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To solve the problem, let's consider the steps and calculations involved in determining the smallest possible area of the smallest rectangle that can fit the four given right-angled triangles with the specified leg lengths. Here are the steps:
### Step 1: Calculate the Hypotenuse
First, let's find the hypotenuse of each triangle using the Pythagorean theorem:
1. For the triangle with sides 1 cm and 4 cm:
[tex]\[ \text{Hypotenuse} = \sqrt{1^2 + 4^2} \approx 4.123 \text{ cm} \][/tex]
2. For the triangle with sides 2 cm and 4 cm:
[tex]\[ \text{Hypotenuse} = \sqrt{2^2 + 4^2} \approx 4.472 \text{ cm} \][/tex]
3. For the triangle with sides 3 cm and 4 cm:
[tex]\[ \text{Hypotenuse} = \sqrt{3^2 + 4^2} \approx 5.0 \text{ cm} \][/tex]
4. For the triangle with sides 4 cm and 4 cm:
[tex]\[ \text{Hypotenuse} = \sqrt{4^2 + 4^2} \approx 5.657 \text{ cm} \][/tex]
### Step 2: Calculate the Area of Each Triangle
Using the formula for the area of a triangle, [tex]\( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \)[/tex]:
1. For the triangle with sides 1 cm and 4 cm:
[tex]\[ \text{Area} = 0.5 \times 1 \times 4 = 2 \text{ cm}^2 \][/tex]
2. For the triangle with sides 2 cm and 4 cm:
[tex]\[ \text{Area} = 0.5 \times 2 \times 4 = 4 \text{ cm}^2 \][/tex]
3. For the triangle with sides 3 cm and 4 cm:
[tex]\[ \text{Area} = 0.5 \times 3 \times 4 = 6 \text{ cm}^2 \][/tex]
4. For the triangle with sides 4 cm and 4 cm:
[tex]\[ \text{Area} = 0.5 \times 4 \times 4 = 8 \text{ cm}^2 \][/tex]
### Step 3: Calculate the Total Area of the Triangles
Sum of the areas:
[tex]\[ \text{Total Area} = 2 + 4 + 6 + 8 = 20 \text{ cm}^2 \][/tex]
### Step 4: Determine the Dimensions of the Smallest Rectangle
To arrange the triangles without overlap, we need to fit them into a rectangle. One efficient way is to align all triangles along their common side (4 cm).
1. Arrange the triangles in a row such that their bases (1 cm, 2 cm, 3 cm, and 4 cm) add up:
[tex]\[ \text{Total Width} = 1 + 2 + 3 + 4 = 10 \text{ cm} \][/tex]
2. Since all triangles share the height of 4 cm:
[tex]\[ \text{Height} = 4 \text{ cm} \][/tex]
### Step 5: Calculate the Area of the Smallest Rectangle
The area of the smallest rectangle that can fit all four triangles:
[tex]\[ \text{Rectangle Area} = \text{Width} \times \text{Height} = 10 \text{ cm} \times 4 \text{ cm} = 40 \text{ cm}^2 \][/tex]
### Conclusion
The smallest possible area of the smallest rectangle that can fit all the given triangles is:
[tex]\[ \boxed{40 \text{ cm}^2} \][/tex]
### Step 1: Calculate the Hypotenuse
First, let's find the hypotenuse of each triangle using the Pythagorean theorem:
1. For the triangle with sides 1 cm and 4 cm:
[tex]\[ \text{Hypotenuse} = \sqrt{1^2 + 4^2} \approx 4.123 \text{ cm} \][/tex]
2. For the triangle with sides 2 cm and 4 cm:
[tex]\[ \text{Hypotenuse} = \sqrt{2^2 + 4^2} \approx 4.472 \text{ cm} \][/tex]
3. For the triangle with sides 3 cm and 4 cm:
[tex]\[ \text{Hypotenuse} = \sqrt{3^2 + 4^2} \approx 5.0 \text{ cm} \][/tex]
4. For the triangle with sides 4 cm and 4 cm:
[tex]\[ \text{Hypotenuse} = \sqrt{4^2 + 4^2} \approx 5.657 \text{ cm} \][/tex]
### Step 2: Calculate the Area of Each Triangle
Using the formula for the area of a triangle, [tex]\( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \)[/tex]:
1. For the triangle with sides 1 cm and 4 cm:
[tex]\[ \text{Area} = 0.5 \times 1 \times 4 = 2 \text{ cm}^2 \][/tex]
2. For the triangle with sides 2 cm and 4 cm:
[tex]\[ \text{Area} = 0.5 \times 2 \times 4 = 4 \text{ cm}^2 \][/tex]
3. For the triangle with sides 3 cm and 4 cm:
[tex]\[ \text{Area} = 0.5 \times 3 \times 4 = 6 \text{ cm}^2 \][/tex]
4. For the triangle with sides 4 cm and 4 cm:
[tex]\[ \text{Area} = 0.5 \times 4 \times 4 = 8 \text{ cm}^2 \][/tex]
### Step 3: Calculate the Total Area of the Triangles
Sum of the areas:
[tex]\[ \text{Total Area} = 2 + 4 + 6 + 8 = 20 \text{ cm}^2 \][/tex]
### Step 4: Determine the Dimensions of the Smallest Rectangle
To arrange the triangles without overlap, we need to fit them into a rectangle. One efficient way is to align all triangles along their common side (4 cm).
1. Arrange the triangles in a row such that their bases (1 cm, 2 cm, 3 cm, and 4 cm) add up:
[tex]\[ \text{Total Width} = 1 + 2 + 3 + 4 = 10 \text{ cm} \][/tex]
2. Since all triangles share the height of 4 cm:
[tex]\[ \text{Height} = 4 \text{ cm} \][/tex]
### Step 5: Calculate the Area of the Smallest Rectangle
The area of the smallest rectangle that can fit all four triangles:
[tex]\[ \text{Rectangle Area} = \text{Width} \times \text{Height} = 10 \text{ cm} \times 4 \text{ cm} = 40 \text{ cm}^2 \][/tex]
### Conclusion
The smallest possible area of the smallest rectangle that can fit all the given triangles is:
[tex]\[ \boxed{40 \text{ cm}^2} \][/tex]
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