BFraction

Relevant wiki: Length and Area Problem Solving

Draw diagonals $AD$ and $BC$ , and let $M$ be its intersection point, and let $N$ be the intersection point of $CP$ , $BQ$ , and $AD$ , as shown below.

Since $CP$ , $BQ$ , and $AM$ are medians of $\triangle ABC$ , $N$ is the centroid, and therefore $MN = \frac{1}{3}AM$ .

This means that $\triangle MNB$ is $\frac{1}{3}$ the area of $\triangle MAB$ , and since $AD$ and $BC$ split the square and the blue portion into $4$ congruent parts, by symmetry the blue portion is $\frac{1}{3}$ the area of square $ABCD$ .

Therefore, the area of the blue portion is $\frac{1}{3}(6 \cdot 6) = \boxed{12}$ .

Relevant wiki: Basic Composite Figures

Let's place the 6 by 6 square on a coordinate grid with $C$ at $(0,0) .$

Notice that the slopes of the line segments connecting midpoints are $\frac{6}{3} ,$ $\frac{3}{6} ,$ $--\frac{6}{3} ,$ and $-\frac{3}{6}$ so they reduce to $\frac{2}{1} ,$ $\frac{1}{2} ,$ $-\frac{2}{1} ,$ and $-\frac{1}{2}$ respectively. This means we know they pass through the marked red points on the grid.

We can now use the formula $\frac{1}{2}bh$ for the area of a triangle to find the area of four of the triangles:

Since the area of the square is $6 \times 6 = 36 ,$ subtracting the triangle areas gets us the central shape area of $36 - 9 - 9 - 3 - 3 = 12$ square units.

Relevant wiki: Basic Composite Figures

Blue region consists of two identical triangles. $\triangle BCF$ has a base $BC=6\sqrt2$ .

From symmetry, line $ED$ is perpendicular to it, so the height of the triangle is $EF$ .

Since $ED, BR,$ and $CS$ are medians, their intersection $F$ splits each of them in $2:1$ ratio and $EF=\frac13\cdot ED=\frac16\cdot AD=\frac16\cdot6\sqrt2=\sqrt2$

The blue area is 2 times the area of triangle $BCF$ , that is $2\cdot\frac12\cdot6\sqrt2\cdot\sqrt2=\boxed{12}$ .

By observation, we can see that the 2 brown triangles make up half the area. Meaning that the parrellogram ADEH is half the area of the original square.

Parallel to DE, construct a line through point C. The intersection of this parallel line and EH is point F. Parallel to DE, construct a line through point G. The intersection of this parallel line and AD is point B. Parallel to CE, construct a line through point B. We see that this parallel line through B intersects EH at point F.

This implies that $AB = BC = CD$ and $HG = GF = FE$

The parallelogram ADEH is now divided into 6 congruent triangles, 4 of which make up the required area.

Hence, area shaded blue $\frac{1}{2}$ * $6^2$ * $\frac{4}{6}$ = 12

Let's draw the vectors $\textbf{u}$ and $\textbf{v}$ . It is clear that these vectors are respectively proportionnal to (2,1) and (1,2), hence $\textbf{u} = c \cdot (2,1)$ and $\textbf{v} = c \cdot (1,2)$ .

We can also see that $\textbf{u} + \textbf{v} = (6,6)$ , hence $c = 2$ .

We eventually get that the area ( $A$ ) is the norm of the cross product of these two vectors in a 3D space, we thus get:

$A = || (4,2,0) \times (2,4,0) || = 12$

Pick's theorem: 9+ $\frac{8}{2}$ -1 = 12

We can easily see, that the blue area is exactly one third of the square.

By shifting the kite shape from the upper left corner to the center and by rotating the two congruent triangles by $\pi$ , we find, that they exactly fill up the blue area. Because of the symmetry, the upper left white, the blue and the lower right white areas thus are the same.

The kite shape and the two triangles exactly fill up the blue area.

Using intuition, (6×6)÷3 = 12

When plotted on a graph, the problem shown has axes of symmetry along $y=x$ and $y=-x+6$ . When cut, it looks like this: We can use one of these cut blue sections to find a partial answer using the formula for the area of a triangle $A=\frac{1}{2}bh$ , then multiplying by 4 to get our final answer. We'll be using the bottom-most section for this solution. The lines that intersect the width of the blue shape can be expressed as $y=\frac{1}{2}x$ and $y=2x-6$ . We can find the point of intersection using simple algebra: $\frac{1}{2}x = 2x-6$ $x = 4x-12$ $3x=12$ $x=4$ then substituting for $x$ : $y=\frac{1}{2}(4)$ $y=2$ By intuition, the point of intersection of the axes of symmetry can be found to be at the center of the square, at $(3,3)$ . Now that we know what the first point of intersection is, we can use the $y$ values of these points to find the area of the cut blue section: $A=\frac{1}{2}(6)(3) - \frac{1}{2}(6)(2)$ $=\frac{1}{2}(6)(3-2)$ $=\frac{1}{2}(6)$ $=3$ Final answer: $3*4=\boxed{12}$

ABDC is a rectangle. PC bisects AB and QB bisects AC.

Draw perpendiculars from N to AB and from N to AC. Label their intersections with AB as T and with AC as U.

Triangles ABQ and TBN are similar (right) triangles, because they share a common angle at B.

Therefore, length(TB) / length(TN) = length(AB) / length(AQ) = 2.

Triangles APC and TPN are also similar (right) triangles, because they share a common angle at P.

Therefore, length(TN) / length(TP) = 2.

Multiplying these ratios, length(TB) / length(TP) = 4.

