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finding an unknown match the width with the given area and length of a …

Question

finding an unknown
match the width with the given area and length of a rectangle.
$l = 7.5\\,\text{cm}, a = 48.75\\,\text{cm}^2$
$l = 9.25\\,\text{cm}, a = 55.5\\,\text{cm}^2$
$l = 5\\,\text{cm}, a = 44\\,\text{cm}^2$
$l = 5\\,\text{cm}, a = 31.25\\,\text{cm}^2$
$w = 6.5\\,\text{cm}$
$w = 6.25\\,\text{cm}$
$w = 8.8\\,\text{cm}$
$w = 6\\,\text{cm}$

Explanation:

Step1: Recall rectangle area formula

The area of a rectangle is $A = l \times w$, so rearranged to solve for width: $w = \frac{A}{l}$

Step2: Calculate width for first pair

Substitute $l=7.5\ \text{cm}, A=48.75\ \text{cm}^2$:
$w = \frac{48.75}{7.5} = 6.5\ \text{cm}$

Step3: Calculate width for second pair

Substitute $l=9.25\ \text{cm}, A=55.5\ \text{cm}^2$:
$w = \frac{55.5}{9.25} = 6\ \text{cm}$

Step4: Calculate width for third pair

Substitute $l=5\ \text{cm}, A=44\ \text{cm}^2$:
$w = \frac{44}{5} = 8.8\ \text{cm}$

Step5: Calculate width for fourth pair

Substitute $l=5\ \text{cm}, A=31.25\ \text{cm}^2$:
$w = \frac{31.25}{5} = 6.25\ \text{cm}$

Answer:

  • $l=7.5\ \text{cm}, A=48.75\ \text{cm}^2$ matches $w=6.5\ \text{cm}$
  • $l=9.25\ \text{cm}, A=55.5\ \text{cm}^2$ matches $w=6\ \text{cm}$
  • $l=5\ \text{cm}, A=44\ \text{cm}^2$ matches $w=8.8\ \text{cm}$
  • $l=5\ \text{cm}, A=31.25\ \text{cm}^2$ matches $w=6.25\ \text{cm}$