Expert please help

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ilovemgmat: Junior | Next Rank: 30 Posts; Posts: 17; Joined: Fri Sep 02, 2011 6:19 pm; Thanked: 1 times

by ilovemgmat » Tue Oct 04, 2011 5:36 pm

A trapezoid has consecutive sides that measure x, 2x, 3x, and 2x. If one of the angles of the trapezoid measures 60 degrees, what is the area of the trapezoid in terms of x?

Answer Choices:
A. 2x^2 (2^(1/2))
B. 3x^2 (2^(1/2))
C. 2x^2 (3^(1/2))
D. 3x^2 (3^(1/2))
E. 4x^2 (3^(1/2))

OA is C

"Whoever one is, and wherever one is, one is always in the wrong if one is rude." ~Maurice Baring
Rudeness and sarcasm won't be entertained!

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Source: Beat The GMAT — Problem Solving | Tue Oct 04, 2011 5:36 pm

nimish: Newbie | Next Rank: 10 Posts; Posts: 8; Joined: Mon Nov 09, 2009 10:38 pm; Thanked: 2 times

by nimish » Tue Oct 04, 2011 6:06 pm

ilovemgmat wrote:A trapezoid has consecutive sides that measure x, 2x, 3x, and 2x. If one of the angles of the trapezoid measures 60 degrees, what is the area of the trapezoid in terms of x?

Answer Choices:
A. 2x^2 (2^(1/2))
B. 3x^2 (2^(1/2))
C. 2x^2 (3^(1/2))
D. 3x^2 (3^(1/2))
E. 4x^2 (3^(1/2))

OA is C

Area of trapezoid -
(1/2)*(sum of parallel sides)*(height).
Here,
sum of parallel sides = (x+3x)=4x
height =
sin 60 = 3^(1/2) = h/2x
so, h = x3^(1/2)
so area = (1/2)*x3^(1/2)*4x= 2x^2(3^(1/2))
so, answers is C

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ilovemgmat: Junior | Next Rank: 30 Posts; Posts: 17; Joined: Fri Sep 02, 2011 6:19 pm; Thanked: 1 times

by ilovemgmat » Tue Oct 04, 2011 6:14 pm

Do we always start from the top when the problem says four consecutive sides? Thank you for the explanation: it helped!

nimish wrote:
ilovemgmat wrote:A trapezoid has consecutive sides that measure x, 2x, 3x, and 2x. If one of the angles of the trapezoid measures 60 degrees, what is the area of the trapezoid in terms of x?

Answer Choices:
A. 2x^2 (2^(1/2))
B. 3x^2 (2^(1/2))
C. 2x^2 (3^(1/2))
D. 3x^2 (3^(1/2))
E. 4x^2 (3^(1/2))

OA is C
Area of trapezoid -
(1/2)*(sum of parallel sides)*(height).
Here,
sum of parallel sides = (x+3x)=4x
height =
sin 60 = 3^(1/2) = h/2x
so, h = x3^(1/2)
so area = (1/2)*x3^(1/2)*4x= 2x^2(3^(1/2))
so, answers is C

"Whoever one is, and wherever one is, one is always in the wrong if one is rude." ~Maurice Baring
Rudeness and sarcasm won't be entertained!

Like this post

sl750: Master | Next Rank: 500 Posts; Posts: 496; Joined: Tue Jun 07, 2011 5:34 amFollowed by: 1 member; Thanked: 38 times

by sl750 » Tue Oct 04, 2011 10:39 pm

Trigonometry is not necessary to solve this problem. Since, it is specified that this is an isosceles Trapezoid. The legs of the Trapezoid are of the same length (2x).

Area of the Trapezoid = Area of Right Triangle 1+Area of Rectangle+Area of Right Triangle 2

Area of Right Triangle1 = Area of Right Triangle 2 = 1/2*x*sqrt(3)*x. You can use the 30-60-90 rule to find the height of the Triangle

Area of the Rectangle = x*sqrt(3)*x

Sum the results = 2*x^2*sqrt(3)

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