Question

what have i learned?
doing the problems in this section will help you to evaluate which types of problems you feel comfortable with and which ones you need more help with.
solve each problem as completely as you can. the table at the end of this closure section provides answers to these problems. it also tells you where you can find additional help and where to find practice problems like them.

cl 1-142. vanson has a bunch of boxes that are all the same. he stacked four boxes, measured the stack, and found that it was three feet high.
a. how high will a stack of 20 of these boxes be?
b. the ceiling in the room where vanson is working is just a little more than 9 feet high. he wants to stack boxes from the floor to the ceiling. how many boxes will fit in one stack?

cl 1-143. find the perimeter and area of jacob’s swimming pool shown in the diagram below. be sure to show all of your work.
diagram of a composite rectangle: 18 m, 6 m, 11 m, 9 m

cl 1-144. tuan is playing a game, but the spinner is incomplete. if the numbers in the sections of the spinner represent the probabilities of spinning each section, help him figure out the fraction for the missing section of the spinner.
spinner diagram with sections \\(\frac{1}{4}\\), \\(\frac{1}{4}\\), \\(\frac{1}{5}\\), and a missing section

Explanation:

CL 1-142a

Step1: Find height of one box

Four boxes are 3 feet high, so one box height is $3\div4 = \frac{3}{4}$ feet.

Step2: Calculate height of 20 boxes

Multiply one box height by 20: $\frac{3}{4} \times 20 = 15$ feet.

Step1: Recall one box height

From part a, one box is $\frac{3}{4}$ feet high.

Step2: Find number of boxes for 9 feet

Divide 9 by $\frac{3}{4}$: $9\div\frac{3}{4}=9\times\frac{4}{3}=12$ boxes. Since ceiling is a little more than 9 feet, 12 boxes (as 12 boxes give 9 feet exactly, and ceiling is more, but we can't have partial boxes, so 12 fits).

Step1: Find perimeter

The shape is a rectangle when we consider the outer sides. The length is $18 + 11 = 29$ m, the width is $6 + 9 = 15$ m. Perimeter of rectangle is $2\times( length + width )=2\times(29 + 15)=2\times44 = 88$ m.

Step2: Find area

We can split the shape into two rectangles. First rectangle: $18\times(6 + 9)=18\times15 = 270$ m². Second rectangle: $11\times9 = 99$ m². Total area: $270 + 99 = 369$ m². (Alternative: Total length 29, total width 15, area $29\times15 = 435$? Wait, no, mistake. Wait, the correct split: the top part is 18m long and 6m + 9m = 15m? No, wait the diagram: the first rectangle is 18m (length) and (6 + 9)m? No, actually, the vertical side: 6m and 9m, so total height is 6 + 9 = 15m. The horizontal side: 18m and 11m, total length 18 + 11 = 29m. But the indent: the area can be calculated as large rectangle minus the missing rectangle. Large rectangle: 29m (length) × 15m (width) = 435 m². Missing rectangle: 11m (length) × 6m (width) = 66 m². So area = 435 - 66 = 369 m². Perimeter: since it's a rectilinear figure, the perimeter is same as the large rectangle, because the indent's horizontal and vertical sides add up. So perimeter: 2×(29 + 15)= 88 m.

Answer:

15 feet

CL 1-142b