To find the amount of wrapping paper needed to wrap the toy in a triangular prism-shaped container, we need to calculate the surface area of the prism.
Step 1: Calculate the Area of the Bases
The base of the prism is a triangle, and we need to calculate the area of that triangular base first.
Area of the Triangular Base
To find the area of the triangular base, we can use the formula for the area of a triangle:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
For our triangular base:
- Base = 27 cm
- Height = 5 cm
\[ \text{Area} = \frac{1}{2} \times 27 \times 5 = \frac{135}{2} = 67.5 \text{ cm}^2 \]
Since there are two triangular bases in the prism, we multiply the area by 2:
\[ \text{Total area of triangular bases} = 2 \times 67.5 = 135 \text{ cm}^2 \]
Step 2: Calculate the Area of the Rectangular Sides
The prism has three rectangular sides. We need to calculate the area of each rectangle:
-
Rectangle 1 (Length x Width):
- Length = 30 cm
- Width = 27 cm
\[ \text{Area} = 30 \times 27 = 810 \text{ cm}^2 \]
-
Rectangle 2 (Length x Side of Triangle): The side of the triangle is calculated using the Pythagorean theorem (assuming a right triangle):
- The two sides of the triangle are the base and the height.
- We can calculate the hypotenuse (side of the triangle) using:
\[ \text{Side} = \sqrt{\left(\frac{27}{2}\right)^2 + 5^2} \] \[ = \sqrt{(13.5)^2 + (5)^2} = \sqrt{182.25 + 25} = \sqrt{207.25} \approx 14.4 \text{ cm} \]
- Rectangle Area:
- Length = 30 cm
- Height = 14.4 cm
\[ \text{Area} = 30 \times 14.4 = 432 \text{ cm}^2 \]
-
Rectangle 3 (Test Side): There's also another side from the top part of the prism which we can assume to be of the same dimension as above.
- Length = 30 cm
- Height (again) = 14.4 cm
\[ \text{Area} = 30 \times 14.4 = 432 \text{ cm}^2 \]
Step 3: Add All Areas Together
Now we combine all areas together for the total surface area:
\[ \text{Total Surface Area} = \text{Area of bases} + \text{Area of Rectangles} \] \[ = 135 + 810 + 432 + 432 \] \[ = 135 + 810 + 864 = 1809 \text{ cm}^2 \]
Final Answer
Rounding \(1809\) to the nearest whole number gives:
\[ \text{Wrapping paper needed} \approx 1810 \text{ cm}^2 \]
So, approximately 1810 cm² of wrapping paper is needed to wrap the toy.