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In rectangle ABCD AE = 3x+3 and BD = 36 what is the value of x

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User Malice
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1 Answer

5 votes

Answer:

Explanation:

Since AE is equal to 3x + 3 and BD is equal to 36, we can set up an equation to solve for x. If we let the width of the rectangle be x, then the length would be 3x + 3.

Area of the rectangle = width * length

36 * (3x + 3) = (3x + 3) * x

Expanding the right side, we get:

36 * (3x + 3) = 3x^2 + 3x + 3x + 9

36 * (3x + 3) = 3x^2 + 6x + 9

Dividing both sides by 3, we get:

12 * (3x + 3) = x^2 + 2x + 3

Expanding the left side, we get:

12 * (3x + 3) = x^2 + 2x + 3

36 * 3x + 36 * 3 = x^2 + 2x + 3

108x + 108 = x^2 + 2x + 3

Subtracting 3 from both sides, we get:

108x + 105 = x^2 + 2x

Expanding the left side, we get:

x^2 + 2x - 108x - 105 = 0

Combining like terms on the left side, we get:

x^2 - 106x - 105 = 0

Using the quadratic formula, we can solve for x:

x = (-b ± √(b^2 - 4ac)) / 2a

where a = 1, b = -106, and c = -105.

Plugging in the values, we get:

x = (-(-106) ± √((-106)^2 - 4 * 1 * -105)) / 2 * 1

x = (106 ± √(11,336 + 420)) / 2

x = (106 ± √11,756) / 2

x = (106 ± 34) / 2

Since x has to be positive, we take the positive square root:

x = (106 + 34) / 2

x = 70 / 2

x = 35.

So the width of the rectangle is 35 units.

answered
User Qbyte
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