Question

A hexagon with an apothem of 14.7 inches is shown. a regular hexagon has an apothem of 14.7 inches and a perimeter of 101.8 inches. what is the area of the hexagon? square inches

Answer 1

The area of the considered regular hexagon which has got 14.7 inches of apothem and a perimeter of 101.8 inches is 748.2 sq. inches.

What is apothem?

Apothem for a regular polygon is a line segment which originates from the center of the regular polygon and touches the mid of one of the sides of the regular polygon. It is perpendicular to the regular polygon's side it touches.

Regular polygons have all side same and that apothem bisects the side in two parts, (provable by symmetry).

Consider the diagram attached below.

The area of the regular hexagon considered = 6 times (area of triangle ABC) (because of symmetry).

Also, we have:

Area of triangle ABC = 2 times (Area of triangle ABD).

Thus, we get:
Area of the considered hexagon = 6×2×(Area of triangle ABD)

Area of the considered hexagon = 12×(Area of triangle ABD)

Perimeter of a closed figure = sum of its sides' lengths.

There are 6 equal sides in a regular hexagon (due to it being regular).

Thus, if each side is of 'a' inch length, then:

Perimeter = 6×a inches

$101.8 = 6a\\\\\text{Dividing both the sides by 6, to get 'a' on one side}\\\\a = \dfrac{101.8}{6} \approx 16.967 \: \rm inches$

This is bisected by the apothem.

Thus, we get:
Length of the line segment BD = |BD| = a/2 ≈ 8.483 inches

Since it is given that the length of the apothem = |AD| = 14.7 inches, therefore, we get:

$\text{Area of ABD} = \dfrac{1}{2} \times \rm base \times height \approx \dfrac{14.7 \times 8.483}{2} \approx 62.35 \: \rm in^2$

Thus, we get:

Area of the considered hexagon = 12×(Area of triangle ABD)

Area of the considered hexagon $\approx 12 \times 62.35 = 748.2 \: \rm in^2$

Thus, the area of the considered regular hexagon which has got 14.7 inches of apothem and a perimeter of 101.8 inches is 748.2 sq. inches.

Learn more about apothem here:

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