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What are the dimensions of a rectangle when one side is 3 cm longer than the other side and its area is twice its perimeter in square cm showing work with answers?
Wiki User
∙ 8y ago
Let the dimensions be x and x+3
Perimeter: 4x+6
Area: x(x+3) = 2(4x+6) => x^2 +3x = 8x+12 => x^2 -5x-12 = 0
Solving the above quadratic equation: x has a positive value of 6.77 rounded to 2dp
Therefore dimensions are: 6.77 cm and 9.77 cm
Wiki User
∙ 8y ago
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What are the dimensions of a rectangle that has a perimeter of 19.8 cm and an area of 24.3 square cm showing work?
Divide the perimeter by 2 then find two numbers that have a sum of 9.9 and a product of 24.3 which will work out as 5.4 and 4.5 by using the quadratic equation formula. Check: 2*(5.4+4.5) = 19.8 cm which is the perimeter Check: 5.4*4.5 = 24.3 square cm which is the area Therefore the dimensions of the rectangle are: 5.4 cm and 4.5 cm
What are the dimensions of a rectangle that has an area of 600 square cm and a perimeter of 1 m showing work?
I suggest that you do the following:* Convert the meters to centimeters, to have compatible units.* Write the equation for the area of the rectangle. Replace the variable "a" (area) with the known area.* Write the equation for the perimeter of a rectangle. Replace the variable for the perimeter with the known perimeter (in cm).* Use any method to solve the simultaneous equations.Another Answer:-Let the dimensions be x and yIf: 2x+2y = 100 then x+y = 50 and x = 50-yIf: xy = 600 then (50-y)y = 600 and so 50y-y2-600 = 0Solving the quadratic equation: y = 20 or y = 30Therefore by substitution the dimensions are: when y = 20 cm then x = 30 cm
What is the area of a rectangle that has a perimeter of 61.18 cm and a diagonal of 226.1 mm showing work and answer?
The area of rectangle is : 13832.797999999999
What is the diagonal length of a rectangle whose perimeter is 452.2 mm with an area of 106.134 square cm showing work and answer to an appropriate degree of accuracy if necessary?
Let the dimensions be x, y and change the perimeter into cm:- Perimeter: 2(x+y) = 45.22 cm => y = 22.61-x Area: xy = 106.134 => x(22.61-x) = 106.134 So it follows: 22.61x-x^2-106.134 = 0 Solving the above quadratic equation: x = 15.96 or x = 6.65 If: x = 6.65 then y = 15.96 Using Pythagoras: 6.65^2+15.96^2 = 298.9441 Square root of 298.9441 = 17.29 cm or 172.9 mm which is the rectangle's length
What is the diagonal length of a rectangle whose area is 212.268 square cm with a perimeter of 61.18 cm showing all work with answer?
Let the dimensions of the rectangle be x and y and divide its perimeter by 2:- So: x+y = 30.59 => y = 30.59 -x Area: xy = 212.268 => x(30.59 -x) = 212.268 It follows that: 30.59x - xsquared -212.268 = 0 Solving the quadratic equation: x = 19.95 or x = 10.64 By substitution: x = 19.95 and y = 10.64 Using Pythagoras: 19.95squared+10.64squated = 511.2121 The square root of 511.2121 is 22.61 cm which is the diagonal length
What are the dimensions of a rectangle that has a perimeter of 19.8 cm and an area of 24.3 square cm showing work?
Divide the perimeter by 2 then find two numbers that have a sum of 9.9 and a product of 24.3 which will work out as 5.4 and 4.5 by using the quadratic equation formula. Check: 2*(5.4+4.5) = 19.8 cm which is the perimeter Check: 5.4*4.5 = 24.3 square cm which is the area Therefore the dimensions of the rectangle are: 5.4 cm and 4.5 cm
What are the dimensions of a rectangle that has an area of 600 square cm and a perimeter of 1 m showing work?
I suggest that you do the following:* Convert the meters to centimeters, to have compatible units.* Write the equation for the area of the rectangle. Replace the variable "a" (area) with the known area.* Write the equation for the perimeter of a rectangle. Replace the variable for the perimeter with the known perimeter (in cm).* Use any method to solve the simultaneous equations.Another Answer:-Let the dimensions be x and yIf: 2x+2y = 100 then x+y = 50 and x = 50-yIf: xy = 600 then (50-y)y = 600 and so 50y-y2-600 = 0Solving the quadratic equation: y = 20 or y = 30Therefore by substitution the dimensions are: when y = 20 cm then x = 30 cm
What is the area of a rectangle that has a perimeter of 61.18 cm and a diagonal of 226.1 mm showing work and answer?
