https://file.aiquickdraw.com/mathbot/file/1728222208135vmzwsits.jpg

Let's break down the problem step by step.

### Part a: How much paint does Josh use?

We know:
- Josh has **5 identical triangular prisms**.
- Each prism's dimensions are given as \( 6 \, \text{m} \times 8 \, \text{m} \times 10 \, \text{m} \) for the triangle, and the length of the prism is \( 12 \, \text{m} \).
- Each square meter needs **32 mL** of paint.

#### Step 1: Surface Area of One Prism
The surface area consists of:
1. **Two triangular faces**: Area of a triangle = \( \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6 \times 8 = 24 \, \text{m}^2 \). There are two such triangles, so their combined area is \( 2 \times 24 = 48 \, \text{m}^2 \).
   
2. **Three rectangular faces**:
   - One rectangle (base) with dimensions \( 12 \, \text{m} \times 6 \, \text{m} = 72 \, \text{m}^2 \).
   - Another rectangle (height) with dimensions \( 12 \, \text{m} \times 8 \, \text{m} = 96 \, \text{m}^2 \).
   - Another rectangle (slant side) with dimensions \( 12 \, \text{m} \times 10 \, \text{m} = 120 \, \text{m}^2 \).

   Total area of the rectangular faces = \( 72 + 96 + 120 = 288 \, \text{m}^2 \).

#### Step 2: Total Surface Area of One Prism
Total surface area of one prism = \( 48 \, \text{m}^2 \) (triangular faces) + \( 288 \, \text{m}^2 \) (rectangular faces) = **336 m²**.

#### Step 3: Paint for 5 Prisms
The total surface area for all 5 prisms is:
\[ 5 \times 336 = 1680 \, \text{m}^2 \]

Since 32 mL of paint is needed per \( 1 \, \text{m}^2 \), the total amount of paint used is:
\[ 1680 \, \text{m}^2 \times 32 \, \text{mL/m}^2 = 53,760 \, \text{mL} = 53.76 \, \text{L} \]

So, Josh uses **53.76 liters of paint**.

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### Part b: How much paint would Josh have saved if he had stuck the prisms together first?

When the prisms are stuck together, the touching surfaces don't need to be painted. Each side rectangle of \( 12 \, \text{m} \times 6 \, \text{m} = 72 \, \text{m}^2 \) is shared between two adjacent prisms. There are 4 internal faces that won't be painted.

#### Step 1: Unpainted Area
The total area of the unpainted surfaces (the ones that overlap) is:
\[ 4 \times 72 = 288 \, \text{m}^2 \]

#### Step 2: Adjusted Total Surface Area
The new total surface area to be painted is:
\[ 1680 \, \text{m}^2 - 288 \, \text{m}^2 = 1392 \, \text{m}^2 \]

#### Step 3: Adjusted Paint Required
The amount of paint needed for the adjusted surface area is:
\[ 1392 \, \text{m}^2 \times 32 \, \text{mL/m}^2 = 44,544 \, \text{mL} = 44.544 \, \text{L} \]

#### Step 4: Paint Saved
Josh would save:
\[ 53.76 \, \text{L} - 44.544 \, \text{L} = 9.216 \, \text{L} \]

So, Josh would save **9.216 liters of paint** if he painted them after sticking them together.

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Let me know if you need any more details or further clarifications!

### Related Questions:
1. What is the formula for the surface area of a triangular prism?
2. How would the calculations change if the prisms were cylinders instead of triangular prisms?
3. How would the cost of the paint change if the price per liter increased by 10%?
4. How much more paint would Josh use if the prisms were slightly larger (e.g., increasing the base of the triangle by 1 meter)?
5. How would the paint usage change if the prisms were positioned in a different way?

### Tip:
Always double-check the areas of individual shapes when calculating the surface area of a 3D object. This can help avoid mistakes in the final calculation!

Josh paints 5 identical prisms then sticks them together, as shown below. Each m² needs 32 mL of paint to cover it. a) How much paint does Josh use? b) How much paint would Josh have saved if he'd stuck the prisms together first and then painted them?

Josh paints 5 identical triangular prisms and calculates the amount of paint required for individual and grouped configurations. This problem involves surface area calculations of prisms, and multiplication of paint required per square meter. Suitable for students in Grades 8-10.

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