Question

1. takeshi made some mistakes when calculating the surface area of the triangular prism shown. describe takeshi’s mistakes and correct them.

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Explanation:

To solve this, we analyze the triangular prism's surface area components: two triangular bases and three rectangular faces.

Step 1: Identify Mistakes

• Triangular Bases: The triangle has base \( b = 4 \), height \( h = 3 \). Area of one triangle is \( \frac{1}{2} \times 4 \times 3 = 6 \), so two triangles should be \( 2 \times 6 = 12 \). Takashi likely miscalculated (e.g., used wrong dimensions or formula).
• Rectangular Faces:
• Face with “16”: Dimensions should be \( 4 \times 4 \)? No—prism length is \( 4 \) (from triangle’s height? Wait, no: the prism’s length (the distance between the two triangles) is \( 4 \)? Wait, the rectangles:
• First rectangle (label “20”): \( 5 \times 4 \)? Wait, no—wait, the triangular prism’s rectangles: the three rectangles have lengths equal to the prism’s length (let’s say \( l = 4 \)) and widths equal to the triangle’s sides (\( 3, 4, 5 \))? Wait, no, the net shows:
• Triangle sides: \( 3, 4, 5 \) (right triangle, \( 3^2 + 4^2 = 5^2 \)).
• Rectangles:
• One with “20”: \( 5 \times 4 \)? Wait, \( 5 \times 4 = 20 \), that’s correct (length \( 4 \), width \( 5 \)).
• One with “16”: \( 4 \times 4 \)? No—wait, the middle rectangle: the triangle’s base is \( 4 \), so the rectangle should be \( 4 \times 4 \)? Wait, no, the prism’s length (the distance between the two triangles) is \( 4 \). So the three rectangles:
• \( 3 \times 4 = 12 \) (correctly labeled “12” for one).
• \( 4 \times 4 = 16 \) (correct? Wait, the triangle’s base is \( 4 \), so the rectangle attached to the base \( 4 \) would be \( 4 \times 4 \), that’s correct.
• \( 5 \times 4 = 20 \) (correctly labeled “20”).

But the triangular bases: area of one triangle is \( \frac{1}{2} \times 3 \times 4 = 6 \), so two triangles should be \( 2 \times 6 = 12 \). If Takashi labeled a triangle as “12” (one triangle), that’s wrong—he should have two triangles (total \( 12 \)) or miscalculated the triangle’s area.

Step 2: Correct Calculation

1. Triangular Bases:

Area of one right triangle: \( \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 3 = 6 \).
Two triangles: \( 2 \times 6 = 12 \).

1. Rectangular Faces:

• Rectangle 1 (side \( 3 \)): \( 3 \times 4 = 12 \).
• Rectangle 2 (side \( 4 \)): \( 4 \times 4 = 16 \).
• Rectangle 3 (side \( 5 \)): \( 5 \times 4 = 20 \).

1. Total Surface Area:

Sum all areas: \( 12 + 12 + 16 + 20 + 6 \)? Wait, no—wait, the net has two triangles and three rectangles. Wait, the net in the image: let’s count the faces. A triangular prism has 2 triangular faces and 3 rectangular faces. So:

• Two triangles: \( 2 \times 6 = 12 \).
• Three rectangles: \( 12 + 16 + 20 = 48 \).
• Total: \( 12 + 48 = 60 \).

Mistake Description

Takashi likely miscalculated the triangular faces (e.g., used \( \frac{1}{2} \times 4 \times 6 \) or counted one triangle as \( 12 \) instead of two triangles summing to \( 12 \)). Or, he misidentified the rectangular face dimensions (e.g., used wrong length/width for a rectangle).

Correct Solution

1. Triangular Faces:

Area of one triangle: \( \frac{1}{2} \times 4 \times 3 = 6 \). Two triangles: \( 2 \times 6 = 12 \).

1. Rectangular Faces:

• \( 3 \times 4 = 12 \), \( 4 \times 4 = 16 \), \( 5 \times 4 = 20 \). Sum: \( 12 + 16 + 20 = 48 \).

1. Total Surface Area:

\( 12 + 48 = 60 \).

Answer:

Takashi’s mistake was likely in calculating the triangular faces (e.g., incorrect formula or dimensions) or mislabeling rectangular faces. The correct surface area is \( \boldsymbol{60} \) (units²).