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How do you find the width and perimeter of a rectangle if the length is 156 cm and the width is 36 more than one half of the length?
L = 156 cm so L/2 = 156/2 = 78 cm and so width = W = 36 cm more = 78+36 = 114 cm Then perimeter = 2*(L+W) = 2*(156 + 114) cm = 2*270 cm = 540 cm
How many different rectangles can you draw with an area of 11 cm squared?
There are infinitely many rectangles. Let K = sqrt(11). Let L be any real number greater than M and let B = 11/L. Then, B < K so that for any two different values of L, the pair (L, B) are distinct even with a swap.The rectangle with length L and breadth B has an area = L*(11/L) = 11 cm2. Since there are infinitely many choices for L, there are infinitely many rectangles.
What are the dimensions of rectangle if the diagonal is 13cm?
The length of the diagonal is not enough to determine the dimensions of the rectangle. Take any length W such that 0<W<sqrt(132/2) that is, 0<W<9.192 cm. And let L = sqrt(132 - W2) cm. Then 9.192<L<13 so that any combination of W and L is unique. There are infinitely many possible values for W and so infinitely many different combinations for W and L. That is, infinitely many rectangles whose width is W cm and length is L cm. And, since, L2 = 132 - W2, L2 + W2 = 132 so the diagonal is 13 cm.
How many cm in 100 meters?
One
What Is the perimeter if a rectangle if the area is 45cm2?
Well, well, well, aren't we getting fancy with our math questions? If the area of a rectangle is 45cm2, and you want to find the perimeter, you need more information. The length and width are required to calculate the perimeter, so without those dimensions, I'm afraid I can't give you a straight answer.
The perimeter of a rectangle is 90 cm. If the length is one and one-half times the width Find the length and the width?
Suppose the length is L cm. Then the width, W, is 2L/3 cm. Perimeter = 2(L+W) = 2(L+2L/3) = 10L/3 = 90 cm So L = 27 cm and W = 2L/3 = 18 cm.
How many different rectangles have a perimeter of 48cm?
Infinitely many. Select any number, L, such that 12 < L < 24. Let W = 24 - L. Then a rectangle with sides of length L cm and width W cm will have a perimeter of 48 cm. And since the choice of L was arbitrary, there are infinitely many possible values of L and thence infnitely many rectangles.
How many cm in l meter?
There are 100cm in a metre.
How many cm3 in 0.75 litres?
1 l = 1000 cm³ → 0.75 l = 0.75 x 1000 cm³ = 750 cm³ (1 cm³ = 1 ml)
On a ray FL how many points are there whose distance from L is 6cm?
If the distance of L from F is ≥ 6 cm then the answer is 2. If the distance of L from F is < 6 cm then the answer is 1.
150 of what length is 66 in?
If 150 / x = 66, then x = 2.2727 cmSuppose the length is L cm.Then 150*L cm = 66 cmSo L cm = 66 cm/150 = 66/150 cm = 0.44 cm = 4.4 mm.
What are 5 ways of making a 20cm squared?
There are infinitely many ways. Pick any number greater than or equal to sqrt(20) = 4.472 cm. Let that be the length of your rectangular area, L. Let the breadth be B = 20/L cm. Then clearly, area = L*B = L*(20/L) = 20 cm2 So, L = 20 cm gives B = 1 cm L = 10 cm gives B = 2 cm L = 5 cm gives B = 4 cm L = 1 metre gives B = 0.2 cm L - 1 kilometre gives B = 0.0002 cm and so on. There is no need to stick with rectangles: yuo could do circles, triangles, other polygons or even irregular shapes. Cut out 20 squares of 1 cm each and arrange them so that each one of them is "connected" side-to-side with another. You can make L-shapes, and F-shapes and all sorts. Its just that rectangles are easiest to calculate and present an answer.
How do you find the width and perimeter of a rectangle if the length is 156 cm and the width is 36 more than one half of the length?
L = 156 cm so L/2 = 156/2 = 78 cm and so width = W = 36 cm more = 78+36 = 114 cm Then perimeter = 2*(L+W) = 2*(156 + 114) cm = 2*270 cm = 540 cm
How many cm in one in?
2 cm
How many liters in one cubic liter?
A litre, as a unit of volume is already a cubic measurement. 1 L = 1000 cm^3
How many different rectangles can you draw with an area of 11 cm squared?
There are infinitely many rectangles. Let K = sqrt(11). Let L be any real number greater than M and let B = 11/L. Then, B < K so that for any two different values of L, the pair (L, B) are distinct even with a swap.The rectangle with length L and breadth B has an area = L*(11/L) = 11 cm2. Since there are infinitely many choices for L, there are infinitely many rectangles.
9 L equals how many cm?
You cannot convert liquid to length.
