Question

this question has two parts. first, answer part a. then, answer part b. part a rectangle rstz has a length that is 3 more units than twice its width. a. draw and label a figure for rectangle rstz on a separate sheet of paper. b. write an algebraic expression for the perimeter of the rectangle. an expression for the perimeter, where w is the width, must be... part b c. find the width if the perimeter is 58 millimeters. explain how you can check that your answer is correct. solving 58 = 2w+__ for w, the width is found to be mm. the sum of all four sides, 23++6 + 6, should equal . d. on a separate sheet of paper, use a ruler to draw and label (overline{pq}), which is congruent to the segment representing the length of rectangle rstz. what is the measure of (overline{pq})? __ mm

Explanation:

Step1: Define variables

Let the width of the rectangle be $w$. Then the length $l = w + 3$.

Step2: Recall perimeter formula

The perimeter $P$ of a rectangle is given by $P=2(l + w)$. Substitute $l = w + 3$ into the formula: $P = 2((w + 3)+w)$.

Step3: Simplify the expression

$P=2(2w + 3)=4w+6$.

Step4: Solve for width when perimeter is 30

Set $P = 30$. So, $30=4w + 6$. Subtract 6 from both sides: $30-6=4w$, which gives $24 = 4w$. Divide both sides by 4: $w = 6$.

Step5: Check the answer

The length $l=w + 3=6 + 3=9$. The sum of all four - sides is $9+6+9+6=30$.

Answer:

Part A: The algebraic expression for the perimeter of the rectangle is $4w + 6$.
Part B: Solving $30=4w + 6$ for $w$, the width is found to be $6$ cm. The sum of all four sides, $9+6+9+6$, should equal $30$. On a separate sheet of paper, using a ruler to draw and label $\overline{PQ}$ (where $\overline{PQ}$ represents the length), if the width is $6$ cm, the length is $9$ cm and the measure of $\overline{PQ}$ is $9$ cm.