(Finding Area Using Triangles and Rectangles MC)

A triangle is shown in the image.

A triangle with a height of 8 inches. The height is perpendicular to the base labeled 32 inches. The side from the top of the perpendicular side to the base is labeled 28 inches.

What is the area of the triangle?

256 in2
224 in2
128 in2
112 in2
Question 2(Multiple Choice Worth 2 points)
(Finding Area Using Triangles and Rectangles MC)

What is the area of a right triangle with a height of six and one fourth yards and a base of 22 yards?

34 yds2
sixty eight and three fourths yds2
132 yds2
one hundred thirty seven and one half yds2
Question 3(Multiple Choice Worth 2 points)
(Finding Area Using Triangles and Rectangles MC)

A family wants to make an addition to a deck that extends off the back of their home. The current deck is 250 ft2. The addition will be 23.25 ft in length and 14 ft wide. What will be the total area of the deck once the addition is complete?

75.5 ft2
325.5 ft2
575.5 ft2
287.25 ft2
Question 4(Multiple Choice Worth 2 points)
(Finding Area Using Triangles and Rectangles MC)

A composite figure is represented in the image.

A six-sided composite figure. A vertical line on the left is labeled 8 meters. The base is labeled 18 meters. There is a small portion from the vertical line that is parallel to the base that is labeled 6 meters. This portion leads to two segments that come to a point, and from that point, there is a height of 8 meters labeled.

What is the total area of the figure?

192 m2
216 m2
288 m2
336 m2
Question 5(Multiple Choice Worth 2 points)
(Finding Area Using Triangles and Rectangles MC)

A composite figure is represented in the image.

A four-sided shape with the base side labeled as 15.4 yards. The height is labeled 9.3 yards. A portion of the top from the perpendicular side to a right vertex is labeled 4.6 yards. A portion of the top from the perpendicular side to a left vertex is labeled 10.8 yards.

What is the total area of the figure?

71.61 yd2
100.44 yd2
121.83 yd2
143.22 yd2
Question 6 (Essay Worth 4 points)
(Finding Area Using Triangles and Rectangles HC)

A community would like to add a brick paver border around their swimming pool. They created the following image to represent the pool with the border.

A large rectangle with a length of 26 feet and a width of 14 feet. Inside of it is another rectangle with a length of 18 feet and a width of 6 feet.

Part A: Find the total area of the brick paver border that surrounds the 6 ft by 18 ft pool. Show your work. (2 points)

Part B: If brick pavers cost $6 per square foot, what is the total cost of the brick pavers needed for this project? Explain. (2 points)

1 answer

Let’s calculate each question step by step.

Question 1:

To find the area of a triangle:

\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]

Given:

  • Height = 8 inches
  • Base = 32 inches

\[ \text{Area} = \frac{1}{2} \times 32 \times 8 = \frac{1}{2} \times 256 = 128 \text{ in}^2 \]

Answer: 128 in²


Question 2:

For the area of a right triangle:

\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]

Given:

  • Height = 6.25 yards (which is \( \frac{25}{4} \))
  • Base = 22 yards

\[ \text{Area} = \frac{1}{2} \times 22 \times 6.25 = 11 \times 6.25 = 68.75 \text{ yds}^2 = 68 \frac{3}{4} \text{ yds}^2 \]

Answer: sixty eight and three fourths yds²


Question 3:

To find the total area of the deck after addition, add the area of the addition to the current area.

Area of the addition:

\[ \text{Area} = \text{length} \times \text{width} = 23.25 \times 14 = 325.5 \text{ ft}^2 \]

Total area:

\[ \text{Total Area} = 250 + 325.5 = 575.5 \text{ ft}^2 \]

Answer: 575.5 ft²


Question 4:

To find the area of the composite figure, one rectangle and one triangle are considered.

  1. Rectangle Area:

    \[ \text{Area}_{\text{rectangle}} = \text{base} \times \text{height} = 18 \times 8 = 144 \text{ m}^2 \]

  2. Triangle Area (base = 6 m, height = 8 m):

    \[ \text{Area}_{\text{triangle}} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6 \times 8 = 24 \text{ m}^2 \]

  3. Total area:

    \[ \text{Total Area} = 144 + 24 = 168 \text{ m}^2 \quad (\text{Note: This does not match the options. Double-check inputs and methods used.}) \]

(In review, the problem appears to involve potentially more fundamental shapes or areas; I'll need further specifics for verification.)


Question 5:

To find the area of a four-sided shape with dimensions defined for height and widths measured, one can calculate the area of a trapezoid:

  • Base1 = 15.4 yards
  • Base2 = (4.6 + 10.8) = 15.4 yards
  • Height = 9.3 yards

\[ \text{Area} = \frac{1}{2} \times (\text{Base1} + \text{Base2}) \times \text{Height} = \frac{1}{2} \times (15.4 + 15.4) \times 9.3 \]

\[ \text{Area} = 15.4 \times 9.3 = 143.22 \text{ yd}^2 \]

Answer: 143.22 yd²


Question 6:

Part A: Calculate the area of the brick paver border:

  1. Area of the outer rectangle:

\[ \text{Area}_{outer} = \text{length} \times \text{width} = 26 \times 14 = 364 \text{ ft}^2 \]

  1. Area of the inner rectangle (pool):

\[ \text{Area}_{inner} = 18 \times 6 = 108 \text{ ft}^2 \]

  1. Area of the border:

\[ \text{Area}{border} = \text{Area}{outer} - \text{Area}_{inner} = 364 - 108 = 256 \text{ ft}^2 \]

(Final calculations should note actual dimensions and overlaps of the shapes involved in this problem)

(Total area of border = 256 ft²)

Part B: To find the total cost of pavers needed:

\[ \text{Total Cost} = \text{Area}_{border} \times \text{Cost per ft}^2 \]

\[ = 256 \times 6 = 1536 \]

Total Cost = $1,536.00.