Answered

A parking lot has two types of parking spaces: rectangles and parallelograms. Plans for the parking spaces are drawn on grid paper.

a. Which parking space covers the greater area, space A or space B? Show your work.
- Area of rectangle = length [tex]$\times$[/tex] breadth
- Length = 6 units
- Breadth = 3 units
- [tex]$6 \times 3 = 18$[/tex] square units
- Therefore, parking space A covers an area of 18 square units.

- Area of parallelogram = base [tex]$\times$[/tex] height
- Base = 4 units
- Height = 5 units
- [tex]$4 \times 5 = 20$[/tex] square units
- Therefore, parking space B covers an area of 20 square units.

Conclusion: Parking space B covers a greater area than parking space A.

b. Check your answer to part (a). Show your work.
- Recalculate the area of space A (rectangle):
- [tex]$6 \times 3 = 18$[/tex] square units

- Recalculate the area of space B (parallelogram):
- [tex]$4 \times 5 = 20$[/tex] square units

Conclusion: The calculations confirm that parking space B covers a greater area than parking space A.



Answer :

GinnyAnswer
### Step-by-Step Solution

Part (a):

Let's start by calculating the areas for both types of parking spaces:

1. Area of the rectangular parking space (Space A):
Given:
- Length of the rectangle = 6 units
- Breadth of the rectangle = 3 units

The area of a rectangle is given by the formula:
[tex]\[ \text{Area of rectangle} = \text{Length} \times \text{Breadth} \][/tex]

Substitute the given values into the formula:
[tex]\[ \text{Area of rectangle} = 6 \, \text{units} \times 3 \, \text{units} = 18 \, \text{square units} \][/tex]

2. Area of the parallelogram parking space (Space B):
Given:
- Base of the parallelogram = 4 units
- Height of the parallelogram = 5 units

The area of a parallelogram is given by the formula:
[tex]\[ \text{Area of parallelogram} = \text{Base} \times \text{Height} \][/tex]

Substitute the given values into the formula:
[tex]\[ \text{Area of parallelogram} = 4 \, \text{units} \times 5 \, \text{units} = 20 \, \text{square units} \][/tex]

From our calculations:
- Area of the rectangular parking space (Space A) = 18 square units
- Area of the parallelogram parking space (Space B) = 20 square units

Since 20 square units (area of Space B) is greater than 18 square units (area of Space A), the parallelogram parking space (Space B) covers the greater area.

Part (b):

Let's verify our calculations:

1. Re-check the area of the rectangular parking space (Space A):
[tex]\[ \text{Area of rectangle} = 6 \, \text{units} \times 3 \, \text{units} = 18 \, \text{square units} \][/tex]

2. Re-check the area of the parallelogram parking space (Space B):
[tex]\[ \text{Area of parallelogram} = 4 \, \text{units} \times 5 \, \text{units} = 20 \, \text{square units} \][/tex]

Given our re-checked and verified calculations:
- The rectangular parking space (Space A) has an area of 18 square units.
- The parallelogram parking space (Space B) has an area of 20 square units.

Therefore, according to both our initial solution and re-checked verification, the parallelogram parking space (Space B) definitively covers the greater area of 20 square units.