Question

the rectangle shown has a perimeter of 84 cm and the given area. its length is 6 more than three times its width. write and solve a system of equations to find the dimensions of the rectangle. the length of the rectangle is \\(\square\\) cm and the width of the rectangle is \\(\square\\) cm. \\(a = 297\\,\text{cm}^2\\)

Explanation:

Step1: Define variables and equations

Let \( L \) be the length and \( W \) be the width. From the problem, we have two equations:

1. Perimeter of a rectangle: \( 2(L + W)=84 \) (simplifies to \( L + W = 42 \))
2. Length - width relationship: \( L = 3W + 6 \)
3. Area (though we can use the first two for perimeter - length - width, and verify with area later): \( L\times W=297 \)

Step2: Substitute \( L \) into perimeter equation

Substitute \( L = 3W + 6 \) into \( L + W = 42 \):
\( (3W + 6)+W = 42 \)
Combine like terms: \( 4W+6 = 42 \)
Subtract 6 from both sides: \( 4W=42 - 6=36 \)
Divide by 4: \( W=\frac{36}{4} = 9 \)

Step3: Find length \( L \)

Substitute \( W = 9 \) into \( L = 3W + 6 \):
\( L=3\times9 + 6=27 + 6=33 \)

Step4: Verify with area

Check \( L\times W=33\times9 = 297 \), which matches the given area.

Answer:

The length of the rectangle is \( \boldsymbol{33} \) cm and the width of the rectangle is \( \boldsymbol{9} \) cm.