+47 1.) What is the Area of a Regulor hexagon whose perimeter is 24 ft.
2.) What is the area of the Shaed region to the nearest tenth of a foot (It is a Rectangel with a circle in it and the left side of the rectangel is 14 ft and the base is 21 ft the shaded region is the inside of the square subtract the circle.
If any of you guys can figure this out for me as soon as possible that would be great!
Thx,
Fox1018
+9673 Hmm.
Area of a regular hexagon = \(\dfrac{3\sqrt3 a^2}{2}\)
where a is the side length.
Therefore area of a regular hexagon whose perimeter is 24 ft. is \(3\sqrt3 \cdot 4^2\cdot \dfrac{1}{2}\\ =24\sqrt3\text{ sq. ft.}\)
For #2, I assume that the circle touches the side whose length is 14ft.
Then the area of shaded region is:
\(14\times 21 - \pi (7^2)\\ =(294-49\pi)\text{ sq. ft.}\)
+129896 1) Area of a hexagon with a perimeter of 24 ft
The side length is 1/6 of this = 4 ft.....
And the area = 6(1/2)(4)^2*sqrt(3)/2 = (3/2)*16*sqrt(3) =
24*sqrt(3) ft^2 ≈ 41.6 ft^2
2) The largesr circle that can be constructed in such a rectangle will be one with a radius of 7.
And the area outside the circle but still inside the rectangle =
Area of the rectangke - the area of the circle =
(14 * 21) - pi (7^2) = 294 - 49pi ≈ 140.1 sq ft
P.S. - I'll provide a pic, if you want.