The area of rectangle is : 13832.797999999999
What is the diagonal length of a rectangle whose perimeter is 452.2 mm with an area of 106.134 square cm showing work and answer to an appropriate degree of accuracy if necessary?
Let the dimensions be x, y and change the perimeter into cm:- Perimeter: 2(x+y) = 45.22 cm => y = 22.61-x Area: xy = 106.134 => x(22.61-x) = 106.134 So it follows: 22.61x-x^2-106.134 = 0 Solving the above quadratic equation: x = 15.96 or x = 6.65 If: x = 6.65 then y = 15.96 Using Pythagoras: 6.65^2+15.96^2 = 298.9441 Square root of 298.9441 = 17.29 cm or 172.9 mm which is the rectangle's length
What is the area of a rectangle which has a diagonal of 17.55 cm and a perimeter of 459 mm showing all work?
The area of rectangle is : 8055.450000000001
What is the diagonal length of a rectangle whose area is 212.268 square cm with a perimeter of 61.18 cm showing all work with answer?
Let the dimensions of the rectangle be x and y and divide its perimeter by 2:- So: x+y = 30.59 => y = 30.59 -x Area: xy = 212.268 => x(30.59 -x) = 212.268 It follows that: 30.59x - xsquared -212.268 = 0 Solving the quadratic equation: x = 19.95 or x = 10.64 By substitution: x = 19.95 and y = 10.64 Using Pythagoras: 19.95squared+10.64squated = 511.2121 The square root of 511.2121 is 22.61 cm which is the diagonal length
What is the area of a rectangle that has a perimeter of 64.6 cm and a diagonal of 24.7 cm showing work?
The area of rectangle is : 1595.62
What is the area of a rectangle that has a perimeter of 61.18 cm and a diagonal of 226.1 mm showing key stages of work?
The area of rectangle is : 13832.797999999999
What is the length and width of a rectangle that has a perimeter of 20 inches and an area of 24.4524 square inches showing work with final answers?
What do we know about the perimeter of a rectangle? perimeter = 2 × (length + width) → 2 × (length + width) = 20 in → length + width = 10 in → length = 10 in - width What do we know about the area of a rectangle: area = length × width → length × width = 24.4524 in² But from the perimeter we know the length in terms of the width and can substitute it in: → (10 in - width) × width = 24.4524 in² → 10 in × width - width² = 24.4524 in² → width² - 10 in × width + 24.4524 in² = 0 This is a quadratic which can be solved by using the formula: ax² + bx + c → x = (-b ±√(b² - 4ac)) / (2a) → width = (-10 ±√(10² - 4 × 1 × 24.4524)) / (2 × 1) in → width = -5 ± ½√(100 - 97.8096) in → width = -5 ±½√2.1904 in → width = -5 ± 0.74 in → width = 4.26 in or 5.74 in → length = 10 in - 4.26 in = 5.74 in or 10 in - 5.74 in = 4.26 in (respectively) By convention the width is the shorter length (though it doesn't have to be) making the width 4.26 in and the length 5.74 in. Thus the rectangle is 5.74 in by 4.26 in
What are the dimensions of a rectangle when one side is greater than the other side by 5 cm and whose area is numerically twice its perimeter showing work?
How many marks do I get from your teacher for this: let the width of the rectangle by w cm Then its length is w + 5 cm → perimeter = 2 x (w + (w + 5)) = 4w + 10 → area = w(w + 5) = w^2 + 5w but area = 2 x perimeter → w^2 + 5w = 2 x (4w + 10) → w^2 + 5w = 8w + 20 → w^2 - 3w - 20 = 0 → w = (3 + √89) ÷ 2 or (3 - √89) ÷ 2 The second leads to a negative length (≈ -3.22) which can't exist → the lengths are (3 + √89) ÷ 2 and (3 + √89) ÷ 2 + 5, ie The rectangle is (3 + √89) ÷ 2 (≈ 6.22 cm) by (13 + √89) ÷ 2 (≈ 11.22 cm)
What is the perimeter of a rectangle which has a diagonal of 8.50 cm and an area of 3000 square mm showing work with answer?
Here is what you are supposed to do: * Convert to consistent units. For example, convert the cm to mm. * Write an equation for the diagonal (in terms of length and width). Replace the known diagonal. * Write an equation for the area, in terms of length and width. * Solve the two equations simultaneously. * Calculate the perimeter.
What is the width of a rectangle that has a length of 7.45 cm and a perimeter of 219.8 mm showing work?
1 Let its width be x and its perimeter change to 21.98 cm2 So: 2(7.45+x) = 21.983 Or: 14.9+2x = 21.984 Therefore: 2x + 21.98-14.9 => 2x = 7.085 Then: x = 7.08/2 => x = 3.546 Width = 3.54 cm
