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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

Q: 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?

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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

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

The area of rectangle is : 13832.797999999999

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

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

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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

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

The area of rectangle is : 13832.797999999999

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

The area of rectangle is : 8055.450000000001

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

The area of rectangle is : 1595.62

The area of rectangle is : 13832.797999999999

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

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)

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

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.