here could be many solutions for this equation..all you have to do is pick a number for the width and do what it says..like so.. L = 2(w) + 7 lets say the width is 5 so…
If the length and width of a rectangle is doubled, it means that both dimensions have increased by a factor of 2. As a result, the area of the rectangle will increase by a factor of 4, because the area is calculated by multiplying the length and width together. Additionally, the perimeter of the rectangle will also increase by a factor of 2, since it is calculated by adding the lengths of all four sides.
I think you mean the width is 6. The answer depends entirely on what the length of the rectangle is. For example, if it is a square, so that the length and width are both 6, the answer is 24. I the length is 10, the perimeter is 32. In general, just add the width (6) to the length and double the sum to get the perimeter.
If by that you mean knowing only the diagonal and the width, then by the formula a2+b2=c2, where a is the length, b the width and c the diagonal. To find the width b, you need to calculate sqrt(c2-a2). For example, the width of a rectangle with length 3 and diameter 5 is sqrt(52-32)=4
There is no such thing as an "irregular rectangle". To calculate the area of a rectangle - if that's what you mean - you multiply length x width.
Do you mean perimeter? If so....2 x length + 2 x width.
The length is 5 times the width
Area of a square/rectangle=its length × its width
There is no "height" of a rectangle, unless it's a rectangular prism. Do you mean the length? If you have the area of the rectangle, the equation should be:A= L x WPlug in the area and the length and solve for the width, or plug in the area the width of the rectangle, and solve for the length.
I think you mean the width is 6. The answer depends entirely on what the length of the rectangle is. For example, if it is a square, so that the length and width are both 6, the answer is 24. I the length is 10, the perimeter is 32. In general, just add the width (6) to the length and double the sum to get the perimeter.
Assuming you mean a golden rectangle as opposed to one which can be split up into squares: length = width × φ = width × (√5 + 1)/2 = 55 × (√5 + 1)/2 ≈ 89.0 units
Badly expressed! Do you mean "what is the area of a rectangle with a length that is 3 more than twice its width?" ? If you have a width, then presumably you have a length, in which case you presumably have a rectangle. But what do we want to know about this rectangle? You haven't said. Its area? Its perimeter? Its color? Its shoe size?
If by that you mean knowing only the diagonal and the width, then by the formula a2+b2=c2, where a is the length, b the width and c the diagonal. To find the width b, you need to calculate sqrt(c2-a2). For example, the width of a rectangle with length 3 and diameter 5 is sqrt(52-32)=4
There is no such thing as an "irregular rectangle". To calculate the area of a rectangle - if that's what you mean - you multiply length x width.
Do you mean perimeter? If so....2 x length + 2 x width.
The length is 5 times the width
It means two dimensional such as the shape of a polygon or a rectangle that has dimensions of length and width.
I don't know what you mean by fixed area. All I know is that the area of a rectangle is the length times the width. As long as you don't change the length or the width, or change it into a different kind of shape, this area will remain fixed.
Assuming you mean 25cm and not 25m...p/4 = 77.5cml = p/4 + 12.5 = 90cmw = p/4 - 12.5 = 65cmCheck:90 - 65 = 25