Question

1. the area of a rectangle is 430 cm². if the length is 8 cm more than three times the width, what are the dimensions?

Explanation:

Step1: Define Variables

Let the width of the rectangle be \( w \) cm. Then the length \( l \) is \( 3w + 8 \) cm (since length is 8 cm more than three times the width).

Step2: Use Area Formula

The area of a rectangle is given by \( A = l \times w \). We know \( A = 430 \, \text{cm}^2 \), so substitute \( l = 3w + 8 \) into the formula:

$$ (3w + 8) \times w = 430 $$

Expand the left - hand side:

$$ 3w^2+8w = 430 $$

Rearrange into standard quadratic form \( ax^2+bx + c = 0 \):

$$ 3w^2+8w - 430=0 $$

Step3: Solve the Quadratic Equation

We can use the quadratic formula \( w=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \), where \( a = 3 \), \( b = 8 \), and \( c=- 430 \).

First, calculate the discriminant \( D=b^{2}-4ac \):

$$ D = 8^{2}-4\times3\times(-430)=64 + 5160=5224 $$

Now, find the square root of the discriminant. \( \sqrt{5224}\approx72.28 \) (we can also simplify \( \sqrt{5224}=\sqrt{4\times1306} = 2\sqrt{1306}\approx2\times36.14 = 72.28 \))

Then, find the two solutions for \( w \):

$$ w=\frac{-8\pm72.28}{2\times3} $$

We have two cases:

• Case 1: \( w=\frac{-8 + 72.28}{6}=\frac{64.28}{6}\approx10.71 \)
• Case 2: \( w=\frac{-8 - 72.28}{6}=\frac{-80.28}{6}\approx - 13.38 \)

Since width cannot be negative, we take \( w\approx10.71 \) cm.

Step4: Find the Length

Now, find the length \( l = 3w+8 \). Substitute \( w\approx10.71 \):

$$ l=3\times10.71 + 8=32.13 + 8=40.13 $$

We can also check the quadratic equation by factoring or completing the square, but since it doesn't factor nicely, the quadratic formula is the way to go. Let's verify the area: \( l\times w\approx40.13\times10.71\approx430 \), which matches the given area.

Answer:

The width of the rectangle is approximately \( 10.71 \, \text{cm} \) and the length is approximately \( 40.13 \, \text{cm} \). If we want to solve it more precisely, we can keep it in exact form or use a calculator for more decimal places. If we solve the quadratic equation \( 3w^{2}+8w - 430 = 0 \) using a calculator for more precision:

The quadratic formula gives \( w=\frac{-8\pm\sqrt{64 + 5160}}{6}=\frac{-8\pm\sqrt{5224}}{6}=\frac{-8\pm4\sqrt{326.5}}{6}=\frac{-4\pm2\sqrt{326.5}}{3} \). Taking the positive root: \( w=\frac{-4 + 2\sqrt{326.5}}{3}\approx\frac{-4+2\times18.07}{3}=\frac{-4 + 36.14}{3}=\frac{32.14}{3}\approx10.71 \), and \( l = 3\times10.71+8 = 40.13 \). So the dimensions are approximately width \( 10.71 \, \text{cm} \) and length \( 40.13 \, \text{cm} \).