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What is the length and width of a rectangle if the perimeter is 18cm?
It is impossible to say. Let L be ANY number such that 4.5 ≤ L < 9 cm and let B = (9 - L) cm. Then for every one of the infinite number of values of L, there will be a different rectangle whose perimeter will be 18 cm.
What are the dimensions of a rectangle if the perimeter is seventy eight centimeters?
There are infinitely many possible answers. Select any number, B, in the interval (0, 19.5) cm or allow 19.5 cm if a square is permitted. Let L = 39 - B cm. Then the perimeter of an L*B rectangle is2*(L + B) = 2*(39 - B + B) = 2*39 = 78 cm.Since the choice of B is arbitraty within that interval, there are infinitely many possible values for B and thus, infinitely many possible dimensions.
Describe the locus of points that are 9 cm from point B?
a circle 9 cm from point b I was co fused by this but you just do a diagram and write this
Rectangular with a perimeter of 9 cm?
There are infinitely many possible rectangles. Suppose A >= 2.25 cm is the length of the rectangle. and B = 4.5 - A cm is the width. Then perimeter = 2*(A + B) = 2*(A + 4.5 - A) = 2*4.5 = 9 cm Also, it is easy to show that A >=B so that A and B cannot swap places. For each of the infinitely many values of A, you have a rectangle with perimeter 9 cm.
How many rectangles have the perimeter of 24cm?
Infinitely many. Let B be any number less than 6 cm, and let L = 12-B cm. Then the perimeter of the rectangle, with length L and breadth B, is 2*[L+B] = 2*[(12-B)+B] = 2*12 = 24 cm. There are infinitely many possible values for B, between 0 and 6 and so there are infinitely many possible rectangles.
What is the length and width of a rectangle if the perimeter is 18cm?
It is impossible to say. Let L be ANY number such that 4.5 ≤ L < 9 cm and let B = (9 - L) cm. Then for every one of the infinite number of values of L, there will be a different rectangle whose perimeter will be 18 cm.
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.
What are the dimensions of a rectangle if the perimeter is seventy eight centimeters?
There are infinitely many possible answers. Select any number, B, in the interval (0, 19.5) cm or allow 19.5 cm if a square is permitted. Let L = 39 - B cm. Then the perimeter of an L*B rectangle is2*(L + B) = 2*(39 - B + B) = 2*39 = 78 cm.Since the choice of B is arbitraty within that interval, there are infinitely many possible values for B and thus, infinitely many possible dimensions.
The base of an isosceles triangle is 3 cm less than the length of the equal sides if the perimeter of the triangle is 36 cm find the length of each sides?
b=l-3 P=2l+b 36=2l+b Sub b=l-3 into 36=2l+b 36=2l+b 36=2l+(l-3) 36=2l+l-3 39=3l l=13 cm Sub l=13 into b=l-3. b=l-3 b=13-3 b=10 cm Therefore, the length of each of the two equal sides is 13 cm and the base is qual to 10 cm.
What is the possible area of a rectangle when one side is 9m and the perimeter is 30m?
Perimeter = 2*(L + B)So 30 = 2*(9 + B)15 = 9 + B so that B = 6And therefore, Area = L*B = 9*6 = 54 sq metres.Perimeter = 2*(L + B)So 30 = 2*(9 + B)15 = 9 + B so that B = 6And therefore, Area = L*B = 9*6 = 54 sq metres.Perimeter = 2*(L + B)So 30 = 2*(9 + B)15 = 9 + B so that B = 6And therefore, Area = L*B = 9*6 = 54 sq metres.Perimeter = 2*(L + B)So 30 = 2*(9 + B)15 = 9 + B so that B = 6And therefore, Area = L*B = 9*6 = 54 sq metres.
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.
Describe the locus of points that are 9 cm from point B?
a circle 9 cm from point b I was co fused by this but you just do a diagram and write this
Rectangular with a perimeter of 9 cm?
There are infinitely many possible rectangles. Suppose A >= 2.25 cm is the length of the rectangle. and B = 4.5 - A cm is the width. Then perimeter = 2*(A + B) = 2*(A + 4.5 - A) = 2*4.5 = 9 cm Also, it is easy to show that A >=B so that A and B cannot swap places. For each of the infinitely many values of A, you have a rectangle with perimeter 9 cm.
How many rectangles have the perimeter of 24cm?
Infinitely many. Let B be any number less than 6 cm, and let L = 12-B cm. Then the perimeter of the rectangle, with length L and breadth B, is 2*[L+B] = 2*[(12-B)+B] = 2*12 = 24 cm. There are infinitely many possible values for B, between 0 and 6 and so there are infinitely many possible rectangles.
How do you draw a rectangle with an area of 12cm2?
Let B be any positive number less than sqrt(12) and let L = 12/B. Then a rectangle with sides of L cm and B cm will have an area of L*B = (12/B)*B = 12 cm2. Since there are infinitely many possible choices for B, there are infinitely many possible rectangles.
What cm is each side if the area of a rectangle is 15cm?
An area cannot be 15 cm since cm is a measure of distance, not area. But suppose the area is 15 cm2. Then there are an infinite number of possible answers: Let b be any number such that 0 < b < 3.87 (there is an infinity of such numbers). Then let l = 15/b Then any rectangle of length l cm and breadth b cm will have an area of 15 cm2.
What is the area of a rectangle with a length of 13.0 cm and a width of 86.0 mm?
Area of rectangle = l X b . Here , l=13 cm . 1 cm =10 mm. so, 13 cm =130 mm. b=86 mm. NOw , area = 11180 mm sq.
