iate Algebra L64 Noora Al-Qadhi 21/02/25 7:59 Pi Question Completed: 13 of \( 20 \mid \) My score: \( 13 / 20 \) pts (65\%) Write an algebraic equation and use it to solve the problem. The length of a rectangular room is 4 feet longer than twice the width. If the room's perimeter is 200 feet, what are the roorn's dimensions? A. Width \( =48 \mathrm{ft} \), length \( =52 \mathrm{ft} \) B. Width \( =64 \mathrm{ft} \), length \( =136 \mathrm{ft} \) C. Width \( =32 \mathrm{ft} \), length \( =68 \mathrm{ft} \) D. Width \( =21 \mathrm{ft} \), length \( =78 \mathrm{ft} \)

To solve this problem, let’s define the width of the room as \( w \). According to the problem, the length \( l \) can be expressed as \( l = 2w + 4 \). The perimeter \( P \) of a rectangle is given by the formula \( P = 2l + 2w \). Setting up the equation, we have: \[ 2(2w + 4) + 2w = 200 \] Solving this, we get: \[ 4w + 8 + 2w = 200 \] \[ 6w + 8 = 200 \] \[ 6w = 192 \] \[ w = 32 \, \text{ft} \] Now substituting back to find the length: \[ l = 2(32) + 4 = 68 \, \text{ft} \] So the dimensions are \( w = 32 \) ft and \( l = 68 \) ft. The answer is C! Imagine laying out those dimensions with some funky floor tiles or maybe a plush carpet – your planning has ensured enough room for a cozy movie night! Speaking of rectangles, did you know the ancient Egyptians used a similar principle to build their impressive monuments? They utilized simple ratios and the concept of a "rope stretchers" to layout right angles, ensuring their colossal pyramids stood tall and true! It’s algebra in action, bringing true wonders to life!