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Q: If s equals 8 inches and a equals 4 square root of 3 find the area of the regular hexagon?
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What is the apothem of a regular hexagon with sides of 16 inches?

We know that the height of an equilateral triangle equals the product of one half of the side length measure with square root of 3.Since in our regular hexagon we form 6 equilateral triangles with sides length of 16 inches, the apothem length equals to 8√3 inches.


What is the area of a regular hexagon if the perimeter is 42 inches?

The area is about 127.3 square inches.


What is the area of a regular hexagon with 24 inch perimeter?

The area of a regular hexagon with a 24-inch perimeter is about 41.57 square inches.


If s 8 inches and a find the area of the regular hexagon. Round the answer to one decimal place?

The area is 166.3 square inches.


What is the area of a hexagon with sides 8in?

Any value from 0 to 166.3 square inches, depending on whether or not it is regular.


What is the area of a regular hexagon whose side lenght is 16 inches and the apothem is 8 square root of 3?

665.1 square units.


If s equals 8 inches and a equals 4 square unit 3 find the area of the area of the regular hexagon?

The answer will depend on what s and a are measures of. And since you have not bothered to provide that crucial bit of information, I cannot provide a more useful answer.


Can you show that you can work out the area of a regular hexagon with a perimeter of LX inches rounding your answer to a suitable degree of accuracy?

Perimeter in inches = LX = 60. A regular hexagon with a perimeter of 60 inches has 6 equal sides of 10 inches. A regular hexagon has 6 equal interior equilateral triangles So in order to find the area of the hexagon we must first find the area of one of its interior triangles and then multiply the result by 6. Area of a triangle = base times perpendicular height divided by 2. An equilateral triangle can be considered as being two right angled triangles joined together. So by halving the length of its base we can find its perpendicular height by using Pythagoras' Theorem: hypotenuse2-base2 = perpendicular height2. 102-52 = 75 square inches. Square root of 75 = 8.660254038 inches. Area of the equilateral triangle = 10*8.660254038/2 = 43.30127019 square inches. Area of the regular hexagon = 43.30127019*6 = 259.8076211 square inches. Therefore the area of a regular hexagon with a perimeter of 60 inches is 260 square inches which is correct to 3 significant figures.


Is ht of hexogon2ht of square?

Not if the hexagon is a regular hexagon with sides of the same length as the sides of the square.


Can a regular hexagon be tessellated?

Yes. A regular tessellation can be created from either an equilateral triangle, a square, or a hexagon.


Area of the regular hexagon whose side length is 16 in and apothem is 8 square root 3 in?

It is 665.1 sq inches.


The area of this rectangle is 24 square inches give the dimensions of three other rectangles that have an area of 24 square inches?

1 inch by 24 inches equals 24 square inches 10 inches by 2.4 inches equals 24 square inches 100 inches by 0.24 inches equals 24 square inches 1000 inches by 0.024 inches equals 24 square inches also: 2 inches by 12 inches equals 24 square inches 2.5 inches by 9.6 inches equals 24 square inches 3 inches by 8 inches equals 24 square inches 3.2 inches by 7.5 inches equals 24 square inches 4 inches by 6 inches equals 24 square inches This list could be very long.