Question

A rectangle has a perimeter of 64 cm. If the length is 6cm more than the width of the rectangle,

what are the dimensions of the rectangle? SHOW ALL work

Answer 1

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To solve this problem, let's denote the width of the rectangle as $w$ (in cm).

According to the given information, the length of the rectangle is 6 cm more than the width, so the length can be represented as $w + 6$ (in cm).

The perimeter of a rectangle is given by the formula:

$$\text{{Perimeter}} = 2(\text{{length}} + \text{{width}})$$

Substituting the values, we have:

$64 = 2((w + 6) + w)$

Now, let's solve for $(w)$:

$64 = 2(2w + 6)$

$64 = 4w + 12$

$4w = 64 - 12$

$4w = 52$

Dividing both sides of the equation by 4, we get:

$w = \frac{{52}}{{4}}$

$w = 13$

So, the width of the rectangle is 13 cm.

Substituting the value of $w$ back into the expression for the length:

$$\text{{Length}} = w + 6 = 13 + 6 = 19$$

Therefore, the dimensions of the rectangle are 13 cm (width) and 19 cm (length).

Answer 2

the length is 19

the width is 13

Explanation:

well since a rectangle has two even sides the width have to be the same and the lengths have to be the same

perimeter is all the sides added together

since the length is the bigger size choose a sensible number (for example 15) and add another 15 to get the length and since it's 6cm smaller take 6 away from 15 which is 9 and add the two 9s together but 15+15+9+9 isn't 64 so you try again with a higher number until you each a total of 64