Mathematical Enigma - VI

Geometry Level 4

Find the area of a nonagon whose corners have the coordinates ( 2 , 9 ) , ( 6 , 11 ) , ( 9 , 8 ) , ( 6 , 7 ) , ( 13 , 6 ) , ( 9 , 2 ) , ( 4 , 3 ) , ( 5 , 8 ) (2,9),(6,11),(9,8),(6,7),(13,6),(9,2),(4,3),(5,8) and ( 3 , 1 ) . (3,1).


The answer is 54.5.

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1 solution

To find the area of the nanogon we integrate (or find the area under the curve) round the perimeter in sequence of the vertices. Let ( x 1 , y 1 ) = ( 2 , 9 ) ; ( x 2 , y 2 ) = ( 6 , 11 ) ; ( x 3 , y 3 ) = ( 9 , 8 ) ; . . . ( x 9 , y 9 ) = ( 2 , 1 ) (x_1, y_1) = (2,9); (x_2, y_2) = (6,11); (x_3, y_3) = (9,8);...(x_9, y_9) = (2,1) and ( x 10 , y 10 ) = ( x 1 , y 1 ) = ( 2 , 9 ) \color{#3D99F6}{(x_{10}, y_{10}) = (x_1, y_1) = (2,9)} . Then the area A A of the nanogon is given by:

A = n = 1 9 ( x n + 1 x n ) ( y n + y n + 1 ) 2 Area of a trapezium. \begin{aligned} A & = \sum_{n=1}^9 \color{#3D99F6}{\frac{(x_{n+1}-x_n)(y_n+y_{n+1})}{2} \quad \quad \small \text{Area of a trapezium.}} \end{aligned}

Using an Excel spreadsheet below, we can easily find that A = 54.5 A=\boxed{54.5} .

You write the most beautiful solutions, thanks for writing one for my problem :)

Swapnil Das - 5 years, 7 months ago

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You are welcome.

Chew-Seong Cheong - 5 years, 7 months ago

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