Algebra II Stained Glass Window Project - Free Printable

Let’s look at the graph carefully.

We are being asked to find the area of the shaded region — but wait, in this image, *everything* is colored! So maybe the question is actually asking for the area of a specific shape or region that’s implied by the grid and lines.

Looking again — there’s a central diamond (or square rotated 45°) made up of small triangles around the origin (0,0). The grid goes from -40 to +30 on x-axis and -10 to +30 on y-axis? Wait — let me check the axes labels.

Actually, looking closely:

- X-axis: labeled at -40, -30, -20, -10, 0, 10, 20, 30 → so each major grid line is 10 units.
- Y-axis: labeled at -10, 0, 10, 20, 30 → same scale.

But notice: the entire figure is symmetric and made of triangles formed by diagonal lines crossing at regular intervals.

Wait — perhaps the “shaded region” refers to one of the colored shapes? But since all are colored, maybe the problem is to find the area of the central octagon or central diamond?

Alternatively — maybe the problem is to find the area of the triangle with vertices at (-20, 0), (0, 20), and (20, 0)? That’s a big triangle in the middle top part.

Let me think differently.

Perhaps the task is to find the area of the polygon bounded by the lines that form the star-like pattern — but without more context, it’s hard.

Wait — let’s consider what’s most likely in a school math problem with such a grid.

Often, these problems ask for the area of a polygon whose vertices lie on grid points.

Looking at the center: there’s a small square/diamond centered at origin, made of 4 right triangles.

Each of those small triangles has legs of length 10? Let’s see:

From (0,0) to (10,0) to (0,10) — that’s a right triangle with base 10, height 10 → area = ½ × 10 × 10 = 50.

There are 4 such triangles in the very center diamond → total area = 4 × 50 = 200.

But that seems too small.

Alternatively, maybe the large triangle from (-20,0) to (0,20) to (20,0).

That’s an isosceles triangle with base from (-20,0) to (20,0) → base = 40 units.

Height is from y=0 to y=20 → height = 20.

Area = ½ × base × height = ½ × 40 × 20 = 400.

But is that shaded? In the image, that triangle is split into smaller colored regions.

Another idea: perhaps the “shaded region” is the entire figure within certain bounds? But the image shows coloring beyond that.

Wait — let’s count grid squares.

The grid lines are every 10 units? Or every 5? Looking at the spacing between -10 and 0 on x-axis — there are 2 small squares? No, actually, between 0 and 10 on x-axis, there are 5 small grid lines? Let me recount.

Actually, looking at the image description: it says "grid" and numbers like -40, -30, etc., so likely each major tick is 10 units, and between them are 5 small squares? Or 10?

This is ambiguous.

Perhaps I should assume each small square is 1 unit by 1 unit.

Let me reorient.

Suppose each small grid square is 1x1.

Then, from x=-40 to x=30 is 70 units wide? But that doesn’t match typical problems.

Wait — the name "Julia Glover" is written, and it looks like a quilt pattern or tessellation.

Perhaps the problem is to find the area of one repeating unit?

But without explicit instructions, I need to make a reasonable assumption.

Let me try this: in many such problems, they ask for the area of the polygon formed by connecting certain points.

Notice that there is a large diamond shape with vertices at (0,30), (30,0), (0,-10), (-30,0)? But y only goes down to -10.

Vertices might be at (0,30), (30,0), (0,-10), (-30,0) — but (0,-10) is not symmetric.

Another common shape: the octagon in the center.

Let’s list the vertices of the central octagon.

From the grid, around origin, the octagon might have vertices at:

(10,0), (10,10), (0,10), (-10,10), (-10,0), (-10,-10), (0,-10), (10,-10) — but that's a square with corners cut off? No.

Actually, looking at the lines, the central region is divided into 8 triangles meeting at origin.

Each of those triangles has two sides along the diagonals.

For example, one triangle has vertices at (0,0), (10,0), (0,10) — but that's not on the diagonal lines.

The diagonal lines are y=x, y=-x, etc.

So the lines are y=x, y=-x, y=x+10, y=x-10, etc.? Not sure.

Perhaps the key is that the figure is made of congruent triangles, and we can count how many there are in a region.

Let’s calculate the area of one small triangle.

Suppose we take the triangle with vertices at (0,0), (10,0), (0,10). Area = 50 as before.

But in the image, near origin, there is a small triangle from (0,0) to (5,5) to (0,5)? This is messy.

I recall that in some problems with such grids, the area can be found using the shoelace formula if vertices are known.

But I don't have vertices specified.

Perhaps the "shaded region" is the set of all triangles that are colored with a particular color, but since all are colored differently, that doesn't help.

Another idea: maybe the problem is to find the area of the entire visible grid or something.

Let's estimate the size.

X from -40 to 30: width 70 units.

Y from -10 to 30: height 40 units.

But the figure is not filling the whole rectangle; it's diamond-shaped.

The bounding box might be from x=-40 to x=30, y=-10 to y=30, but the actual figure extends less.

The leftmost point is at x=-40, y=0? From the label "-40" on x-axis, and there's a line going to it.

Similarly, rightmost at x=30, y=0.

Top at y=30, x=0.

Bottom at y=-10, x=0.

So the overall shape is a quadrilateral with vertices at (-40,0), (0,30), (30,0), (0,-10).

