First time poster so let me know if I've done anything wrong formatting wise. I believe the answer to be

45%

Justification:

Observe that all of the triangles are similar to each other (base angles are equal by alternate angles). Further the $n$th triangle ($n$ corresponds to how many bands the triangle contains) has side length ratio $n:1$ with the top triangle, using the fact that the bands are evenly spaced. We can use the fact that the ratio of areas of similar triangles is equal to the square of the ratio of side lengths to obtain:$$\frac{T_n}{T} = \frac{n^2}{1^2}$$ Where $T$ is the area of the top triangle and $T_n$ is the area of the $n$th triangle. Rearranging yields:$$T_n = n^2T$$ To find the area of the $n$th band from the top we can take the difference of $T_n$ and $T_{n-1}$:$$A_n = T_n - T_{n-1} = n^2T - (n-1)^2A = (2n-1)A$$That is, the area of the $n$th band is equal to $2n-1$ times the area of the top triangle. We can then compute the area of the shaded parts as they are simply the odd bands:$$A_{shaded} = \sum_{i \in \{1, 3, 5, 7, 9\}}A_{i} = \sum_{i \in \{1, 3, 5, 7, 9\}}(2i-1)T = 45T$$We can also compute the area of all regions by summing all the shaded areas:$$A_{all} = \sum_{i = 1}^{10}A_{i} = \sum_{i = 1}^{10}(2i-1)T = 100T$$Then taking the ratio gives:$$\frac{A_{shaded}}{A_{all}} = \frac{45T}{100T} = 0.45$$Giving the answer of 45%

I believe the answer to be

45%

Justification:

Observe that all of the triangles are similar to each other (base angles are equal by alternate angles). Further the $n$th triangle ($n$ corresponds to how many bands the triangle contains) has side length ratio $n:1$ with the top triangle, using the fact that the bands are evenly spaced. We can use the fact that the ratio of areas of similar triangles is equal to the square of the ratio of side lengths to obtain:$$\frac{T_n}{T} = \frac{n^2}{1^2}$$ Where $T$ is the area of the top triangle and $T_n$ is the area of the $n$th triangle. Rearranging yields:$$T_n = n^2T$$ To find the area of the $n$th band from the top we can take the difference of $T_n$ and $T_{n-1}$:$$A_n = T_n - T_{n-1} = n^2T - (n-1)^2A = (2n-1)A$$That is, the area of the $n$th band is equal to $2n-1$ times the area of the top triangle. We can then compute the area of the shaded parts as they are simply the odd bands:$$A_{shaded} = \sum_{i \in \{1, 3, 5, 7, 9\}}A_{i} = \sum_{i \in \{1, 3, 5, 7, 9\}}(2i-1)T = 45T$$We can also compute the area of all regions by summing all the shaded areas:$$A_{all} = \sum_{i = 1}^{10}A_{i} = \sum_{i = 1}^{10}(2i-1)T = 100T$$Then taking the ratio gives:$$\frac{A_{shaded}}{A_{all}} = \frac{45T}{100T} = 0.45$$Giving the answer of 45%

First time poster so let me know if I've done anything wrong formatting wise. I believe the answer to be

45%

Justification:

Observe that all of the triangles are similar to each other (base angles are equal by alternate angles). Further the $n$th triangle ($n$ corresponds to how many bands the triangle contains) has side length ratio $n:1$ with the top triangle, assuming the bands are equally spaced (not sure if this is possible without the assumption). We can use the fact that the ratio of areas of similar triangles is equal to the square of the ratio of side lengths to obtain:$$\frac{T_n}{T} = \frac{n^2}{1^2}$$ Where $T$ is the area of the top triangle and $T_n$ is the area of the $n$th triangle. Rearranging yields:$$T_n = n^2T$$ To find the area of the $n$th band from the top we can take the difference of $T_n$ and $T_{n-1}$:$$A_n = T_n - T_{n-1} = n^2T - (n-1)^2A = (2n-1)A$$That is, the area of the $n$th band is equal to $2n-1$ times the area of the top triangle. We can then compute the area of the shaded parts as they are simply the odd bands:$$A_{shaded} = \sum_{i \in \{1, 3, 5, 7, 9\}}A_{i} = \sum_{i \in \{1, 3, 5, 7, 9\}}(2i-1)T = 45T$$We can also compute the area of all regions by summing all the shaded areas:$$A_{all} = \sum_{i = 1}^{10}A_{i} = \sum_{i = 1}^{10}(2i-1)T = 100T$$Then taking the ratio gives:$$\frac{A_{shaded}}{A_{all}} = \frac{45T}{100T} = 0.45$$Giving the answer of 45%

First time poster so let me know if I've done anything wrong formatting wise. I believe the answer to be

45%

Justification:

Observe that all of the triangles are similar to each other (base angles are equal by alternate angles). Further the $n$th triangle ($n$ corresponds to how many bands the triangle contains) has side length ratio $n:1$ with the top triangle, using the fact that the bands are evenly spaced. We can use the fact that the ratio of areas of similar triangles is equal to the square of the ratio of side lengths to obtain:$$\frac{T_n}{T} = \frac{n^2}{1^2}$$ Where $T$ is the area of the top triangle and $T_n$ is the area of the $n$th triangle. Rearranging yields:$$T_n = n^2T$$ To find the area of the $n$th band from the top we can take the difference of $T_n$ and $T_{n-1}$:$$A_n = T_n - T_{n-1} = n^2T - (n-1)^2A = (2n-1)A$$That is, the area of the $n$th band is equal to $2n-1$ times the area of the top triangle. We can then compute the area of the shaded parts as they are simply the odd bands:$$A_{shaded} = \sum_{i \in \{1, 3, 5, 7, 9\}}A_{i} = \sum_{i \in \{1, 3, 5, 7, 9\}}(2i-1)T = 45T$$We can also compute the area of all regions by summing all the shaded areas:$$A_{all} = \sum_{i = 1}^{10}A_{i} = \sum_{i = 1}^{10}(2i-1)T = 100T$$Then taking the ratio gives:$$\frac{A_{shaded}}{A_{all}} = \frac{45T}{100T} = 0.45$$Giving the answer of 45%