Three right-angled triangles are shown below. Each triangle is divided into an even number of sections of equal width, some of which are shaded.
a) For each triangle, find the ratio of the shaded area to the unshaded area.
Give each answer in its simplest form.
b) Using your answers to part a), suggest a general rule for this ratio when the triangle is split into 2n sections.
Your answer should be in its simplest form.

The triangle is divided into 2 equal-width sections. The rightmost section forms a smaller, similar shaded triangle. The base and height of the shaded triangle are each half those of the larger triangle.

The area of a triangle is half the product of its base and height. So, the area of the shaded triangle is:

[tex]\textsf{Shaded area}=\dfrac12\left(\dfrac12b\cdot\dfrac12h\right)=\dfrac18bh[/tex]

The unshaded area can be calculated by subtracting the shaded area from the area of the larger triangle:

[tex]\textsf{Unshaded area}=\dfrac12bh-\dfrac18bh=\dfrac38bh[/tex]

So, the ratio of the shaded area to the unshaded area is:

[tex]\dfrac18bh:\dfrac38bh=\boxed{1:3}[/tex]

[tex]\dotfill[/tex]

Question a)ii)

The triangle is divided into 4 equal-width sections. The rightmost section forms a smaller, similar shaded triangle. The base and height of the shaded triangle are each a quarter of those of the larger triangle.

The area of the smallest shaded triangle is:

[tex]\textsf{Shaded area 1}=\dfrac12\left(\dfrac14b\cdot\dfrac14h\right)=\dfrac{1}{32}bh[/tex]

The unshaded area to the left of the smallest shaded triangle is the difference between the area of the smallest shaded triangle and a similar triangle with half the base and height of the larger triangle.

[tex]\textsf{Unshaded area 1}=\dfrac12\left(\dfrac12b\cdot \dfrac12h\right)-\dfrac{1}{32}bh=\dfrac{3}{32}bh[/tex]

The shaded area to the left of unshaded area 1 is found by subtracting the area of a triangle with half the base and height of the larger triangle from the area of a triangle with 3/4 the base and height of the larger triangle.

[tex]\textsf{Shaded area 2}=\dfra12\left(\dfrac34b\cdot\dfrac34h\right)-\dfrac12\left(\dfrac12b\cdot \dfrac12h\right)=\dfrac{5}{32}bh[/tex]

The unshaded area to the left of shaded area 2 is found by subtracting the area of a triangle with 3/4 the base and height of the larger triangle from the area of the larger triangle:

[tex]\textsf{Unshaded area 2}=\dfrac12bh-\dfrac12\left(\dfrac34b\cdot\dfrac34h\right)=\dfrac{7}{32}bh[/tex]

So, the total shaded area is:

[tex]\textsf{Total shaded area}=\dfrac{1}{32}bh+\dfrac{5}{32}bh=\dfrac{3}{16}bh[/tex]

The total unshaded area is:

[tex]\textsf{Total unshaded area}=\dfrac{3}{32}bh+\dfrac{7}{32}bh=\dfrac{5}{16}bh[/tex]

So, the ratio of the shaded area to the unshaded area is:

[tex]\dfrac{3}{16}bh:\dfrac{5}{16}bh=\boxed{3:5}[/tex]

[tex]\dotfill[/tex]

Question a)iii)

The triangle is divided into 6 equal-width sections. The rightmost section forms a smaller, similar shaded triangle. The base and height of the shaded triangle are each a sixth of those of the larger triangle.

(Please see attachments for full explanation of this part).

[tex]\textsf{Shaded area 1}=\dfrac12\left(\dfrac16b\cdot\dfrac16h\right)=\dfrac{1}{72}bh[/tex]

[tex]\textsf{Unshaded area 1}=\dfrac12\left(\dfrac26b\cdot\dfrac26h\right)-\dfrac{1}{72}bh=\dfrac{1}{24}bh[/tex]

[tex]\textsf{Shaded area 2}=\dfrac12\left(\dfrac36b\cdot\dfrac36h\right)-\dfrac12\left(\dfrac26b\cdot\dfrac26h\right)=\dfrac{5}{72}bh[/tex]

[tex]\textsf{Unshaded area 2}=\dfrac12\left(\dfrac46b\cdot \dfrac46h\right)-\dfrac12\left(\dfrac36b\cdot\dfrac36h\right)=\dfrac{7}{72}bh[/tex]

[tex]\textsf{Shaded area 3}=\dfrac12\left(\dfrac56b\cdot\dfrac56h\right)-\dfrac12\left(\dfrac46b\cdot\dfrac46h\right)=\dfrac18bh[/tex]

[tex]\textsf{Unshaded area 3}=\dfrac12bh-\dfrac12\left(\dfrac56b\cdot\dfrac56h\right)=\dfrac{11}{72}bh[/tex]

So, the total shaded area is:

[tex]\textsf{Total shaded area}=\dfrac{1}{72}bh+\dfrac{5}{72}bh+\dfrac18bh=\dfrac{5}{24}bh[/tex]

The total unshaded area is:

[tex]\textsf{Total unshaded area}=\dfrac{1}{24}bh+\dfrac{7}{72}bh+\dfrac{11}{72}bh=\dfrac{7}{24}bh[/tex]

So, the ratio of the shaded area to the unshaded area is:

[tex]\dfrac{5}{24}bh:\dfrac{7}{24}bh=\boxed{5:7}[/tex]

[tex]\dotfill[/tex]

Question b)

The ratios of the shaded area to the unshaded area when the triangle is split into 2n sections is:

[tex]n=1\implies 1:3[/tex]

[tex]n=2\implies 3:5[/tex]

[tex]n=3\implies 5:7[/tex]

The pattern for the ratios is that the first number is always an odd number, and the second number is always 2 greater than the first.

So, the general rule for this ratio is:

[tex]\boxed{(2n - 1):(2n + 1)}[/tex]