What is the Area of an octagon with sides 7.5 in. and diameter of 8.25 in.?

Answer 1

What is the "diameter" of an octagon?

For a circle it is the length of the chord between two points on the circumference which passes through the centre of the circle. This can also be described as the length of a straight line between two points on the circumference of the circle which also passes through the point which is the centre of the circle.

For an octagon you cannot choose a point such that the length of the straight line between two points on opposite sides of the octagon which passes through that point is a constant.

There is a circumcircle which is the circle which passes through all vertices of the octagon. When all sides of the octagon are equal in length this can only occur if the octagon is regular (ie every interior angle of the octagon is the same [at 135°]). In this case the diameter of this circle for an octagon with side length 7.5 in is approx 19.6 in

There is also the inscribed circle which is the largest circle that can be drawn inside the octagon. If it touches every side, the octagon must be a regular octagon and this circle has a diameter of approx 18.1 in.

If you are referring to a circle with a diameter of 8.25 in drawn between two opposite sides (which are parallel) then this does not define a unique octagon as the octagon will be rather squashed and the three sides at each end of the opposite sides which are 8.25 in apart can be placed in up to two different configurations which lead to 4 different possible areas for each skew position of the sides 8.25 in apart.

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If you are asking for the area of the regular octagon with side length 7.5 in then:

Consider the circumcircle of the octagon and its centre O.

Join every vertex to the centre O.

The area of the octagon is the sum of the areas of these triangles.

As two sides of each the triangle are radii of the circle, they are isosceles triangles.

As the third side of each triangle is the same, these isosceles triangles are congruent.

The area of the regular octagon is thus 8 times the area of one of these triangles.

As there are 8 triangles all around the centre O, the angle at the vertex at the point O is:

Angle at O = 360° ÷ 8 = 45°

Drop a perpendicular from this angle to the side of the octagon to get the height of the triangle

As the triangle is an isosceles triangle, this perpendicular line bisects the angle at O, and the base of the triangle (side of the octagon).

You now has a right angle triangle with its hypotenuse the radius of the circumcircle, one side half the length of the side of the octagon and its third side the height of the isosceles triangle.

Using the tangent ratio, the height of the isosceles triangle can be found:

tangent = opposite/adjacent

→ tan 22½° = half_octagon_side_length / height_of_triangle

→ height_of_triangle = (7.5 in ÷ 2) / tan 22½° ≈ 9.05 in

→ area triangle ≈ ½ × 7.5 in × 9.05 in ≈ 33.95 in²

→ area octagon = 8 × area triangles ≈ 8 × 33.95 in² ≈ 271.6 in²

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This can be generalised into the area of ANY regular polygon:

area_regular_polygon = side² × ¼ × number_of_sides ÷ tan(180° ÷ number_of_sides)

For an octagon with side length 7.5 in, this gives:

area = (7.5 in)² × ¼ × 8 ÷ tan(180° ÷ 8) ≈ 271.6 in² as before.

The height of the isosceles triangles is called the apothem of the [regular] polygon; it is the radius of the inscribed circle. This means the formula for the area of the regular polygon becomes:

area_regular_polygon = number_of_sides × ½ × side × apothem