Inside perimeters

The perimeter of the large, outer triangle (see the figure on the right) is composed of 3 three red lines, three green lines, three orange lines and three violet lines. The total perimeter of the three smaller rectangles (see the figure on the left;exclude the blue rectangle) is 12+6+6+8=32 and it is composed of four red lines, four green lines, four orange lines and four violet lines. So to get the perimeter of the biggest rectangle we need to subtract one red line, one green line, one orange line and one violet line. And that is exactly the perimeter of the blue rectangle (see left figure). Therefore, the desired answer is 32-4=28.

Notice that we can identify each green/red/orange/purple side with a black side, so the perimeters of the green/red/orange/purple rectangles cover the perimeter of the black rectangle. However, there are 4 colored sides left over - but they each can be identified with a side of the blue rectangle. So the black rectangle's perimeter is: $(6 + 12 + 6 + 8) - 4 = \boxed{28}$

Let the side lengths be assigned as $a$ , $b$ , $c$ , $d$ , $e$ and $f$ as shown in the figure. Then the parameter of the the larger, outer rectangle is given by $p = 2(a+b+c+d+e+f)$ and we have:

$\begin{cases} 2a+2e = 12 & ...(1) \\ 2b+2d = 6 & ...(2) \\ 2b+2e = 4 & ...(3) \\ 2b+2f = 8 & ...(4) \\ 2c+2e = 6 & ...(5) \end{cases}$

Adding $(1)+(2)+(3)+(4)+(5)$ , we have:

$\begin{aligned} 2a+6b+2c+2d+6e+2f & = 12+6+4+8+6 = 36 \\ 2a+2b+2c+2d+2e+2f + \color{#3D99F6} 4b+4e & - 36 \\ 2(a+b+c+d+e+f) & = 36 - 2\color{#3D99F6}(2b+2e) & \small \color{#3D99F6} (3): \ 2b+2e = 4 \\ p & = 36 - 2{\color{#3D99F6}(4)} = \boxed{28} \end{aligned}$

Dumbest, yet the fastest method: This response is based on an assumption, which is that the rectangle with a perimeter of 4 has side lengths of 1, therefore making it a square. Hence, you can easily find the side length and width of the larger rectangle being 8 and 6 respectively. Add them up and multiply by 2, and you'll get 28.

Two side lengths of a rectangle add to half the perimeter of that rectangle, so:

$B+D=6\div2=3$

$A+E=12\div2=6$

$C+E=6\div2=3$

$B+F=8\div2=4$

Add these up to get $A+2B+C+D+2E+F=16$ .

Also, note that the two blue rectangle side lengths are $B$ and $E$ .

$B+E=4\div2=2$

Subtract this new equation from the last to get:

$A+B+C+D+E+F=14$

The total perimeter is $2\times14=\boxed{28}.$

So we are looking for $2(x + y + z) + 2(a + b + c)$ which is also equivalent to $2a + 2b + 2c + 2x + 2y + 2z$

$2y + 2a = 12$

$2x + 2b = 6$

$2y + 2b = 4 \Rightarrow 4y + 4b = 8$

$2z + 2b = 8$

$2c + 2y = 6$

Adding all the equations, we get:

$2a + 6b + 2c + 2x +6y + 2z = 36$

And we are looking for $2a + 2b + 2c + 2x + 2y + 2z$ , so we subtract $4y + 4b$ from $2a + 6b + 2c + 2x +6y + 2z$ :

$2a + 2b + 2c + 2x + 2y + 2z = (2a + 6b + 2c + 2x +6y + 2z) - (4y + 4b)$

$2a + 2b + 2c + 2x + 2y + 2z = 36 - 8$

$2a + 2b + 2c + 2x + 2y + 2z = \boxed{28}$

I start with the inner rectangle. I can see it has, as one solution, equal sides of 1 unit each. I can then determine the remaining lengths:

(6-1-1)/2=2

(8-1-1)/2=3

(12-1-1)/2=5

I can now calculate the larger perimeter:

2((5+1+2) + (2+1+3))=28

This result can be checked by assigning different values to the original rectangle.

Lol, I didn't think of rearranging the walls and I just proved it mathematicly XD x+y= $\frac{1}{3}$ (x+a)= $\frac{2}{3}$ (x+b) Therefore: E = 3(x+y)+x+y+ $\frac{3}{2}$ (x+y)= ... = $\frac{15}{16}$ x+ $\frac{11}{2}$ b AND x+y= $\frac{2}{3}$ (y+c)= $\frac{1}{2}$ (y+d) Therefore: F = $\frac{3}{2}$ (x+y)+x+y+2(x+y)= ... = $\frac{9}{2}$ + $\frac{3}{2}$ b Perimeter = 2E+2F=5x+11y+9x+3y=14x+14y If 4=2x+2y then 2=x+y and y=2-x Perimeter=14x+14(2-x)=14x-14x+28= 28

Perimeter of Rectangle = 2(l+b)

It could be several different answers. The perimeters are given but it doesn't say the sides must be whole. What if the middle rectangle is .5x1.5? The answer could be 30. If the drawing is in perfect scale, there is no way all of the sides are whole numbers.

The heights for all the middle "rectangles" are 0. Those are lines. The scaling for this diagram is atrocious.