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In the figure above, PQRS and QTUV are squares,
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20 May 2021, 09:29
20 May 2021, 09:29
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85% (02:36) correct
14% (01:47) wrong based on 7 sessions
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In the figure above, ๐๐๐ ๐ and ๐๐๐๐ are squares, ๐ and ๐ are the midpoints of \(\overline{PQ}\) and \(\overline{RS}\) respectively, and ๐๐ = ๐๐. If ๐ ๐ = \(\frac{1}{2}\), what is the length of \(\overline {UV}\)?
A. \(\frac{\sqrt{5}-1}{2}
\)
B. \(\frac{\sqrt{3}-1}{2}
\)
C. \(\frac{\sqrt{5}}{2}\)
D. \(\frac{\sqrt{3}}{2}\)
E. \(\frac{2}{3}\)
Re: In the figure above, PQRS and QTUV are squares,
[#permalink]
20 May 2021, 19:23
20 May 2021, 19:23
1
PQRS and QTUV are squares
๐ ๐ = \(\frac{1}{2}\) = \(SX \)
So \(PQRS\) is a square with a side length of \(1\)
Right triangle SPW, \(SP^2 + PW^2 = SW^2\)
\(SW^2 = 1 + \frac{1}{4} = \frac{5}{4}\)
\(SW\) โ= \(\frac{\sqrt{5}}{2}\) = \(TW\)
\(TQ = TW - QW\)
\(QW = \frac{1}{2}\)
So TQ = \(\frac{\sqrt{5}}{2}\) - \(\frac{1}{2}\)
\(\overline {TQ}\) = \(\overline {UV}\) = \(\frac{\sqrt{5}-1}{2}\)
Answer A
๐ ๐ = \(\frac{1}{2}\) = \(SX \)
So \(PQRS\) is a square with a side length of \(1\)
Right triangle SPW, \(SP^2 + PW^2 = SW^2\)
\(SW^2 = 1 + \frac{1}{4} = \frac{5}{4}\)
\(SW\) โ= \(\frac{\sqrt{5}}{2}\) = \(TW\)
\(TQ = TW - QW\)
\(QW = \frac{1}{2}\)
So TQ = \(\frac{\sqrt{5}}{2}\) - \(\frac{1}{2}\)
\(\overline {TQ}\) = \(\overline {UV}\) = \(\frac{\sqrt{5}-1}{2}\)
Answer A
gmatclubot
Re: In the figure above, PQRS and QTUV are squares, [#permalink]
20 May 2021, 19:23
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