3456 units squared
The surface area of a cuboid is 2*(L*B + B*H + H*L) where L = length, B = breadth and H = height. In a cube, L = B = H and so surface area = 6*L2
I will rephrase your question: What happens to the surface area of a cube when the volume doubles. Ans. Surface area becomes 1.5876 times larger. Explanation: Let L = the length of the side of the original cube and h x L the length of the cube that is double the volume. Now: Vol= L^3 x 2 = (h x L)^3 or h = 2^(1/3) = 1.2599, so the length will be 1.2599 times larger. Surface area = 6 x L^2 for original cube and 6 x L^2 x 1.2599^2 for the cube with twice the volume. 1.2599^2 = 1.5876 If you are asking what happens to the surface area when the sides double, then the larger cube has surface area = 6 * 2^2 * L^2 , so 6 * 2^2 = 24. Each side is 4 times larger so the total surface area is 24 times larger.
For a cuboid, the total surface area = 2*(L*B + B*H + H*L) square units where L = length, B = breadth and H = height.In the case of a cube, L = B = H and so the total surface area is 6*L^2 square units.
Define h = height w = width d = depth Then, the surface area is 2hw+2hd+2wd.
If you cut a cube of jello in half, it will still have the same total volume. The only thing that will change is the total surface area. Assuming that the piece is a perfect cube, and that it has been divided into two equal pieces, the net surface area of the two resulting cubes would be: Original: SA= 6(h^2) New: SA= 2[2(h^2) + 1/2 (4)(h^2)] Difference: [2(h^2) + 1/2 (4)(h^2)] - 6(h^2) = 8(h^2) - 6(h^2) = 2(h^2) Where: SA = Surface Area h = the length of each side So, if the original cube was 2x2x2 cm, then it's surface area would be 24 cm^2; when it is divided into two, the net surface area of the two pieces together would be 32 cm^2
If you cut a cube of jello in half, it will still have the same total volume. The only thing that will change is the total surface area. Assuming that the piece is a perfect cube, and that it has been divided into two equal pieces, the net surface area of the two resulting cubes would be: Original: SA= 6(h^2) New: SA= 2[2(h^2) + 1/2 (4)(h^2)] Difference: [2(h^2) + 1/2 (4)(h^2)] - 6(h^2) = 8(h^2) - 6(h^2) = 2(h^2) Where: SA = Surface Area h = the length of each side So, if the original cube was 2x2x2 cm, then it's surface area would be 24 cm^2; when it is divided into two, the net surface area of the two pieces together would be 32 cm^2
The surface area is 600 sq. inches. The formula for surface area of a cube is 6s^2 where s is the side length. The general surface area formula is 2B+Ph where B is the area of the base, P is the perimeter of the base and h is the height.
6 cmgiven that a cube has L=W=H,the surface area = 6x² (6 from 6 sides, and x² because, as mentioned, L=W=H)216 = 6x²36 = x²x = 6
you have to multiply lxw to get the answer
The formula for Surface Area of a cube is EASY because it's bassically (2xbxh) + (2xbxw) + (2xhxw). It sounds complicated, but it is not even close to hard once you learn it. b=base w=width h=height x= multiplication sign
v=b/h/w=area
Concept The surface-area-to-volume ratio is calculated by dividing the surface area by the volume of any object. If you know the formula for the surface area and the volume of an object, then simply compute (surface area) / (volume) to calculate the surface-area-to-volume ratio. The actual surface-area-to-volume ratio of any object depends upon that object's shape and geometry. Cube Consider a cube with equal sides of length x. The cube has six faces (top, bottom, left, right, front, back), and each face has a surface area of x2, so the total surface area of the cube is 6x2. The volume of the cube is x3. So the surface-area-to-volume ratio for a cube is 6x2 / x3, which can be reduced to 6/x. This surface-area-to-volume ratio, 6/x, holds true for all cubes. Let's test this ratio. Consider a cube that has a 1 cm length on all sides. The surface area is 6 sides of 1 cm x 1 cm (6 cm2), and the volume is 1 cm x 1 cm x 1 cm (1 cm3). Dividing the surface area by the volume gives a surface-area-to-volume ratio of 6 (which is 6/1). If the length of the cube sides is 6 cm, then the surface area is 6 sides of 6 cm x 6 cm (216 cm2) and the volume is 6 cm x 6 cm x 6 cm (216cm3), so the surface-area-to-volume ratio is 216/216, or 1 (which is 6/6). If the length of the cube side is 12 cm, then the surface area is 6 sides of 12 cm x 12 cm (864 cm2) and the volume is 12 cm x 12 cm x 12 cm (1728 cm3), so the surface-area-to-volume ratio is 864/1728, or 0.5 (which is 6/12). We can empirically verify that this surface-area-to-volume ratio for a cube is therefore 6/x. Sphere Consider a sphere (a round ball) of radius r. The surface area is 4 PIr2, whereas the volume is (4/3)PIr3. So the surface-area-to-volume ratio of a sphere is (4 PIr2) / [(4/3)PIr3], which can be reduced to 3/r. As in the cube, the surface-area-to-volume ratio of 3/r holds true for all spheres. In the previous description, the symbol 'PI' is meant to represent Pi, or 3.1415 ... T Irregular Objects For irregular objects, such as a rectangular prism (a box) with different lengths in each dimension, the surface-area-to-volume ratio must be calculated for each shape. Consider a box with dimensions of l (length), w (width), and h (height). Like the cube, the box has six faces, but it is easier to consider it as three face pairs (front/back, left/right, and top/bottom). The surface area of both faces in a pair are the same (the front face has the same surface area as the back face). So the surface area of the box is: A = 2(l x w) + 2(w x h) + 2(l x h), or 2( (l x w) + (w x h) + (l x h) ). The volume is: V = l x w x h So the surface area to volume ratio (A/V) of a box is: 2( (l x w) + (w x h) + (w x h) ) / (l x w x h). The surface-area-to-volume ratio of a cylinder (like a soup can) is: ( (2 PI r2) + (2 PI r h) ) / (PI r2h) Where r is the radius of the circle on the top and bottom of the cylinder, h is the height of the cylinder, and. PI is Pi, or 3.1415 ... Unlike regular objects, such as the cube or sphere, no further simplification of the box's or cylinder's surface-area-to-volume ratio equation exists. The above appropriate equations must be applied to each box or cylinder separately.