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Find the length of a square playground whose area is equal to the area of rectangular field of dimensions 72m and 388m?
First of all you need to calculate the area of the rectangular field.
You do this by multiplying the length by the width.
72m x 388m = 27,936m2
If a square (which has four equal sides of course) has the same area then we can calculate the length of a side by finding the square root of the above.
27,936m1/2 = 167.14m (to 2 decimal places).
What else can I help you with?
What is the actual area if a rectangular playground is 50 times the dimensions of a scale drawing and the scale drawing is 6 square feet?
area is length times width so the actual area is 50 x 50 or 2500 times higher, or 2500 x 6 = 15000 sq ft
What is dimensions of a square?
The dimensions are: The dimensions of the square are LW Length x width (srry about the last one)
How do you find the diagonal of a rectangular prism?
If the prism is a rectangular parallelepiped (that is, all the angles are right angles), just add the squares of the 3 dimensions (length, width, and height) together, and take the square root. This involves 2 applications of the Pythagorean Theorem.
What are the dimensions of a square room that has 9000 square feet?
To get the length of a side, take the square root of 9000.
What is the perimeter of 24 square ft?
To know the perimeter, you must know the dimensions (length and width). But there are many possible dimensions if you have 24 square feet.
The dimensions of a rectangular playground are 50 times the dimensions of a scale drawing the playground The area of the scale drawing is 6 square feet What is the area of the actual playground?
If the dimensions of the actual playground are 50 times those of the scale drawing, then the length and width of the actual playground can be represented as 50 times the length and width of the scale drawing. The area of a rectangle is calculated by multiplying length by width. Since the area of the scale drawing is 6 square feet, the area of the actual playground will be ( (50 \times \text{length}) \times (50 \times \text{width}) = 2500 \times \text{(length} \times \text{width)} ). Therefore, the area of the actual playground is ( 2500 \times 6 = 15,000 ) square feet.
What is the actual area if a rectangular playground is 50 times the dimensions of a scale drawing and the scale drawing is 6 square feet?
area is length times width so the actual area is 50 x 50 or 2500 times higher, or 2500 x 6 = 15000 sq ft
What is dimensions of a square?
The dimensions are: The dimensions of the square are LW Length x width (srry about the last one)
How many dimensions in a square?
There are 2 dimensions in a square, length and width, which are always the same, due to the nature of a square.
Can a cuboid have a square face?
Yes but its length is rectangular
How do you find the volume of a square or rectangular object?
If it is a cube (basically a square) or a rectangular prism multiply the top, the length, and the base.
How do you find the diagonal of a rectangular prism?
If the prism is a rectangular parallelepiped (that is, all the angles are right angles), just add the squares of the 3 dimensions (length, width, and height) together, and take the square root. This involves 2 applications of the Pythagorean Theorem.
What are the Dimensions of a square with a length and width?
Length and width must have equal proportions.
How do you find the length of a rectangular box with given width and diameter?
Subtract the square of the width from the diameter. The square root of that is the length.
How is a square root related to the dimensions of a square?
the length of a side of a square is the square root of the area of the square.
How many cm is a square?
That will depend on the dimensions of the square in question. If the square has a length of 1cm, then 1x1=1cm2. If the square has a length of 2cm, then 2x2=4cm2.
How do you find the dimensions of height length and width from square feet?
Square footage is two-dimensional, the product of length times width. Unless the shape is a square, the area by itself is not sufficient to definitively ascertain the dimensions.
