Question

a rectangle is drawn so the width is 8 inches longer than the height. if the rectangle 30 inches, find the height. give your answer rounded to 1 decimal place. inches add work > next question

Explanation:

Step1: Define variables

Let the height of the rectangle be \( h \) inches. Then the width is \( h + 8 \) inches. (Assuming the "30 inches" refers to the perimeter, since area is not specified and perimeter is a common rectangle measurement. If it's perimeter, the formula for the perimeter \( P \) of a rectangle is \( P = 2\times(\text{width}+\text{height}) \). So we set up the equation: \( 2(h + (h + 8))=30 \)

Step2: Simplify the equation

First, simplify the left - hand side: \( 2(2h + 8)=30 \)
Then, divide both sides by 2: \( 2h+8 = 15 \)

Step3: Solve for h

Subtract 8 from both sides: \( 2h=15 - 8=7 \)
Then divide both sides by 2: \( h=\frac{7}{2}=3.5 \)? Wait, no, maybe the 30 is the area. Let's re - consider. If the area \( A \) of a rectangle is \( A=\text{width}\times\text{height} \), then \( h(h + 8)=30 \)
Expand the left - hand side: \( h^{2}+8h - 30 = 0 \)
Use the quadratic formula \( h=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \) for the quadratic equation \( ax^{2}+bx + c = 0 \). Here, \( a = 1 \), \( b = 8 \), \( c=-30 \)
Calculate the discriminant \( \Delta=b^{2}-4ac=(8)^{2}-4\times1\times(-30)=64 + 120 = 184 \)
Then \( h=\frac{-8\pm\sqrt{184}}{2}=\frac{-8\pm2\sqrt{46}}{2}=-4\pm\sqrt{46} \)
Since height can't be negative, we take the positive root: \( h=-4+\sqrt{46}\approx - 4 + 6.782=2.782\approx2.8 \) (Wait, maybe the 30 is the perimeter. Let's go back to perimeter. If \( P = 30 \), \( 2(h+(h + 8))=30\Rightarrow2(2h + 8)=30\Rightarrow4h+16 = 30\Rightarrow4h=14\Rightarrow h = 3.5 \). But the problem statement is a bit unclear about whether 30 is perimeter or area. Since the first approach with perimeter gives a nice answer, but maybe the user made a typo. Wait, let's check the quadratic formula again for the area case. \( \sqrt{46}\approx6.782 \), so \( h=-4 + 6.782 = 2.782\approx2.8 \). But maybe the 30 is the perimeter. Let's assume it's perimeter.

Wait, the original problem says "If the rectangle 30 inches" - probably a typo, maybe "If the rectangle has a perimeter of 30 inches" or "If the rectangle has an area of 30 square inches". Let's assume it's area. Then:

For the quadratic equation \( h^{2}+8h - 30=0 \)
\( a = 1 \), \( b = 8 \), \( c=-30 \)
\( h=\frac{-8\pm\sqrt{64+120}}{2}=\frac{-8\pm\sqrt{184}}{2}=\frac{-8\pm2\sqrt{46}}{2}=-4\pm\sqrt{46} \)
\( \sqrt{46}\approx6.782 \), so \( h=-4 + 6.782=2.782\approx2.8 \) (rounded to 1 decimal place)

Answer:

\( 2.8 \)