QUESTION IMAGE
Question
match the width with the given area and length of a rectangle.
w = 6.5 cm w = 8.8 cm w = 6 cm w = 6.25 cm
l = 5 cm; a = 31.25 cm²
l = 7.5 cm; a = 48.75 cm²
l = 9.25 cm; a = 55.5 cm²
l = 5 cm; a = 44 cm²
To find the width \( w \) of a rectangle, we use the formula for the area of a rectangle: \( A = l \times w \), which can be rearranged to \( w=\frac{A}{l} \). We will apply this formula to each case.
Step 1: For \( l = 5\space\text{cm} \) and \( A=31.25\space\text{cm}^2 \)
We substitute \( l = 5 \) and \( A = 31.25 \) into the formula \( w=\frac{A}{l} \).
\[
w=\frac{31.25}{5}=6.25\space\text{cm}
\]
Step 2: For \( l = 7.5\space\text{cm} \) and \( A = 48.75\space\text{cm}^2 \)
Substitute \( l=7.5 \) and \( A = 48.75 \) into \( w=\frac{A}{l} \).
\[
w=\frac{48.75}{7.5} = 6.5\space\text{cm}
\]
Step 3: For \( l=9.25\space\text{cm} \) and \( A = 55.5\space\text{cm}^2 \)
Substitute \( l = 9.25 \) and \( A=55.5 \) into \( w=\frac{A}{l} \).
\[
w=\frac{55.5}{9.25}=6\space\text{cm}
\]
Step 4: For \( l = 5\space\text{cm} \) and \( A=44\space\text{cm}^2 \)
Substitute \( l = 5 \) and \( A = 44 \) into \( w=\frac{A}{l} \).
\[
w=\frac{44}{5}=8.8\space\text{cm}
\]
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- \( l = 5\space\text{cm}, A = 31.25\space\text{cm}^2 \): \( w = 6.25\space\text{cm} \)
- \( l = 7.5\space\text{cm}, A = 48.75\space\text{cm}^2 \): \( w = 6.5\space\text{cm} \)
- \( l = 9.25\space\text{cm}, A = 55.5\space\text{cm}^2 \): \( w = 6\space\text{cm} \)
- \( l = 5\space\text{cm}, A = 44\space\text{cm}^2 \): \( w = 8.8\space\text{cm} \)