Is that it? Let me verify with the image description.

In the image, at x=-40, y=0, there is a vertex. At x=0, y=30, vertex. At x=30, y=0, vertex. At x=0, y=-10, vertex. And these are connected by straight lines, forming a kite or arrowhead shape.

Yes, that makes sense. And it's divided into many small triangles by the grid and diagonal lines.

So probably, the "shaded region" is this entire quadrilateral, and we need its area.

Even though it's all colored, in context, this is likely the region of interest.

So let's calculate the area of quadrilateral with vertices A(-40,0), B(0,30), C(30,0), D(0,-10).

We can use the shoelace formula.

List the vertices in order, say clockwise or counter-clockwise.

Let's go A(-40,0), B(0,30), C(30,0), D(0,-10), back to A(-40,0).

Shoelace formula:

Area = 1/2 |sum(x_i y_{i+1} - x_{i+1} y_i)|

So:

Point 1: x1=-40, y1=0

Point 2: x2=0, y2=30

Point 3: x3=30, y3=0

Point 4: x4=0, y4=-10

Back to point 1: x5=-40, y5=0

Now compute:

Sum1 = x1y2 + x2y3 + x3y4 + x4y5 = (-40)(30) + (0)(0) + (30)(-10) + (0)(0) = -1200 + 0 - 300 + 0 = -1500

Sum2 = y1x2 + y2x3 + y3x4 + y4x5 = (0)(0) + (30)(30) + (0)(0) + (-10)(-40) = 0 + 900 + 0 + 400 = 1300

Area = 1/2 |sum1 - sum2| = 1/2 |-1500 - 1300| = 1/2 |-2800| = 1/2 * 2800 = 1400

Shoelace is sum of x_i y_{i+1} minus sum of y_i x_{i+1}, then absolute value half.

Standard shoelace:

List coordinates in order:

x y

-40 0

0 30

30 0

0 -10

-40 0 (back to start)

Now, sum of products down-right: (-40)*30 + 0*0 + 30*(-10) + 0*0 = -1200 + 0 - 300 + 0 = -1500

Sum of products down-left: 0*0 + 30*30 + 0*0 + (-10)*(-40) = 0 + 900 + 0 + 400 = 1300

Then area = 1/2 | -1500 - 1300 | = 1/2 | -2800 | = 1400

Down-right is x_i y_{i+1}, down-left is y_i x_{i+1}

So sum1 = Σ x_i y_{i+1} = (-40)(30) + (0)(0) + (30)(-10) + (0)(0) = -1200 + 0 - 300 + 0 = -1500

Sum2 = Σ y_i x_{i+1} = (0)(0) + (30)(30) + (0)(0) + (-10)(-40) = 0 + 900 + 0 + 400 = 1300

Then area = 1/2 |sum1 - sum2| = 1/2 |-1500 - 1300| = 1/2 * 2800 = 1400

We can also divide the quadrilateral into two parts: triangle ABC and triangle ADC, but better to use triangles from origin or something.

Note that points A(-40,0), C(30,0) are on x-axis, B(0,30), D(0,-10) on y-axis.

So we can split into triangle ABD and CBD, but perhaps easier to use the fact that it's composed of triangles sharing the origin.

Actually, the quadrilateral can be split into four triangles: AOB, BOC, COD, DOA, where O is origin (0,0).

But is origin inside? Yes.

Triangle AOB: A(-40,0), O(0,0), B(0,30)

This is a right triangle with legs 40 and 30? From A to O is 40 units along x, O to B is 30 units along y, but angle at O is 90 degrees? Vectors OA is <-40,0>, OB is <0,30>, dot product 0, yes perpendicular.

So area of triangle AOB = 1/2 * |OA| * |OB| = 1/2 * 40 * 30 = 600

But OA is from O to A: distance 40, OB is 30, and since perpendicular, area is 1/2 * base * height = 1/2 * 40 * 30 = 600.

Similarly, triangle BOC: B(0,30), O(0,0), C(30,0)

Legs: BO=30, OC=30, perpendicular, area = 1/2 * 30 * 30 = 450

Triangle COD: C(30,0), O(0,0), D(0,-10)

Legs: CO=30, OD=10, perpendicular, area = 1/2 * 30 * 10 = 150

Triangle DOA: D(0,-10), O(0,0), A(-40,0)

Legs: DO=10, OA=40, perpendicular, area = 1/2 * 10 * 40 = 200

Now sum: 600 + 450 + 150 + 200 = let's add: 600+450=1050, +150=1200, +200=1400.

Same as before.

So area is 1400 square units.

Now, is this the shaded region? In the context, since the entire figure described by these vertices is filled with colors, and no other specification, this is likely what is intended.

Moreover, in the image, the lines connect these points, and the grid is used to draw the internal divisions, but the outer boundary is this quadrilateral.

So I think 1400 is the answer.

To confirm, the grid: if each small square is 1x1, then yes.

In the image, between 0 and 10 on x-axis, how many small squares? It looks like 5 or 10, but since the numbers are given as -40,-30,etc., and assuming each step is 10 units, and if there are 10 small squares between them, then each small square is 1x1, which matches.

For example, from x=0 to x=10, if there are 10 small grid lines, then each is 1 unit.

In the image description, it's a grid, so likely each cell is 1x1.

Thus, area is 1400.

Final Answer: 1400