Question

The equilateral triangle ABC has side lengths of 8 cm. D is the midpoint of line BC and E is the midpoint of line AC. Find the area of the white section of the triangle.

What is the area of the white region?

Group of answer choices

20.9cm2


22.3cm2


19.3cm2


23.1cm2

Answer 1

The area of the white section of the triangle is equal to the area of the whole triangle minus the area of the three smaller triangles that are shaded.

To find the area of the whole triangle, we can use the formula $$A = \frac{\sqrt{3}}{4}s^2$$, where $$s$$ is the side length of the equilateral triangle. Plugging in $$s = 8$$ cm, we get $$A = \frac{\sqrt{3}}{4}(8)^2 = 16\sqrt{3}$$ cm$$^2$$.

To find the area of each smaller triangle, we can use the same formula, but with half the side length. That is, $$s = 4$$ cm. Then, the area of each smaller triangle is $$A = \frac{\sqrt{3}}{4}(4)^2 = 4\sqrt{3}$$ cm$$^2$$.

Therefore, the area of the white section of the triangle is $$16\sqrt{3} - 3(4\sqrt{3}) = 4\sqrt{3}$$ cm$$^2$$.

Answer 2

To find the area of the white region, we can consider the equilateral triangle ABC and subtract the areas of three smaller triangles (ADE, BDE, and CEF).

Since D is the midpoint of BC and E is the midpoint of AC, the segments AD, DE, and EC are also midpoints of the sides of the triangle, dividing it into four congruent smaller triangles. Each of these small triangles is an equilateral triangle.

The area of the white region is then the area of the large equilateral triangle ABC minus the sum of the areas of the three smaller equilateral triangles.

The area of an equilateral triangle with side length \(s\) can be calculated using the formula:

\[ \text{Area} = \frac{\sqrt{3}}{4} \times s^2 \]

For the large triangle ABC:

\[ \text{Area}_{ABC} = \frac{\sqrt{3}}{4} \times 8^2 \]

And for each of the smaller triangles (ADE, BDE, CEF):

\[ \text{Area}_{small} = \frac{\sqrt{3}}{4} \times \left(\frac{8}{2}\right)^2 \]

Now subtract the sum of the areas of the three smaller triangles from the area of the large triangle to get the area of the white region.

\[ \text{Area}_{white} = \text{Area}_{ABC} - 3 \times \text{Area}_{small} \]

\[ \text{Area}_{ABC} = \frac{\sqrt{3}}{4} \times 8^2 = 16\sqrt{3} \, \text{cm}^2 \]

\[ \text{Area}_{small} = \frac{\sqrt{3}}{4} \times \left(\frac{8}{2}\right)^2 = 4\sqrt{3} \, \text{cm}^2 \]

Now, calculate the area of the white region:

\[ \text{Area}_{white} = \text{Area}_{ABC} - 3 \times \text{Area}_{small} \]

\[ \text{Area}_{white} = 16\sqrt{3} - 3 \times 4\sqrt{3} = 16\sqrt{3} - 12\sqrt{3} = 4\sqrt{3} \, \text{cm}^2 \]

Now, approximate this value. The closest value among the provided choices is \( 4\sqrt{3} \approx 20.8 \, \text{cm}^2 \).

Therefore, the closest match is \( \boxed{20.9 \, \text{cm}^2} \).