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What is the perimeter and area of a rectangle with 5 in length and 11 in width?
Perimeter = 32 inches. Area = 55 square inches.
How many square feet is 18' by 55'?
990 square feet. To calculate square feet simply multiply length by width - the result is the area in square feet.
If ABCD is a rectangle and mADB equals 55 what is the value of x?
I'm having some trouble understanding exactly what 'm' and 'x' have to dowith the rectangle. A look at the drawing would really help.the answer is 70 -FlyBoi
One tenth of a line is 5 and a half cm What is the line's total length?
If one tenth of a line is 5 and a half cm, then the total length of the line would be 10 times that amount, which is 55 cm. This is because one tenth represents 1/10th of the total length, so to find the total length, you would need to multiply by 10. Therefore, the line's total length is 55 cm.
Basic unit of length in the metric system?
The basic unit of length in the metric system is the meter. Grams are used to measure weight and liter is used to measure liquid capacity.
The area of a rectangle is 55 square meters how do you find the length and width of the rectangle if its length is 6 meters greater than its width?
Prime factors of 55 are 5 and 11, which differ by 6, so length 11m, width 5m!
What is the area of a rectangle that has a length of 10 cm and a width of 5.5 cm?
Well, isn't that just a happy little rectangle! To find the area, you simply multiply the length by the width. So, for this lovely rectangle with a length of 10 cm and a width of 5.5 cm, the area would be 55 square centimeters. Just imagine all the beautiful landscapes you could paint on that canvas!
What is the area of a square that has a length of 55 and a width of 45?
this is not a square, but a rectangle, first off. second, the rectangle's area is 2475 square feet.
What is the perimeter and area of a rectangle with 5 in length and 11 in width?
Perimeter = 32 inches. Area = 55 square inches.
Find the area of a rectangle whose diagonal is 10 feet long and makes a 55 degree angle with one of the sides?
A rectangle = LW If the diagonal of length of 10 ft makes a 55⁰ angle with length L, then the length of width W equals to 10 sin 55⁰ ft, and the length of L equals to 10 cos 55⁰ ft. So that A = LW = (10 sin 55⁰ ft)(10 cos 55⁰ ft) ≈ 47 ft2
What is the perimeter of a rectangle length 45meters and width 10 meters?
A rectangle has two pairs of equal sides. So to calculate the perimeter, add two known sides and multiply by two. 45 + 10 = 55 55 x 2 = 110 Perimeter - 110 metres.
The width of a rectangle is 20 cm less than twice the length The perimeter is at least 55 cm What are the smallest possible dimensions of the rectangle if each dimension is an integer?
16 and 12
What is the wide is a 55 gallon fish tank?
length and width for a 55 gallon fish tank
The perimeter of a rectangle is 110 cm more than twice the width find the demensions of the rectangle?
The perimeter of a rectangle is twice the width added to twice the length -> 2(W) + 2(L) We are told that for this rectangle the perimeter = 2(W) + 110 So for this rectangle 2(L) = 110 Therefore L = 55 But this does not tell us enough to work out the width ..... if the width is 1 cm then the perimeter would be 2(1) + 110 and if the width is 300 cm then the perimeter would be 2(300) + 110 In each case the perimeter meets the requirement of being 110 cm more than twice the width.
What is the formula used to find the volume of a rectangular object?
Height x Width x Length. Measure the height, width and depth of the object, and be sure to do it in the same units. Then multiply them together. H x W x L = Volume of the rectangular solid The answer should be in units cubed. As an example, a rectangle has these dimensions: Length = 55 mm Width = 30 mm Length = 40 mm 55 X 30 X 40 = 66,000 cubic millimeters (mm3)
The area of a room is 55 square feet The length is 6 feet longer than the width Find the length and width of the room?
11 and 5
What is the length and width size of 500 sqft?
There is not enough information to answer the question. First of all, the shape of the area is unknown. Even if it were a rectangle, all that can be said is that one of the sides must be at least sqrt(500) = 22.36 ft (approx). Take any number L greater than 22.36 and let W = 500/L Consider a rectangle with Length = L ft and Width = W ft. The area of such a rectangle is L*W = L*(500/L) = 55 sq ft. But L is ANY length more than 22.36 feet. There are infinitely many possible values for L.
