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Show that the area of a regular 8gon is equal to the product of its longest diagonal and its shortest diagonal.
 2 years ago
 2 years ago
Show that the area of a regular 8gon is equal to the product of its longest diagonal and its shortest diagonal.
 2 years ago
 2 years ago

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s_martinBest ResponseYou've already chosen the best response.0
dw:1318922329389:dw the octagon is regular giving it equal sides and angles thus all the triangles have the same area and both rectangles have the same area. now, we have 8 triangles with area= 1/2ab giving, total area of the triangles =8(1/2)ab =4ab also we have 2 rectangles with area =bc giving total area of the rectangles =2bc thus total area of the octagon =4ab+2bc ***** now, the length of the longest diagonal call it X = a+c+a =2a +c the length of the shortest diagonal call it Y = 2b multiplying X and Y gives XY=(2a +c)(2b)=4ab+2bc =area of octagon end of proof
 2 years ago
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