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A square has side length (2x - 1) cm. If each side is increased by 4 cm, what is the increase in area, in cm2?

Mathematics

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Answer

16x+8 cm216x + 8 \text{ cm}^2

Step-by-step solution

5 steps

  1. Determine the original length of the square.

    The original length of the square is L1=(2x−1)L_1 = (2x - 1) cm.

  2. Calculate the original area of the square.

    The area of a square is given by the formula A=L2A = L^2.

    A1=(2x−1)2A_1 = (2x - 1)^2

    Expand the expression:

    A1=(2x)2−2(2x)(1)+(1)2A_1 = (2x)^2 - 2(2x)(1) + (1)^2 A1=4x2−4x+1 cm2A_1 = 4x^2 - 4x + 1 \text{ cm}^2

  3. Determine the new length of the square after the increase.

    The dimensions of the square are each increased by 4 cm.

    L2=(2x−1)+4L_2 = (2x - 1) + 4 L2=2x+3 cmL_2 = 2x + 3 \text{ cm}

  4. Calculate the new area of the square.

    A2=(2x+3)2A_2 = (2x + 3)^2

    Expand the expression:

    A2=(2x)2+2(2x)(3)+(3)2A_2 = (2x)^2 + 2(2x)(3) + (3)^2 A2=4x2+12x+9 cm2A_2 = 4x^2 + 12x + 9 \text{ cm}^2

  5. Calculate the increase in the area of the square.

    The increase in area is the new area minus the original area.

    Increase in Area=A2−A1\text{Increase in Area} = A_2 - A_1 Increase in Area=(4x2+12x+9)−(4x2−4x+1)\text{Increase in Area} = (4x^2 + 12x + 9) - (4x^2 - 4x + 1)

    Distribute the negative sign:

    Increase in Area=4x2+12x+9−4x2+4x−1\text{Increase in Area} = 4x^2 + 12x + 9 - 4x^2 + 4x - 1

    Combine like terms:

    Increase in Area=(4x2−4x2)+(12x+4x)+(9−1)\text{Increase in Area} = (4x^2 - 4x^2) + (12x + 4x) + (9 - 1) Increase in Area=0+16x+8\text{Increase in Area} = 0 + 16x + 8 Increase in Area=16x+8 cm2\text{Increase in Area} = 16x + 8 \text{ cm}^2

    The final answer is 16x+8 cm216x + 8 \text{ cm}^2.

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