Question

1. a rectangles length is three meters less than twice of width. its area is 24 m². find the perimeter of the rectangle nearest tenth.

Explanation:

Step1: Define variables

Let the width of the rectangle be \( w \) meters. Then the length \( l \) is \( 2w - 3 \) meters (since length is three meters less than twice the width).

Step2: Use area formula

The area of a rectangle is \( A = l \times w \). We know \( A = 24 \, m^2 \), so substitute \( l = 2w - 3 \) into the area formula:

$$ (2w - 3)w = 24 $$

Expand to get a quadratic equation:

$$ 2w^2 - 3w - 24 = 0 $$

Step3: Solve quadratic equation

Use the quadratic formula \( w=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} \) for \( ax^2+bx+c = 0 \). Here, \( a = 2 \), \( b=-3 \), \( c = -24 \).
First, calculate the discriminant \( D=b^2 - 4ac=(-3)^2-4\times2\times(-24)=9 + 192 = 201 \)
Then,

$$ w=\frac{3\pm\sqrt{201}}{4} $$

We take the positive root (since width can't be negative):

$$ w=\frac{3+\sqrt{201}}{4}\approx\frac{3 + 14.177}{4}\approx\frac{17.177}{4}\approx4.294 \, m $$

Step4: Find length

Substitute \( w\approx4.294 \) into \( l = 2w - 3 \):

$$ l=2\times4.294-3=8.588 - 3 = 5.588 \, m $$

Step5: Calculate perimeter

The perimeter \( P \) of a rectangle is \( P = 2(l + w) \). Substitute \( l\approx5.588 \) and \( w\approx4.294 \):

$$ P = 2(5.588 + 4.294)=2\times9.882 = 19.764\approx19.8 \, m $$

Answer:

The perimeter of the rectangle is approximately \( \boldsymbol{19.8} \) meters.