Which Area is the Largest?

We can count the number of smaller triangular tiles, each one has the same area, so the section with more tiles is the largest

The diagram shows three similar triangles, of heights 3, 5, and 6 respectively. Their areas are proportional to the heights; thus we have $A_1 : A_2 : A_3 = 3^2 : 5^2 : 6^2 = 9 : 25 : 36.$ The blue area is the difference between the second and first triangle; the red area is the difference between the third and second triangle. Thus $\text{orange}\ :\ \text{blue}\ :\ \text{red} = 9 : (25 - 9) : (36 - 25) = 9 : 16 : 11.$ Thus the orange area is smallest, the red is slightly bigger, and the $\boxed{\text{blue area}}$ is almost twice as large.

Generalization

Suppose the triangle were divided into $n$ trapezoidal sections, with height ratios $n : (n-1) : (n-2) : \dots : 2 : 1$ . Then the total area of the first $k$ sections would be proportional to $\begin{aligned} A_1 + \cdots + A_k & \propto {\big(\normalsize n + (n-1) + (n-2) + \cdots + (n-k+1)\big)\normalsize}^2 \\ & = {\big(\normalsize\tfrac12n(n+1) - \tfrac12(n-k)(n-k+1)\big)\normalsize}^2 \\ & = {\big(\normalsize \tfrac12(2n-k+1)k\big)\normalsize}^2. \end{aligned}$ Subtracting, we get $\begin{aligned} A_k & \propto {\big(\normalsize \tfrac12(2n-k+1)k\big)\normalsize}^2 - {\big(\normalsize \tfrac12(2n-k+2)(k-1)\big)\normalsize}^2 \\ & = k^3 - 3k^2(n+1) + k(2n^2 + 5n + 3) - (n+1)^2. \end{aligned}$ To see which of the areas $A_k$ is maximal, consider the difference $\begin{aligned} A_k - A_{k-1} & \propto 3k^2 - 3k(2n+3) + (2n^2 +8 n + 7) \\ & = 3\left(n + \frac32 - k\right)^2 - \left(n+ \tfrac12\right)^2 + \tfrac 12. \end{aligned}$ This expression is positive for $1 \leq k < k_0$ and negative for $k_0 < k \leq n$ , where $k_0 = n + \frac 32 - \sqrt{\frac{(n + \tfrac12)^2 - \tfrac12}3}.$ This means that $A_k$ grows as long as $k < k_0$ but shrinks after that. Thus the maximum value of $A_k$ is obtained for the smallest integer $k$ less then $k_0$ , i.e. $A_k\ \text{is maximal}\ \ \ \ \Longrightarrow\ \ \ \ \ k = \left\lfloor n + \frac 32 - \sqrt{\frac{(n + \tfrac12)^2 - \tfrac12}3} \right\rfloor.$

The expression for $k_0$ may be approximated as $k_0 \approx n + \tfrac 32 - \frac{n + \tfrac12}{\sqrt 3} = \left(1 - \frac 1{\sqrt 3}\right) n + \left(\frac 3 2 - \frac 1{2\sqrt 3}\right) \approx 0.42265 n + 1.21132.$ This means that the section with the largest area is that with index $k = \left\lfloor 0.42265 n + 1.21132 \right\rfloor,$

For the situation $n = 3$ in this problem, we obtain $k = 2$ , which means that the second section (the blue section) is largest.

Use Cavalieri's principle to transform the given triangle to a right triangle, and then flip the triangle over to get a rectangle.

From the image below it's clear that the blue area is the greatest.

Let $b_1$ be the base of the red area, $b_2$ the base of the blue area, and $b_3$ the base of the orange area.

By similar triangles, $\frac{b_2}{3 + 2} = \frac{b_1}{3 + 2 + 1}$ , so $b_2 = \frac{5}{6}b_1$ . Also, $\frac{b_3}{3} = \frac{b_1}{3 + 2 + 1}$ , so $b_3 = \frac{1}{2}b_1$ .

The orange area is the orange triangle with an area of $A_o = \frac{1}{2}b_3 \cdot 3$ , and since $b_3 = \frac{1}{2}b_1$ , $A_o = \frac{1}{2}(\frac{1}{2}b_1)3 = \frac{3}{4}b_1$ .

The blue area is the blue trapezoid with an area of $A_b = \frac{1}{2}(b_3 + b_2)2$ , and since $b_2 = \frac{5}{6}b_1$ and $b_3 = \frac{1}{2}b_1$ , $A_b = \frac{1}{2}(\frac{1}{2}b_1 + \frac{5}{6}b_1)2 = \frac{4}{3}b_1$ .

