Shuriken

Let $A_B$ be the area of the blue region and $A_Y$ be the area of the yellow region. Then,

$A_B=\dfrac{1}{2}(4)(1)(4)=\dfrac{16}{2}=$ $\boxed{8}$

$A_Y=(4)(4)-8=16-8=$ $\boxed{8}$

$\color{#D61F06}\boxed{\large Conclusion:They~have~equal~areas.}$

We can split the Shuriken into 4 small equal regions as shown on the figure on the right, above.

So, by looking at the right region, we can just determine whether which of the areas (blue versus yellow) has a larger region.

The area of the blue region is $\dfrac12 xy$ .
The area of the blue region and the yellow region is $\dfrac12 x(2y) = xy$ .
So the area of the yellow region is just the difference of the areas we found above, $xy - \dfrac12 xy =\dfrac12 xy$ .

Since the area of the blue region and the yellow region are equal, then they both covered the same amount of area.

Relevant wiki: Area of Triangles - Problem Solving - Easy

Let us denote the square and the Shuriken vertices with alphabets over the $4\times 4$ squares as shown below:

Considering the blue region, clearly there are four congruent triangles: $ABF$ , $BGC$ , $CHD$ , & $DEA$ .

And each has an area of $\dfrac{1}{2}\times 4\times 1 = 2$ .

Thus, the whole blue area = $4\times 2\ = 8$ . Then the yellow area = $16 - 8 = 8$ .

As a result, both colors have the same area.

Alternatively, if we designate the new points $I$ & $J$ over the grid as shown below, considering triangle $AIF$ with the base $IF$ , its height is $1$ unit, so it will have the same area as the triangle $JIF$ , which also has the same base $IF$ and the height of $1$ .

Repeating the same method for other congruent triangles, we will reach the final picture of the new yellow square, which has half area of the big square because its area = $\dfrac{1}{2}\times 4\times 4 = 8$ .

The area of each of the purple triangle can be easily calculated... 4*1/2=2

Therefore, the total area of the purple region is 8.

Since the purple area is half the area of the square, yellow=purple.

Actually, the blue region is made up of 8 right triangles. Each right triangle has an area of 1 unit, so the area of the blue region is 8 units. And the area of the square is 16 units. By subtracting the area of blue region from the area of the square we will get the area of yellow region. And they are indeed equal!

We can use Pick's Theorem. The number of boundary points (8) of the star and the number of interior points (5) is used for Area = B/2 + I - 1.

(8/2) + 5 - 1 = 8;

8 is half of 16, which shows that the two areas are equivalent.

fluke ha ha ha......!!!!!!

The area of the square is 4x4 = 16

The area of one blue triangle is

A = bh/2

A = (1)(4)/2 = 2

The total area of 4 blue triangles is 2(4) = 8

Knowing the area of the square is 16 and and the blue region is 8, it means that blue and yellow region are equal.

Divide the square into 8 triangles by cutting diagonal and orthogonal diameters. Each component triangle will consist of two triangles (one blue, one yellow) with L = 1 and H = 2. Therefore each will have an area of $\frac{L*H}{2}$ = $\frac{1*2}{2}$ = 1.

The area of the entire square is $4 \times 4 = 16$ . Since the dots are all equally spaced, we can find the area of the blue region by multiplying the area of each isosceles triangle (blue) by 4. This gives $4 \times 4 \times 1 \times \dfrac{1}{2} = 8$ . The area of the Shuriken is now $16 - 8 = 8$ . Hence, the colored regions have equal area.

Since the dots are equally spaced by one unit each, we can use isosceles triangles for the first part of this problem. When connected, the dots on the concave edges of the shuriken (throwing star) form a square of side length $sqrt(2)$ because they are the hypotenuse of the four isosceles triangles that can be formed in the square of leg length $1$ .

This means that the area of this center square is $sqrt2*sqrt2$ , which is equal to an area of $2$ . Now, since we have that the side lengths of the square are $sqrt(2)$ , this means that the four congruent triangles which make up the rest of the shuriken (throwing star) have base length $sqrt2$ . Their height is therefore $sqrt2+sqrt2/2=3sqrt2/2$ . To find the area of one of these triangles, we simply use the following formula: $Bh/2$ . $(sqrt2*3sqrt2/2)/2=(3*2/2)/2=(6/2)/2=3/2$ .

Now we have that the area of one of the triangles equals $3/2$ . Since there are four triangles, we multiply $(3/2)*4=12/2=6$ . Adding this to the area of our previous center square gives us $6+2=8$ . So the area of the shuriken (throwing star) is $8$ .

Since the larger square, the $4*4$ grid, has an area of $16$ , we can subtract $8$ from $16$ which gives us that the area of both the shuriken (throwing star) and the blue area of the $4*4$ grid equals $8$ . Therefore, both of the colored regions (yellow and blue) have the same area.