Problem Solving

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

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

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

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)