The width of a rectangle is 2 less than twice its length. If the area of the rectangle is 182 cm 2, what is the length of the diagonal?

20.93cm

Step-by-step explanation:

The shape of the solid is rectangle

Area = length * width

But the width (w) = 2 times the length minus 2

W = 2L - 2

A = L * (2L - 2)

Area of the rectangle = 182cm²

A = 2L² - 2L

182 = 2L² - 2L

2L² - 2L - 182 = 0

L² - L - 91 = 0

Solving the equation using quadratic formula,

A = 1, b = - 1, c = - 91

L = [-b±√ (b² - 4ac) ]/2a

L = [ - (-1) ± √ ((-1) ² - 4*1 * (-91)) ] / 2 (1) ²

L = [1 ± √ (1² + 364) ] / 2

L = [1 ± √ (365) ] / 2

L = (1 ± 19.10) / 2

Note : Since we're working with length, we can't have a negative value as a solution hence we would not test both sides

L = (1 + 19.10) / 2 = 10.05cm

Width = 2L - 2

W = 20.1 - 2 = 18.10cm

Length = 10.05cm

Width = 18.10cm

Diagonal of the rectangle = ?

To find the diagonal, we'll use pythagorean theorem,

A = diagonal

B = length

C = width

a² = b² + c²

a² = (10.05) ² + (18.10) ²

a² = 110.25 + 327.61

a² = 437.86

a = √ (437.86)

a = 20.93cm

The diagonal of the rectangle is 20.93cm