The red area is the red trapezoid with an area of $A_r = \frac{1}{2}(b_2 + b_1)1$ , and since $b_2 = \frac{5}{6}b_1$ , $A_r = \frac{1}{2}(\frac{5}{6}b_1 + b_1)1 = \frac{11}{12}b_1$ .

Since $\frac{4}{3}b_1 > \frac{11}{12}b_1 > \frac{3}{4}b_1$ , the blue area has the largest area.

Trapezoid area is height times average of top side and bottom side.

1x(6+5)/2=5.5 (red),

2x(5+3)/2=8 (winner),

3x(3+0)/2=4.5 (Top triangle)

If we graph the triangle in the Oxy-plane, where the most left corner of the triangle is at the origin $O(0, 0)$ and label the top of the triangle as $T(t, 1+2+3)=(t, 6)$ and label the most right corner of the triangle as $A(a, 0)$ , we will get the following: In this picture, I named the red line $f(x)$ , the purple line $g(x)$ and the green line is just the line $y=0$ . To get to a formula for $f(x)$ , we see the line goes through the origin and the point $T(t, 6)$ , this implies that $f(x)= \frac{\Delta y }{\Delta x} \cdot x = \frac{6-0}{t-0} \cdot x = \frac{6x}{t}$ . With the same strategy in mind, we can find that $g(x)= \frac{-6(x-t)}{a-t} + 6$

Now we want to find the coordinates of the intersections of $f(x)$ and $g(x)$ with the lines $y=1$ and $y=1+2=3$ , so we want to know the values of x that saturate the equations $f(x)=1, g(x)=1, f(x)=3$ and $g(x)=3$ . Isolating x in the 4 equations, will give us $f(\frac{a}{6})=1, g(\frac{t}{6}+\frac{5a}{6})=1, f(\frac{t}{2})=3$ and $g(\frac{t}{2}+\frac{a}{2})=3$ .

With this information it is clear that the line $y=1$ between the two intersections has a length of $\frac{5a}{6}$ and that the line $y=3$ has a length of $\frac{a}{2}$ . This implies firstly that the red area is equal to $\frac{a+\frac{5a}{6}}{2} \cdot 1=\frac{11a}{12}$ , secondly that the blue area is equal to $\frac{\frac{5a}{6}+\frac{a}{2}}{2} \cdot 2 = \frac{8a}{6}$ and at last that the yellow area is equal to $\frac{\frac{a}{2}}{2} \cdot 3 = \frac{3a}{4}$ . Because a is strictly positive, the blue area is the biggest. So, $\boxed{Blue}$ is the right answer.

Let the base of the big triangel be a. Then the area of the big triangel is T=a 6/2=3a. The area of the similar orange triangel is O=(1/2) (a/2) 3=(9/12) a; the orange and the blue together O+B=1/2 (5/6) a 5=(25/12) a hence B=(16/12) a. The red area R=T-(B+O)=3a-(25/12) a=(11/12)*a. B is largest.

draw a perpendicular through the triangle which will be of 6 units. then we will have two pairs of three similar triangles( both will have the height as a side).as they are similar, the ratio of heights and bases will be the same. calculate the areas of each color in the two triangles(we divided it into two with the height) then observe.BLUE HAS MAX AREA.

As you can see in the diagram above, the triangle is separated into 3 sections. Now, if you consider the base of each individual section as the value a then you can easily see, even at first glance, that placing each section through it's corresponding shape's area equation would place the parallelogram with height '1' to be the one with the one having the least area; hence having the colored area which is the smallest. So this eliminates the bottom section.

Now we are left with the two larger sections above. The one with height '2' being a parallelogram as well and the top most section with height '3' being a triangle. Now if you want to give values to the upper and lower horizontal lines for the parallelogram, do that at your own risk because every value is not suitable for every side of every shape. But if you need to do that at any cost, then please figure out the ratios of both these lines before giving ANY values to them.

Now looking at the figure, the triangle being at the top, has the smallest base which can also be imagined as 1/3 a , therefore it is smaller than the parallelogram which can have an imagined base of 1/2 a . But since the parallelogram is a quadrilateral it will always have an area that is greater than a triangle of equal size. So there is really no need to look at the base at all. For example a triangle of size 1x1 or 2x2 would have less area when compared to a quadrilateral of size 1x1x1 or 2x2x2, and since the triangle is already smaller than the parallelogram there is no possible way for it to have a greater colored area than the parallelogram. In either case the middle section has a greater area than the bottom or top section, which comes to the conclusion that it is the largest colored area within the large triangle.

