knew
the figure was symmetrical, which we dont.

Thanks.

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.

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Re: Quadrilateral RSTU shown above is a site plan for a parking lot in [#permalink]
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. Sufficient
Statement 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 [#permalink]

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.

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
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Re: Quadrilateral RSTU shown above is a site plan for a parking lot in [#permalink]
Bunuel

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.

Attachment:
trapezoid.png

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 [#permalink]
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dave13
Bunuel

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.

Attachment:
trapezoid.png

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 dave13

The 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 [#permalink]
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Re: Quadrilateral RSTU shown above is a site plan for a parking lot in [#permalink]
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