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29 January, 08:27

A triangle with area 45 square inches has a height that is two less than four times the width. Find the width and height of the triangle.

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Answers (2)
  1. 29 January, 08:33
    0
    The width is 5 inches and the height is 18 inches.

    Step-by-step explanation:

    The area of a triangle can be calculated as:

    A = w*h/2

    where w is the width of the base, and h is the height.

    then we have that:

    w*h/2 = 45 in^2

    and the height is 2 less than 4 times the width.

    h = 4*w - 2in

    then we have a system with 2 equations:

    w*h/2 = 45 in^2

    h = 4*w - 2in

    We can replace the second one in the first one and solve it for w.

    w * (4*2 - 2in) / 2 = 45 in^2

    2w^2 - w = 45in^2

    2w^2 - w - 45in = 0

    then we have to solve the quadratic equation:

    the solutions are:

    w = (1 + - √ (1^2 - 4*2 * (-45)) / (2*2) = (1 + - √361) / 4

    where we have two solutions for w, one for each sign of the square root.

    we obviously need to take the positive solution (because we can not have a negative length)

    w = (1 + √361) / 4 = 5in

    now, we can replace it in the equation "h = 4*w - 2in" to get the height.

    h = 4*5in - 2in = 18in
  2. 29 January, 08:47
    0
    Step-by-step explanation:

    45 = (b (4b - 2)) / 2

    90 = 4b2 - 2b

    4b2 - 2b - 90 = 0

    (2b + 9) (2b - 10) = 0

    2b + 9 = 0

    2b = - 9

    b = - 9/2

    If we were looking for zeroes, this would be an answer, but since we're looking for measurement, we don't use negative numbers.

    90 = 4b2 - 2b

    4b2 - 2b - 90 = 0

    (2b + 9) (2b - 10) = 0

    2b + 9 = 0

    2b = - 9

    b = - 9/2
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