Let $b$ represent the base length of the triangle. Now imagine two points starting at either side of the base and moving inward at different rates so that the height from both points straight up to the sides of the triangle is always the same. These points can move inward until they meet beneath the top vertex, where the maximum height is reached. Call the combined movement inward by both points $x$ , and the height of the triangle's sides directly above both points after this movement $y$ . These are related by a constant ratio since the sides of the triangle have constant slope. When the points meet beneath the top vertex, their combined movement is equal to the base length, $b$ , and the height above them is 6, since this is the height of the vertex above the base according to the diagram. So, the constant ratio $\frac{x}{y}$ is $\frac{b}{6}$ . Thus, $x = \frac{b}{6}y$ . Using this fact, when $y = 1$ (top of the red region), $x = \frac{b}{6}$ , while when $y = 3$ (top of the blue region), $x = \frac{b}{2}$ . The new base length at a given height is equal to $b - x$ , since $x$ is the distance moved inward, so $b_1 = \frac{5}{6}b$ and $b_3 = \frac{b}{2}$ , where $b_1$ and $b_3$ denote the base lengths at $y = 1$ and $y = 3$ , respectively. Using the triangle area formula $A = \frac{1}{2}bh$ , the total area of the triangle ( $A_0$ ) as well as the areas from $y = 1$ up ( $A_1$ ) and from $y = 3$ up ( $A_3$ ) can be calculated:

$A_0 = \frac{1}{2} \cdot 6b = 3b$ $A_1 = \frac{1}{2} \cdot \frac{5}{6} \cdot 5b = \frac{25}{12}b$ $A_3 = \frac{1}{2} \cdot \frac{1}{2} \cdot 3b = \frac{3}{4}b$

The area of the red region is the total area of the triangle ( $A_0$ ) minus the area above the red region $A_1$ . The area of the blue region is the area above the red region minus the area of the orange region, and the area of the orange region is simply $A_3$ . So, the calculations are:

$A_{red} = A_0 - A_1 = 3b - \frac{25}{12}b = \frac{36}{12}b - \frac{25}{12}b = \frac{11}{12}b$ $A_{blue} = A_1 - A_3 = \frac{25}{12}b - \frac{3}{4}b = \frac{25}{12}b - \frac{9}{12}b = \frac{16}{12}b$ $A_{orange} = A_3 = \frac{3}{4}b$

$A_{blue}$ is the largest of these.

First, we can divide the triangle into 6 sections in which in each step to the top, it is 1/6th of lesser than before. (as in each level 1/6 is deducted, we assume that the ratio between each level is consistent)

Using the formula to find the area of a trapezoid A=h(a+b/2), we can try to estimate each section of the triangle.

1. With the first slab being 1(6/6 + 5/6)/2 = 11/12
2. On to the second slab: 2(5/6 + 3/6)/2 = 4/3 (2 and 2 cancel out, leaving 8/6 which is 4/3 )
3. Third slab: (3/6 x 3)/2 = 3/4 (Since this is a triangle, not a trapezoid)

We can see that the second section (the blue one) is bigger.

My logic goes a little something like this:

-The horizontal lines only represent the level of each slab out of 6 total level. In other words the height only. (1/6 is 1 out of 6 levels)

-First, there's a total of 6 levels that's proportionally stacked on top. (in other words, the height is 6)

-It is a triangle that is slightly slanted to the right, otherwise, there is no discernible difference. (I'm sure there are some math and explanation to this in which I will not go into because I'm just a rookie)

-I assumed that at each level, the ratio is 1/6 lesser than the level beneath it.

Knowing that the centre of gravity of a triangle occurs at a third from the base in the vertical direction, and that a third of the total height in this case is 2, using intuition it's pretty safe to assume that it occurs in the blue region which has a height of two and covers the centre of gravity (where the most mass - in this case area - is distributed).

It’s because 2x2>1x3.

I superimposed a portion of the middle segment onto the top segment. You can see that the very top triangle that sticks out will not have enough area to cover the difference of the superimposed portion and the top segment.. For the next comparison the portion of the middle segment that is not covered by the superimposed rectangle is enough to cover not only the difference between the two but also the two tiny traingles that stick out from the bottom.

Total height of the triangle is 6. If the base is called B, we know (can be proven using trigonometry) that the base of the orange triangle is $\frac{B}{2}$ and the base of the triangle combined by the blue and the orange area is $\frac{5B}{6}$ . Thus the red area is $\frac{11B}{12}$ , the blue area is $\frac{16B}{12}$ and the orange area is $\frac{9B}{12}$ .

I just looked at it