Shaded Area B4E95F

1. **State the problem:** We have triangle ABC with base $AB = 12$ cm and height $h = 10$ cm. The height is 2 cm shorter than the base, which matches since $12 - 2 = 10$. We need to find the area of the shaded portions, which are two smaller triangles on the left and right sides inside ABC. 2. **Understand the figure:** The base $AB$ is divided into three segments: $3$ cm on the left, $6$ cm in the middle, and $3$ cm on the right. The middle segment corresponds to the base of the unshaded central triangle. 3. **Calculate the total area of triangle ABC:** $$\text{Area}_{ABC} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 12 \times 10 = 60 \text{ cm}^2.$$ 4. **Calculate the height of the central unshaded triangle:** Since the base of the central triangle is $6$ cm and the height of ABC is $10$ cm, the height of the central triangle is $10$ cm (same altitude line from $C$ to $AB$). 5. **Calculate the area of the central unshaded triangle:** $$\text{Area}_{central} = \frac{1}{2} \times 6 \times 10 = 30 \text{ cm}^2.$$ 6. **Calculate the combined area of the shaded portions:** $$\text{Area}_{shaded} = \text{Area}_{ABC} - \text{Area}_{central} = 60 - 30 = 30 \text{ cm}^2.$$ 7. **Verify the shaded areas:** The shaded portions are two triangles with bases $3$ cm each and height $10$ cm. Area of one shaded triangle: $$\frac{1}{2} \times 3 \times 10 = 15 \text{ cm}^2.$$ Two such triangles: $$2 \times 15 = 30 \text{ cm}^2,$$ which matches the previous calculation. **Final answer:** The total area of the shaded portions is $30$ cm$^2$.