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manjus_mailme
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The hexagon ABCDEF is regular.That means all its sides are the same length and all its interior angles are the same size.Each side of the Hexagon is 2 feet.What is the area of the rectangle BCEF ?The answer is 4(square root of 3)
The solution given in the book is that Given BC and EF are each 2 feet.Since area of the rectangle is length times width ,you must find the length(CE or BF).Look at Triangle ABF .It has two equal sides(AB=AF) ,So the perpendicular from A to the line BF divides ABF into two congruent right triangles AHF and AHB each with hypotenuse 2.
The angle FAB is 120 since the total of the angles of the hexagon is 720 .So each of the two triangles is 30-60-90 right triangle with hypotenuse 2.So AH=1,anf FH=HB =Square root of 3.There for BF=2*^3 and the area is 2*2^3=4^3 square feet.
I did not understand how they arrived at AH=1 from the angles of the triangles and hypotenuse?If anyone can pls explain to me.Thanks[/img]
The solution given in the book is that Given BC and EF are each 2 feet.Since area of the rectangle is length times width ,you must find the length(CE or BF).Look at Triangle ABF .It has two equal sides(AB=AF) ,So the perpendicular from A to the line BF divides ABF into two congruent right triangles AHF and AHB each with hypotenuse 2.
The angle FAB is 120 since the total of the angles of the hexagon is 720 .So each of the two triangles is 30-60-90 right triangle with hypotenuse 2.So AH=1,anf FH=HB =Square root of 3.There for BF=2*^3 and the area is 2*2^3=4^3 square feet.
I did not understand how they arrived at AH=1 from the angles of the triangles and hypotenuse?If anyone can pls explain to me.Thanks[/img]
