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

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.

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)

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

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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