Therefore, length(TP) + length(PB) = 4 length(TP), so length(PB) = 3 length(TP).

Because length(AB) = 2 length(PB), length(AB) = 6 length(TP).

Because length(AB) = 2 length(AP), length(AB) = 3 length(AT).

By the same reasoning, length(AC) = 3 length(AU).

Therefore, using formulas for areas of squares and right triangles, area(ABDC) = 9 area(ATNU) = 9 area(TBN) = 9 area(UCN).

Adding the areas of non-overlapping figures, area(ABDC) = 3 area(ABNC).

By the same reasoning, area(ABDC) = 3 area(CDBM).

Thus, because the blue region in the problem does not overlap the two white regions, area(ABDC) = 3 area(CNBM).

In this case, area(CBNM) = 1/3 of (6 x 6), or 12.

This solution is based only on proving triangles with equal angles are similar and on the equality of ratios of lengths of sides of similar triangles. It does not depend on rectangle(ABDC) being a square. In fact, this solution generalizes to any parallelogram, and possibly to every convex quadrilateral, by constructing the line segments NT and NU parallel to the sides of the parallelogram or quadrilateral and adding the formula for the area of a parallelogram.

The area of $\triangle ACI$ is equal to $\triangle ACD - (\triangle CEI + \triangle ADE )$ .

Since the area of $\triangle ACD = 18$ and area of $\triangle ADE = 9$ , we need to find the area of $\triangle CEI$

The $\angle DCG = \arctan(3/6) = \alpha$ .

As $\triangle CDG$ is rectangle, we know that $\angle DGC = \pi/2 - \arctan(3/6) = \pi/2 -\alpha$ . Then $\angle CEI = \pi - \angle DGC = \pi/2 + \alpha$

From sinus law in triangle we can deduce: $S = \frac{|EC|^2 \sin(\alpha)\sin(\pi/2+\alpha)}{2\sin(\pi/2+2\alpha)} = \frac{3.6}{1.2} = 3$

Then we conclude than the area of the blue portion is: $2 (18 - 9 - 3) = 12$

If a is the area of a triangle then we can write this square as 12 triangle .then 12a=36 implies a=3.Now the blue region can be wrriten as 4 triangle then the area of region is 4*a=12

$\frac{L-x}{x} = 2 \Leftrightarrow x=\frac{L}{3}$

Where L is the length side. The blue area is then:

$2(A+B)=2(\frac{L}{2} \times (\frac{x}{2} + (\frac{L}{2}-x)))=L \times \frac{L - x}{2}$

Substituting $x$ :

$\text{Area} =L \times \frac{L - L/3}{2} = \frac{L^2}{3}$

For a length side $L=6$ :

$\text{Area} = \frac{6^2}{3} = 12$

None except one of the answer is a multiple of 36. The blue are looks about a third of the square, so I guessed 12.

I imagined moving N to A and M to D, thus the blue area would be 100% of the square.

12 is the only divisor of 36 in the possível solutions list.

Area of ΔADF is 9. ΔAGJ and ΔGDJ have the same area because their bases AG and GD are congruent and they have the same height (segment JK). ΔGDJ ≅ ΔFDJ so ΔAGJ, ΔGDJ, and ΔFDJ all have the same area and their total area is 9 (ΔADF). Therefore each small triangle has an area of 3, and there are 8 small triangles (ΔAGJ, ΔGDJ, ΔFDJ, ΔCFJ, ΔCEI, ΔEBI, ΔHBI, and ΔAHI) for a total area of 24. The complete square has an area of 36.

36 - 24 = 12

In general, the area of shaded portion i.e. rhombus

$=\frac13(\text{area of square)}=\frac13(6^2)=12$

with the help of autocad drawing i could easily solve this problem. the ans is 12

Let A, B , C and D be the four corners of the square,

W, X, Y and Z be the centre of two corners (W = centre of AB, X = centre of BC, and so on),

And P and Q be the two corners of the blue region that is not touching the square. (P nearer to W and Z and Q nearer to X and Y)

The area of the blue region is four times as big as a triangle formed by - a corner of the square (e.g. A) - a point in the centre of two corners (e.g. W) - and the nearest corner of the blue region to them (e.g. P)

Parallelogram BYCZ is six times th triangle mentioned up there, and its area is equal to 3 × 6 which is 18.

Thus, the area of the blue region = 18 ÷ 6 × 4 = 12

6×6=36. A square has 4 sides the middle of the square would have to be an even number divisible by 6

No symbols solution: You can fit the top triangle, left triangle and bottom right quadrilateral in the central piece just with translation. That means that one white region has the same area as the blue region. Hence the blue region is a third of the whole area. 36/3=12

Let's start by drawing the diagonal $AJ$

Since the equation of $DC$ is $y = \frac{x}{2} + \frac{1}{2}$ and the equation of $GB$ is $y = 2x$

Set them equal to each other to find that they intersect at the value of $\frac{1}{3}$

From this we can see that $AE = EI = IJ$

Break the figure into 6 triangles $\bigtriangleup AEG$ $\bigtriangleup EGI$ $\bigtriangleup GIJ$ $\bigtriangleup ACE$ $\bigtriangleup CEI$ $\bigtriangleup CIJ$ , since they all have the same base length and height their areas must be equal. Since this figure is made up of 6 of these triangles one of these triangles is $\frac{1}{6}$ the area of the total square, since we have two triangles we have $\frac{2}{6}$ of the area of the total square.

In this case the area is $36$ so we multiply that by $\frac{2}{6}$ and get the answer $\boxed{12}$