To find the areas of the original triangle and the dilated triangle, we start with the original triangle.
The area \( A \) of a triangle is given by the formula: \[ A = \frac{1}{2} \times \text{base} \times \text{height} \] For the original triangle:
- Base = 8 cm
- Height = 3 cm
Calculating the area: \[ A = \frac{1}{2} \times 8 , \text{cm} \times 3 , \text{cm} = \frac{1}{2} \times 24 , \text{cm}^2 = 12 , \text{cm}^2 \]
Now, the triangle is dilated by a scale factor of \( k = 3 \).
When a shape is dilated by a scale factor, the area of the dilated shape is scaled by the square of the scale factor. Therefore, the area \( A' \) of the dilated triangle can be calculated as: \[ A' = k^2 \times A \] Substituting the known values: \[ A' = 3^2 \times 12 , \text{cm}^2 = 9 \times 12 , \text{cm}^2 = 108 , \text{cm}^2 \]
Now we can compare the areas of the original triangle and the dilated triangle:
- Original triangle area: \( 12 , \text{cm}^2 \)
- Dilated triangle area: \( 108 , \text{cm}^2 \)
To determine how many times the area of the original triangle fits into the area of the dilated triangle: \[ \text{Ratio} = \frac{A'}{A} = \frac{108 , \text{cm}^2}{12 , \text{cm}^2} = 9 \]
Thus, the original triangle has an area of \( 12 , \text{cm}^2 \) and the dilated triangle has an area of \( 108 , \text{cm}^2 \); therefore, the dilated triangle has an area 9 times the size of the original triangle.
The correct statement based on the calculations is:
- The original triangle has an area of 12 cm² and the dilated triangle has an area of 108 cm²; therefore, the dilated triangle has an area 9 times the size of the original triangle.
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