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20 July, 12:21

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?

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  1. 20 July, 12:44
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    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
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