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# Two roads intersect as shown in the figure above. If RS = ST = TU = UR

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Math Expert
Joined: 02 Sep 2009
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Two roads intersect as shown in the figure above. If RS = ST = TU = UR [#permalink]

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05 Oct 2017, 01:33
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Two roads intersect as shown in the figure above. If RS = ST = TU = UR = 5 meters, what is the straight-line distance, in meters, from S to U?

(A) 5√3/2
(B) 5√2
(C) 15/2
(D) 5√3
(E) 10

[Reveal] Spoiler:
Attachment:

2017-10-04_1124_003.png [ 5.96 KiB | Viewed 422 times ]
[Reveal] Spoiler: OA

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Kudos [?]: 129160 [0], given: 12194

Senior Manager
Joined: 25 Nov 2015
Posts: 254

Kudos [?]: 71 [0], given: 245

Location: India
GPA: 3.64
WE: Engineering (Energy and Utilities)
Two roads intersect as shown in the figure above. If RS = ST = TU = UR [#permalink]

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05 Oct 2017, 07:11
RSTU forms a rhombus with each side =5
Join the diagonals RT and SU.
The diagonals of a rhombus are perpendicular bisectors.
Suppose they bisect at V.
Triangle RUV is a right angled triangle with RU=5, RV=5/2
By pythagoras theorem, $$\sqrt{5^2-(5/2)^2}$$ = $$\sqrt{75/4}$$
=5$$\sqrt{3}$$/2
The length of the diagonal is 5$$\sqrt{3}$$/2 x 2=5$$\sqrt{3}$$

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Last edited by souvonik2k on 05 Oct 2017, 22:11, edited 1 time in total.

Kudos [?]: 71 [0], given: 245

Director
Joined: 18 Aug 2016
Posts: 512

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GMAT 1: 630 Q47 V29
Re: Two roads intersect as shown in the figure above. If RS = ST = TU = UR [#permalink]

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05 Oct 2017, 12:10
souvonik2k wrote:
RSTU forms a rhombus with each side =5
Join the diagonals RT and SU.
The diagonals of a rhombus are perpendicular bisectors.
Suppose they bisect at V.
Triangle RUV is a right angled triangle with RU=5, UV=5/2
By pythagoras theorem, $$\sqrt{5^2-(5/2)^2}$$ = $$\sqrt{75/4}$$
=5$$\sqrt{3}$$/2
The length of the diagonal is 5$$\sqrt{3}$$/2 x 2=5$$\sqrt{3}$$

Hi souvonik2k

how did u get this UV value?
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Thanks
Luckisnoexcuse

Kudos [?]: 140 [0], given: 123

Director
Joined: 22 May 2016
Posts: 823

Kudos [?]: 266 [1], given: 553

Two roads intersect as shown in the figure above. If RS = ST = TU = UR [#permalink]

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05 Oct 2017, 12:12
1
KUDOS
Bunuel wrote:

Two roads intersect as shown in the figure above. If RS = ST = TU = UR = 5 meters, what is the straight-line distance, in meters, from S to U?

(A) 5√3/2
(B) 5√2
(C) 15/2
(D) 5√3
(E) 10

[Reveal] Spoiler:
Attachment:
The attachment 2017-10-04_1124_003.png is no longer available

Attachment:

rrrrr.png [ 25.02 KiB | Viewed 201 times ]

This can be done in one minute with some basic geometry; it is good but not necessary to know that diagonals of a rhombus are perpendicular bisectors. (I did not use that fact directly.)

If all four sides of a quadrilateral are equal, it is a rhombus. A square is a rhombus; this rhombus does not have right angles. It is not a square (so, for example, its four angles are not all equal).

You do need to know, about a rhombus:

a) opposite angles are equal
b) its diagonals are angle bisectors
c) like any quadrilateral, sum of interior angles is 360 degrees

NOTE: I am using capital letters S, T, U, and R as the names of the angle measures INSIDE the figure -- that is, as the name of the angles of the vertices

1) Find angle measures in degrees of S, T, U, and R

S = 60 (S's vertical angle = 60)
U = 60 (opposite angles of rhombus are equal)

S + U = 120
(360 - 120) = 240 degrees remaining

Opposite angles T and R are equal, so their measures are:
$$\frac{240}{2} = 120$$ degrees each

2) Draw the diagonals, note degree measure of all 4 bisected angles (see diagram)

S and U are now composed of two 30-degree angles
T and R and now composed of two 60-degree angles

Intersecting diagonals create 4 triangles

3) Assess the triangles whose side lengths, added, equal the length of $$SU$$

∆ SAT and ∆ UAT have sides lengths that sum to length of $$SU$$

In both triangles, one angle = 30, one angle = 60. The third must be 90. They are both right triangles

Further, they are 30-60-90 triangles with side ratio $$x: x√3: 2x$$

4) Find side length of SA, and multiply by 2. That is the answer

For 30-60-90 ∆ SAT, the corresponding side lengths from above ratio are:

$$2x = 5$$

$$x = \frac{5}{2}$$

$$x√3 = (\frac{5}{2})√3 = SA$$

$$SA = UA$$, and $$SA + UA = SU$$

Hence multiply $$SA$$ by 2

$$(\frac{5}{2})√3 * (2) = 5√3$$

Kudos [?]: 266 [1], given: 553

Senior Manager
Joined: 25 Nov 2015
Posts: 254

Kudos [?]: 71 [0], given: 245

Location: India
GPA: 3.64
WE: Engineering (Energy and Utilities)
Two roads intersect as shown in the figure above. If RS = ST = TU = UR [#permalink]

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05 Oct 2017, 22:10
Luckisnoexcuse wrote:
souvonik2k wrote:
RSTU forms a rhombus with each side =5
Join the diagonals RT and SU.
The diagonals of a rhombus are perpendicular bisectors.
Suppose they bisect at V.
Triangle RUV is a right angled triangle with RU=5, UV=5/2
By pythagoras theorem, $$\sqrt{5^2-(5/2)^2}$$ = $$\sqrt{75/4}$$
=5$$\sqrt{3}$$/2
The length of the diagonal is 5$$\sqrt{3}$$/2 x 2=5$$\sqrt{3}$$

Hi souvonik2k

how did u get this UV value?

Hi Luckisnoexcuse
Thanks for pointing out, it is a typo.
Actually it should be RV instead of UV in the highlighted portion.
Corrected in my earlier post.
_________________

When the going gets tough, the tough gets going...

Kudos [?]: 71 [0], given: 245

Two roads intersect as shown in the figure above. If RS = ST = TU = UR   [#permalink] 05 Oct 2017, 22:10
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