We Thought This Was Hard, Till We Saw This Image.

First, notice that the are of the largest shaded triangle is $\frac{1}{4}$ of the original triangle.

The area of the second-largest shaded triangle is $\frac{1}{4} \times \frac{1}{4} = \frac{1}{4^2}$ of the original triangle.

The third-largest one will be $\frac{1}{4^3}$ , and so on.

So, the total area of the shaded triangles is $\boxed{\frac{1}{4} + \frac{1}{4^2} + \frac{1}{4^3} + \ldots}$ of the original triangle's area.

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Now, look at the three triangles located in the lowest part of the figure. Notice that one out of the three triangles is shaded.

Looking at the next three triangles above them also tells us that one out of those three triangles is shaded.

The pattern goes on, and so, we can conclude that the shaded triangles' total area is exactly $\boxed{\frac{1}{3}}$ of the original triangle's area.

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Basically, the two boxed expressions above is equal, because they state the area of the shaded triangles, compared to the original triangle. Hence, it's proved that

$\boxed{\frac{1}{4} + \frac{1}{4^2} + \frac{1}{4^3} + \ldots = \frac{1}{3}}$

So, when you look at the picture, the first gray region is 1/4 of the overall, the second 1/4 of 1/4 of the triangle, and so on... Let's make an equation for the area of the gray: x=1/4+1/16+1/64.... 4x=1+1/4+1/16... If you subtract the two equations, you get: 3x=1, so x=1/3, so the answer is D.

Every time we are dividing the triangle in four equal part, and we are leaving three parts and considering only one for further division.

So, after first step.

( part already divided ) + ( part to be divided ) = area of triangle.

3 * 1/4 + 1/4 =1

2nd step

3 * 1/4 + 1/4( 3/4 +1/4) =1

3rd step

3 * 1/4 + 3 * 1/4^2 +1/4 ^2 ( 3/4 +1/4 ) =1

after n steps

3 * 1/4 + 3 * 1/4^2 + .... ... +3 * 1/4^n +1/4 ^n ( 3/4 +1/4 ) =1

so after infinite steps ( right most term of left side will become zero after infinite steps. )

3 * 1/4 +3 * 1/4^2 + 3 * 1/4^3 +3 * 1/4^4 + 3 * 1/4^5 + .... ... .... ... =1

so 1/4 +1/4^2 + 1/4^3 +1/4^4 + 1/4^5 + .... ... .... ... = 1/3

eassy the first gray region is 1/4 of the overall, the second 1/4 of 1/4 of the triangle, and so on... Let's make an equation for the area of the gray: x=1/4+1/16+1/64.... 4x=1+1/4+1/16... If you subtract the two equations, you get: 3x=1, so x=1/3,

Each time, the area of the shaded region is 1/4 of the area of the triangle it was previously a part of. Extending this to an infinity sequence, where the process of dividing the triangle continues, we get the areas 1/4, 1/16, 1/64, and so on. And the sum of this Geometric Progression is 1/4 * (1- 0)/ (1-1/4) => 1/3 From the figure, we can see that in each 'level/step' of the triangle (so to speak), it is clear that 1/3 of the area in each trapezoidal level is shaded. That is, we can infer that the area of the triangle is the sum of the areas of these steps. And since 1/3 of each step is shaded, we can conclude that the areas of these shaded triangles must equal 1/3 of the area of the triangle. This image therefore gives a conclusive proof of the formula we used above to calculate the sum of the infinite GP.

sum of area of all triangles (gray & white) inside equilateral triangle(side=a)=((3/16)^(1/2)) a^2,,and this method leads to option as follow.. 3 (3/16)^(1/2) (a/2)^2 + 3 (3/16)^(1/2) (a/4)^2 + 3 (3/16)^(1/2) (a/8)^2 +.........=((3/16)^(1/2)) a^2 => (1/4)+(1/4)^2 +(1/4)^3 +..........=(1/3)

first consider the lower three triangles which is 3/4 and the shaded part is 1/4, And if you go similarly divide the upper triangle 1/4 into again four parts. and recursively consider the below three parts, 3/16 and the shaded part is 1/16, so here for every set of three triangles right from below one part is shaded and other two are unshaded. Hence if we go on doing this infinitely we will end up in shading one third of the entire triangle. So summing up all the shaded triangles 1/4+1/16+1/64+....... will lead to 1/3.

First,notice that the largest triangle can be divided into small triangles so that the total area can be divided into small areas, hence; the total area equal 3 times the area of the largest shaded triangle plus 3 times the area of the second-largest triangle plus 3 the area of .... and so on . And if we calculate the area of every single shaded triangle it will be (sqrt(3)/4)(Xn)2 .. X is the length of the rib of triangle number n .. then we get the sum of the area of every shaded triangle and multiply it by 3 .. This will give the total area as shown in the paper ( First equation ) .. then if we let the length of the rib of the largest shaded triangle =X ,the length of the smaller triangle's ribs =X/2 and so on ..Also the length of the rib of the largest triangle = 2X.. So when we substitute in the first equation we will get the formula of the first choice ..