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Identify the area of a regular hexagon with side length 17 in. rounded to the nearest tenth.

User Angelin Calu
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2 Answers

20 votes
20 votes

Answer:

750.8in squared

Explanation:

The area of a regular polygon is one-half times the length of the apothem times the perimeter.

Notice that the formula combines the areas for the six triangles that fit inside the hexagon.

The perimeter is equal to the sum of the six triangle bases and the apothem is the height of each triangle.

The interior angles of the hexagon have a measure of 120°

, so the angles at the base of the triangle (and the third angle) have measure 60°

.The height forms a 30°−60°−90°

triangle, which has one leg equal to half of the hexagon side, or 8.5 in.

, and the other leg equal to 8.53–√ in

The figure shows a regular hexagon. Each angle of the hexagon measures 120 degrees. There is a triangle inside the hexagon. The base of the triangle coincides with one of the sides of the hexagon and one of the vertices of the triangle is the center of the hexagon. The length of the apothem is 8.5 times square root of 3 inches. The angles at the base of the triangle measure 60 degrees. The angle between the apothem and the side of the triangle measures 30 degrees. The apothem divides the base of the triangle into two parts. One of the parts is 8.5 inches long.

Find the perimeter of the hexagon.

P=6(17)

Simplify.

P=102in.

The formula for the area is A=12aP

. Substitute the known values and simplify.

A=12(8.53–√)(102)

=433.53–√in2

Round to the nearest tenth as 750.8 in

Therefore, the area of the regular hexagon with side length 17in.

is A≈7508. in2

.

User Mikebloch
by
2.9k points
6 votes
6 votes
750.8 square inches
User Kriyeta
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2.9k points