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Sub 505 (Easy)|   Geometry|      
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Bunuel
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What is the value of xy? 17 Jul 2019, 00:45
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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5% (low)

Question Stats:

89% (01:06) correct 11% (01:31) wrong based on 101 sessions

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What is the value of xy?

A. 2400
B. 3200
C. 4800
D. 6300
E. 7200

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E
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What is the value of xy? 17 Jul 2019, 00:58
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The angles of an octagon are all equal and is 120 degrees each

Complementary angle sum is 180
So outside angle is 60 deg

X= 120
Y = 60

Xy = 7200 = E

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What is the value of xy? 17 Jul 2019, 01:23
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Bunuel

What is the value of xy?

A. 2400
B. 3200
C. 4800
D. 6300
E. 7200

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Attachment:
#GREpracticequestion What is the value of xy.jpg
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Since y =60
x = 180-60 =120
xy = 120 *60= 7200

IMO E
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What is the value of xy? 17 Jul 2019, 01:27
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We can divide the figure in Hexagon and a triangle.
For Hexagon, each interior angle x= \(\frac{(6-2)}{6}\) * 180 = 120
For Triangle, each interior angle y = 60

Thus, x*y = 120*60 = 7200

IMO the answer is E.

Please hit kudos if you like the solution.
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What is the value of xy? 17 Jul 2019, 02:12
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x is the interior angle of a regular hexagon. So x=120

y is the interior angle of an equilateral triangle. So y=60

xy=120*60=7200

Answer is (E)

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What is the value of xy? 17 Jul 2019, 02:24
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Since all the angles of the triangle are equal therefore value of y=60.Now sum of suplimentery angle are 180 therefore x=120
So correct answere must be 7200(E)
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What is the value of xy? 17 Jul 2019, 03:18
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The top part of the figure is a regular hexagon. Each angle of a hexagon is 120.

So, x = 120
y = 60 (linear pair)

xy = 120 x 60 = 7200.

Hence, the answer is E.


Note: The sum of interior angles of an n-sided polygon = (n - 2)180

Please give kudos if you liked the solution. :)


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What is the value of xy? 17 Jul 2019, 08:07
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angles of regular hexagon = x=120 each
Sum of angle of triangle =180
each angle y =60
hence xy =7200
D
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What is the value of xy? 21 Aug 2020, 21:35
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Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

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