Mathematics question

Question

Given a regular octagon ABCDEF GH with side length 3. By drawing the four diagonals AF,

BE, CH, and DG, the octagon is divided into a square, four triangles, and four rectangles.

Find the sum of the areas of the square and the four triangles.

Answer 1

Find the sum of the areas of the square and the four triangles.

The square has the side length of 3, equal to that of the octagon. Its area is 9.

Each of the four triangles is an isosceles right triangle with hypotenuse length of 3.

Let is combine these four triangles by moving them towards the center, so that their right angles “meet” in one point.

Together these four triangles (or their hypotenuses) form a square with a side length of 3. Its area is also 9.

Hence the combined area of the square and 4 triangles is 9+9=18

Answer 2

octagon area = 2(1+sqr2)s^2

=2(1+sqr2)3^2= about 43.46

=about 2(2.828)9

=about 43.45584412

square area = 3x3 =9

rectangle = 3x3/sqr2= 9/sqr2

4 rectangles=18sqr2

triangle = (1/2)(9/2)=9/4

4 triangles= 9

sum of square, 4 triangles and 4 rectangles =9+18sqr2+9= about 43.46

sum of square + 4 triangles = 9+9 = 18