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the trapezoid ABCD

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by sanju09 » Wed Nov 16, 2011 4:40 am
What is the area of the trapezoid ABCD if AB = BC = CD = 6, and AD = 12?
(A) 6√6
(B) 18√2
(C) 18√3
(D) 27√2
(E) 27√3
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Source: — Problem Solving |

by neelgandham » Wed Nov 16, 2011 5:14 am
Trapezoid ABCD looks like the one in the attachment

Area of ABCD = Area of the rectangle(in red) + 2*(Area of the triangle in green)
Height of the triangle = 0.5* base * height = 0.5*3*( square root((6^2)-(3^2)) = 0.5*3*3*square root(3)
Area of ABCD = 6*3*square root(3) + 2*0.5*3*3*square root(3)= 27*square root(3)

IMO E
p.s: Not sure of a formula !
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by shankar.ashwin » Wed Nov 16, 2011 7:24 am
Formula for area of trapezoid = 1/2 * Sum of parallel sides * height

Sum of parallel sides = 6+12 = 18

Height = √(36-9) = 3√(3)

Area = 1/2 * 18 * 3 √3 = 27 √3 E
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by MBA.Aspirant » Wed Nov 16, 2011 4:21 pm
sanju09 wrote:What is the area of the trapezoid ABCD if AB = BC = CD = 6, and AD = 12?
(A) 6√6
(B) 18√2
(C) 18√3
(D) 27√2
(E) 27√3
height = 3 √3

area = (6+12)/2 * 3 √3= 27 √3
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