Three triangles and a figure!

We know that $\frac{a}{b}=\frac{c}{d}=\frac{e}{f}$ , $ab=6$ , $cd=4$ and $ef=2$

$\Rightarrow \frac{a}{\frac{6}{a}}=\frac{c}{\frac{4}{c}}=\frac{e}{\frac{2}{e}}$

$\Rightarrow a^{2}=\frac{3c^{2}}{2}\Rightarrow a=\frac{\sqrt{6}c}{2}$

$\Rightarrow e^{2}=\frac{c^{2}}{2}\Rightarrow e=\frac{\sqrt{2}c}{2}$

Similarly, we can derive $b=\frac{\sqrt{6}d}{2}$ and $f=\frac{\sqrt{2}d}{2}$ .

Then, $P_{blue figure}=P_{large triangle}-3-2-1$

$P_{blue figure}=\frac{(a+c+e)(b+d+f)}{2}-6$

$P_{blue figure}=\frac{\sqrt{2}c\times (\sqrt{3}+1+\sqrt{2})\times \sqrt{2}d\times (\sqrt{3}+1+\sqrt{2})}{8}-6=\frac{2cd\times (\sqrt{3}+1+\sqrt{2})^{2}}{8}-6$ , which simplifies to

$P_{blue figure}=(\sqrt{3}+1+\sqrt{2})^{2}-6\approx 11.19$ .

The interesting thing is that the area is found without ever determining the dimensions of any triangle.

Since the triangle is 2-dimensional, the ratio of the area is equal to the square of the sides, therefore, the base of the triangle of area 2 is equal to the base of the triangle of area 3 multiplied by $\sqrt{\frac{2}{3}}$ . Let's substitute the base of the triangle of area 3 as $l$ . Now, we can write the base of the triangle of area 2 as $l \cdot \sqrt{\frac{2}{3}}$ . Now, doing the same to solve for the base of the triangle with area 1, we get $l \cdot \sqrt{\frac{2}{3}} \cdot \sqrt{\frac{1}{2}}=l \cdot \sqrt{\frac{1}{3}}$ Adding these and then finding the ratio in relation to $l$ will allow us to find the common ratio between the triangle of area 3 and the whole triangle (since they're similar). $l + (l \cdot \sqrt{\frac{2}{3}}) + (l \cdot \sqrt{\frac{1}{3}}) \\ l \cdot (1+\sqrt{\frac{2}{3}} + \sqrt{\frac{1}{3}})$ From this, we learn that the ratio is $1+\sqrt{\frac{2}{3}}+\sqrt{\frac{1}{3}}$ . Now, to find the area of the whole triangle, we must square this and then multiple it to the original area, $3$ , since area is 2-dimensional. $(1+\sqrt{\frac{2}{3}}+\sqrt{\frac{1}{3}})^2 \approx 5.7305 \\ 3 \cdot 5.7305 \approx 17.19$ Now, we simply subtract the area of the smaller triangles $17.19-1-2-3=\boxed{11.19}$

1) Model the large triangle as an isosceles right angled triangle and mirror along the hypotenuse to create a square.

The areas of the three smaller triangles double to give the areas of three small squares, 6,4 and 2 and the blue area is doubled also.

3) From this, you can calculate the length of of sides for each of the smaller squares, root(6), 2 and root(2) respectively.

4) From this we know the length of the sides of the large square created in step one has length root(6)+2+root(2)= 5.863703305

5) Square this to find the area of the large square, = 34.38301615

6) Subtract the areas of the smaller squares to get 34.38...-(6+4+2)=22.38... i.e. to large blue area

7) Remember the blue area was double from the original triangle, so half it to get 11.19 to 2 d.p.

To solve this problem, I used numerical values for the side lengths of the smallest triangle, and found the side lengths of the other two triangles using the theorem which states that the ratio between the areas of two similar polygons is the square of the ratio between the side lengths.

First, I said that the bottom of the triangle with area 1 was of length 2 and the other leg was of length 1. If so, then the corresponding sides of the triangle with area 2 have lengths 2 $\sqrt{2}$ and $\sqrt{2}$ , respectively, and the corresponding sides of the triangle with area 3 have lengths 2 $\sqrt{3}$ and $\sqrt{3}$ , respectively.

Then, I split up the blue area into two rectangles with dimensions of 2 by $\sqrt{2}$ and $\sqrt{3}$ by 2 $\sqrt{2}$ + 2 . Multiplying and adding yields

2 $\sqrt{2}$ + 2 $\sqrt{6}$ + 2 $\sqrt{3}$

Plug it into a calculator, and you get a decimal starting with 11.19

Second triangle is sqrt(2) longer than the first triangle, third triangle is sqrt(3) longer than the first triangle.

Take the length of the master triangle to be the sum of all the little triangles, square that value to get the total area.

Then just subtract 6. (The area of the client triangles)

Let $x =$ hypotenuse of the triangle of area 1. The large triangle has a hypotenuse of $x +\sqrt 2 x+\sqrt 3 x$ , or $4.146 x$ . Square that and subtract 6 for the white triangles.

Make triangle into square by adding one more triangle at the top. You will get the squares each of areas 6,4 and 2. So the sides will be sqr(6),2 and sqr(2). The answer will be 2(sqr(6)+sqr(3)+sqr(2))=11.19

My solution is based on the four obviously similar triangles, and the fact that the areas of similar figures scale in proportion to their lengths squared.

Suppose the small triangle with area 1 has base of length $x$ .

Then the triangles with areas 2 and 3 have bases of length $\sqrt{2}x$ and $\sqrt{3}x$

So the big uncut triangle has base $(1+\sqrt{2}+\sqrt{3})x$

Comparing with the triangle of side $x$ and area 1, we see that the area of this big triangle is $(1+\sqrt{2}+\sqrt{3})^2=17.19$

Now take away the area of the three cut outs to get the answer $17.19-6=\boxed{11.19}$

Area of largest triangle is .5 b h = 3 so bh = 6. Area of next triangle is .5kb kh = 2 so bh = 4/k^2 =>k^2 = 2/3 => k = sqrt(2/3). Area of smallest triangle is .5mb mh = 1 so bh = 1/m^2 =>m^2 = 1/3 =>m = sqrt(1/3).

As a result of this the area of the triangle is =.5 (1+k+m)b (1+k+m)h = .5bh(1+k+m)^2 = 3(1+k+m)^2 = 17.19.

To get the blue area we just compute 17.19 - (1+2+3) = 11.19

sqr(1+sqrt(2)+sqrt(3)) - (1+2+3) = 11.189; triangle area is proportional to square of hypotenuse length, so the large triangle has hypotenuse length as the first term before square