The area of a rectangle is 12 square inches. The length is 5 more than twice the width. Find the length and width.

Since the equation for area of a rectangle is A = l×w, and the length is 5 more than twice the width, making it 2w + 5, you have to plug in the quantities. A = l×w 12 = (2w+5)×w 12 = 2w²+5w 0 = 2w²+5w-12 0 = (2w-3)(w+4) Since the width can't be a negative number, the answer is: 2w-3 = 0 2w = 3 w = 3/2 And now, since length is 2w+5, you plug in the value of the width into the equation. 2(3/2)+5 3+5 8

Answered by CamilleEaton210 | 2024-06-10

Let the width of the triangle be x. Therefore length = 2x + 5. (5 more than twice its width).
Area = L * W = (2x + 5) x = 12. 2x^2 + 5x = 12 2x^2 + 5x - 12 = 0. This is a quadratic equation. (2)*(-12) = -24. We think of two numbers whose product is -24 and it sum is +5. Those two numbers are 8 and -3. So we replace the middle term of quadratic with (8 -3). 2x^2 + 5x - 12 = 0. 2x^2 + 8x-3x - 12 = 0. Factorize. 2x(x + 4) - 3(x +4) = 0 (2x-3)(x+4) = 0. (2x-3) = 0 or (x+4) = 0 2x = 3. x = 0 -4 x = 3/2 = 1.5 x = -4. (x can't be negative, since we are solving for lengths) x = 1.5 is only valid solution.
width = x = 1.5 Length = (2x + 5) = 2*1.5 +5 = 3 + 5 =8. Therefore length = 8, and width = 1.5 Cheers.

Answered by olemakpadu | 2024-06-10

The width of the rectangle is 1.5 inches, and the length is 8 inches. We found this by setting up the area equation, solving a quadratic equation, and then substituting back to find the length. Finally, we confirmed that the equations used align with the given area and relationship between length and width.
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Answered by CamilleEaton210 | 2024-12-27