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What Is the perimeter if a rectangle if the area is 45cm2?
Any number greater than or equal to 4*sqrt(45) cm or 26.83 cm (approx). To see that, suppose L ≥ sqrt(45) cm be the length of the rectangle and let B = 45/L cm Then Area = L*B = L*(45/L) = 45 cm2 If L = 9cm, then B = 45/9 = 5 cm and then perimeter = 2*(L+B) = 28 cm If L = 90 cm, then B = 0.5 cm and P = 181 cm If L = 900 cm, then B = 0.05 cm and P = 1800.1 cm If L = 9000000 cm then B = 0.000005 cm and P = 18000000.00001 cm As can be seen, the perimeter can be increased without limit.
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
How many cm in l meter?
There are 100cm in a metre.
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.
What Is the perimeter if a rectangle if the area is 45cm2?
Any number greater than or equal to 4*sqrt(45) cm or 26.83 cm (approx). To see that, suppose L ≥ sqrt(45) cm be the length of the rectangle and let B = 45/L cm Then Area = L*B = L*(45/L) = 45 cm2 If L = 9cm, then B = 45/9 = 5 cm and then perimeter = 2*(L+B) = 28 cm If L = 90 cm, then B = 0.5 cm and P = 181 cm If L = 900 cm, then B = 0.05 cm and P = 1800.1 cm If L = 9000000 cm then B = 0.000005 cm and P = 18000000.00001 cm As can be seen, the perimeter can be increased without limit.
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 liters in one cubic liter?
A litre, as a unit of volume is already a cubic measurement. 1 L = 1000 cm^3
How many cm in one in?
2 cm
How many foot in in 45 cm?
L = ( 45 cm ) ( 1.0 in / 2.54 cm ) ( 1.0 ft / 12 in ) = 1.48 ft
How many cm are there in one inch?
2.54 cm
How many cm. are in a in.?
There are about 2.54 cm in one inch
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.
