Question
What is the surface area of the net of the Triangle prism?
Illustration shows unfolded triangular prism. Has one Triangle on each side. The right Triangle shows 4cm in broken line down the middle of Triangle. Base of Triangle is 5 cm. The middle is made up of 3 rectangles length of 10 cm each end of rectangle is 5 cm
Illustration shows unfolded triangular prism. Has one Triangle on each side. The right Triangle shows 4cm in broken line down the middle of Triangle. Base of Triangle is 5 cm. The middle is made up of 3 rectangles length of 10 cm each end of rectangle is 5 cm
Answers
Answered by
GPT-4o mini
To calculate the surface area of the net of a triangular prism, we need to consider the areas of the triangular bases and the rectangular lateral faces.
### Step 1: Calculate the area of the triangular bases.
The triangular prism has two triangular bases. The area \( A \) of a triangle can be calculated using the formula:
\[
A = \frac{1}{2} \times \text{base} \times \text{height}
\]
From your description, the base of the triangle is 5 cm and the height (which is the broken line down the middle of the triangle) is 4 cm. Thus, the area of one triangular base is:
\[
A = \frac{1}{2} \times 5 \, \text{cm} \times 4 \, \text{cm} = \frac{1}{2} \times 20 \, \text{cm}^2 = 10 \, \text{cm}^2
\]
Since there are two triangular bases, their total area is:
\[
\text{Total area of triangular bases} = 2 \times 10 \, \text{cm}^2 = 20 \, \text{cm}^2
\]
### Step 2: Calculate the area of the rectangular faces.
The prism has three rectangular lateral faces. The dimensions of the rectangles are:
- Two rectangles have a height of 10 cm and a width equal to the sides of the triangular base (which is the base of 5 cm).
- The third rectangle has a height of 10 cm and a width equal to the length of the flute (the hypotenuse of the triangle).
To find the hypotenuse \( c \) of the right triangle, we can use the Pythagorean theorem:
\[
c = \sqrt{(\text{base})^2 + (\text{height})^2} = \sqrt{(5 \, \text{cm})^2 + (4 \, \text{cm})^2} = \sqrt{25 + 16} = \sqrt{41} \approx 6.4 \, \text{cm}
\]
Now we have the dimensions of the lateral faces:
1. **First rectangle**: \( 10 \, \text{cm} \times 5 \, \text{cm} \)
2. **Second rectangle**: \( 10 \, \text{cm} \times 5 \, \text{cm} \)
3. **Third rectangle**: \( 10 \, \text{cm} \times \sqrt{41} \, \text{cm} \)
Calculating the areas:
1. Area of the first rectangle:
\[
A_1 = 10 \, \text{cm} \times 5 \, \text{cm} = 50 \, \text{cm}^2
\]
2. Area of the second rectangle:
\[
A_2 = 10 \, \text{cm} \times 5 \, \text{cm} = 50 \, \text{cm}^2
\]
3. Area of the third rectangle:
\[
A_3 = 10 \, \text{cm} \times \sqrt{41} \, \text{cm} \approx 10 \, \text{cm} \times 6.4 \, \text{cm} \approx 64 \, \text{cm}^2
\]
### Step 3: Total area of the rectangular faces:
\[
\text{Total area of rectangles} = 50 + 50 + 64 \approx 164 \, \text{cm}^2
\]
### Step 4: Calculate the total surface area of the triangular prism:
Adding the areas of the triangular bases and the rectangular faces:
\[
\text{Total Surface Area} = \text{Area of triangular bases} + \text{Total area of rectangles}
\]
\[
\text{Total Surface Area} = 20 \, \text{cm}^2 + 164 \, \text{cm}^2 = 184 \, \text{cm}^2
\]
Thus, the total surface area of the triangular prism is approximately \( 184 \, \text{cm}^2 \).
### Step 1: Calculate the area of the triangular bases.
The triangular prism has two triangular bases. The area \( A \) of a triangle can be calculated using the formula:
\[
A = \frac{1}{2} \times \text{base} \times \text{height}
\]
From your description, the base of the triangle is 5 cm and the height (which is the broken line down the middle of the triangle) is 4 cm. Thus, the area of one triangular base is:
\[
A = \frac{1}{2} \times 5 \, \text{cm} \times 4 \, \text{cm} = \frac{1}{2} \times 20 \, \text{cm}^2 = 10 \, \text{cm}^2
\]
Since there are two triangular bases, their total area is:
\[
\text{Total area of triangular bases} = 2 \times 10 \, \text{cm}^2 = 20 \, \text{cm}^2
\]
### Step 2: Calculate the area of the rectangular faces.
The prism has three rectangular lateral faces. The dimensions of the rectangles are:
- Two rectangles have a height of 10 cm and a width equal to the sides of the triangular base (which is the base of 5 cm).
- The third rectangle has a height of 10 cm and a width equal to the length of the flute (the hypotenuse of the triangle).
To find the hypotenuse \( c \) of the right triangle, we can use the Pythagorean theorem:
\[
c = \sqrt{(\text{base})^2 + (\text{height})^2} = \sqrt{(5 \, \text{cm})^2 + (4 \, \text{cm})^2} = \sqrt{25 + 16} = \sqrt{41} \approx 6.4 \, \text{cm}
\]
Now we have the dimensions of the lateral faces:
1. **First rectangle**: \( 10 \, \text{cm} \times 5 \, \text{cm} \)
2. **Second rectangle**: \( 10 \, \text{cm} \times 5 \, \text{cm} \)
3. **Third rectangle**: \( 10 \, \text{cm} \times \sqrt{41} \, \text{cm} \)
Calculating the areas:
1. Area of the first rectangle:
\[
A_1 = 10 \, \text{cm} \times 5 \, \text{cm} = 50 \, \text{cm}^2
\]
2. Area of the second rectangle:
\[
A_2 = 10 \, \text{cm} \times 5 \, \text{cm} = 50 \, \text{cm}^2
\]
3. Area of the third rectangle:
\[
A_3 = 10 \, \text{cm} \times \sqrt{41} \, \text{cm} \approx 10 \, \text{cm} \times 6.4 \, \text{cm} \approx 64 \, \text{cm}^2
\]
### Step 3: Total area of the rectangular faces:
\[
\text{Total area of rectangles} = 50 + 50 + 64 \approx 164 \, \text{cm}^2
\]
### Step 4: Calculate the total surface area of the triangular prism:
Adding the areas of the triangular bases and the rectangular faces:
\[
\text{Total Surface Area} = \text{Area of triangular bases} + \text{Total area of rectangles}
\]
\[
\text{Total Surface Area} = 20 \, \text{cm}^2 + 164 \, \text{cm}^2 = 184 \, \text{cm}^2
\]
Thus, the total surface area of the triangular prism is approximately \( 184 \, \text{cm}^2 \).