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584.6 inches2

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Q: What is the area of a hexagon if the length is 15 inches?
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Related questions

What is the area of a square with a side length of 15 inches?

Area = 15*15 = 225 square inches


What is the length of a rectangle when its area is 240 square inches and the width is 15 inches?

240/15 = 16 inches


What is the area of a rectangle with the length of 5 inches and width of 3 inches?

15 square inches


What is the area of a rectangle whose length equals 15 inches and width equals 2 inches?

Area = L*W = 15 * 2 = 30 square inches.


What is the width of a rectangle with length 25 in and area 375in2?

15 inches. Divide the known area by the known length !


What is the width of a rectangle with length 25inand area 375in?

area = length x width width = area/length = 375/25 = 15 inches


A regular hexagon has a side length of 15cm. What is the perimeter of the hexagon?

The perimeter of the hexagon is 6 times 15 = 90 cm


The length of rectangle is 2 inches more than its width The area of the rectangle is 15 square incheswhats the width and length?

They are: 3 inches and 5 inches respectively


What is the width of a rectangle with the length of 8 inches greater than its width find the width of the rectangle if its area is 105 square inches?

The dimensions are: length = 15 inches and width = 7 inches Check: 15*7 = 105 square inches


How do you get the area of a square?

You multiply Length times width. Lets say all sides are 15 inches, you multiply 15 times 15 and get 225 inches squared.


What is the length of a 5 inch by 3 inch rectangular?

Probably 5 inches. ****** Assuming it should be area (not length!): 5 x 3 = 15 square inches


A triangular pyramid has a height of 7 inches. The base length is 6 inches and the area of the base is 15 square inches. What is the surface area of the pyramid?

If the base length is 6 inches then the base area will be 9*sqrt(3) = 15.6 cm2, which rounds to 16, not 15. So some of the information provided is patently incorrect and therefore the question has no sensible answer.