Math worksheet for calculating the area of irregular shapes with a scoring guide.

To find the area of each irregular shape, we break them down into simpler shapes like rectangles and triangles. Then we calculate the area of each part and add them together.

Here is the step-by-step solution for each problem:

1. Shape 1:
* This shape looks like a rectangle with a triangle on top.
* Rectangle part: The width is 4 cm and the height is 3 cm. Area = $4 \times 3 = 12$ sq cm.
* Triangle part: The base is the same as the rectangle's width (4 cm). The total height is 5 cm, so the triangle's height is $5 - 3 = 2$ cm. Area = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 2 = 4$ sq cm.
* Total Area: $12 + 4 = 16$ sq cm.

2. Shape 2:
* This shape can be split into a large rectangle and two small triangles on the sides, or viewed as a central rectangle plus side triangles. Let's look at the dimensions.
* It looks like a central rectangle of $4 \text{ cm} \times 6 \text{ cm}$? No, let's look closer. The vertical side is 6 cm. The horizontal protrusion is 2 cm. The top/bottom flat parts are 4 cm.
* Let's split it vertically into three parts: a left triangle, a middle rectangle, and a right triangle.
* Wait, the arrows point outwards. Let's split it horizontally.
* Top triangle: Base = 4 cm (top edge) + 2 cm (left stick out) + 2 cm (right stick out)? No, the diagram shows the top horizontal segment is 4cm. The vertical segment connecting to the arrow tip is not given directly, but the total height is 6cm.
* Let's try splitting it into a central rectangle and two triangles on the ends.
* Central Rectangle: Width = 4 cm. Height = ? The side label says 6 cm for the vertical part between the arrow tips? Or is 6 cm the total height? Usually, in these diagrams, the labeled dimension covers the extent shown. Let's assume the vertical distance between the top and bottom parallel lines is 6 cm? No, the label "6 cm" is next to the vertical line segment. The label "4 cm" is next to the top horizontal segment. The label "2 cm" is next to the horizontal arrow projection.
* Let's decompose into: One central rectangle and two triangles.
* Actually, a easier way is to see it as a large rectangle minus corners? No.
* Let's split it into a middle rectangle and two side triangles.
* Middle Rectangle: The vertical side is labeled 6 cm. The horizontal width of the main body is 4 cm. So, Area = $4 \times 6 = 24$ sq cm.
* Left Triangle: Base = 2 cm (the arrow part). Height = 6 cm (matches the vertical side). Area = $\frac{1}{2} \times 2 \times 6 = 6$ sq cm.
* Right Triangle: Base = 2 cm. Height = 6 cm. Area = $\frac{1}{2} \times 2 \times 6 = 6$ sq cm.
* Total Area = $24 + 6 + 6 = 36$ sq cm.

3. Shape 3:
* This is an L-shape. We can split it into two rectangles.
* Vertical Rectangle: Width = 3 cm, Height = 8 cm. Area = $3 \times 8 = 24$ sq cm.
* Horizontal Rectangle: The total bottom width is 7 cm. The vertical part takes up 3 cm. So the remaining width is $7 - 3 = 4$ cm. The height of this bottom section is 2 cm. Area = $4 \times 2 = 8$ sq cm.
* Total Area: $24 + 8 = 32$ sq cm.
* *Alternative check:* Top horizontal part is $7-3=4$ wide? No, let's split horizontally.
* Bottom rectangle: $7 \text{ cm} \times 2 \text{ cm} = 14$ sq cm.
* Top rectangle: Width = 3 cm. Height = $8 - 2 = 6$ cm. Area = $3 \times 6 = 18$ sq cm.
* Total: $14 + 18 = 32$ sq cm. Matches.

4. Shape 4:
* This is a kite or diamond shape inside a square? No, it's a quadrilateral.
* We can split it into two triangles using the horizontal diagonal.
* Top Triangle: Base = 6 cm. Height = 3 cm. Area = $\frac{1}{2} \times 6 \times 3 = 9$ sq cm.
* Bottom Triangle: Base = 6 cm. Height = 3 cm. Area = $\frac{1}{2} \times 6 \times 3 = 9$ sq cm.
* Total Area: $9 + 9 = 18$ sq cm.

5. Shape 5:
* This is a parallelogram-like shape or composed of triangles. Let's look at the labels.
* It looks like two triangles joined at a vertex? Or a large triangle with a chunk missing?
* Let's split it into two triangles horizontally? No.
* Let's look at the coordinates implied.
* Left vertical side: 4 cm.
* Top horizontal side: 3 cm.
* Bottom horizontal side: 5 cm.
* Right vertical drop: 2 cm?
* Let's split it into a rectangle and triangles.
* Draw a vertical line down from the top-right corner of the top segment.
* This creates a rectangle of $3 \text{ cm} \times 4 \text{ cm}$? No, the right side goes down 2cm then slants.
* Let's try splitting it into a trapezoid and a triangle?
* Let's use the "box method". Imagine a bounding box.
* Total Width = 5 cm. Total Height = 4 cm.
* Let's split into simple shapes.
* Shape A (Left Rectangle): Width 3 cm, Height 4 cm. Area = 12. But the top right connects to a lower point.
* Let's look at the vertices.
* (0,0) to (0,4) is the left side. Length 4.
* (0,4) to (3,4) is the top side. Length 3.
* (3,4) to (5,2)? The label "2 cm" is on the vertical drop on the right. So from y=4 to y=2.
* Then from (5,2) to (5,0)? No, the bottom is length 5. So (0,0) to (5,0).
* And there is a diagonal from (5,2) to somewhere? Or is the right side a single slant?
* The diagram shows a vertical segment of 2cm on the right, then a slant to the bottom right corner?
* Let's re-read the shape. It looks like a pentagon.
* Left side: 4 cm.
* Top side: 3 cm.
* Right side has a vertical part labeled 2 cm.
* Bottom side: 5 cm.
* There is a diagonal connecting the end of the top side to the start of the bottom side? No.
* Let's assume the shape is composed of a rectangle and a triangle.
* Split vertically at x=3.
* Left part: Rectangle $3 \times 4$. Area = 12.
* Right part: This is tricky. The bottom is 5, so the right part width is $5-3=2$. The height on the right is labeled 2. Is it a trapezoid?
* Let's look at the diagonal. It connects the inner corner to the outer corner?
* Actually, usually these are simpler. Let's look at it as a large rectangle ($5 \times 4$) minus a missing piece.
* Missing piece is a trapezoid or triangle on the top right?
* Let's try splitting into two triangles and a rectangle.
* Rectangle: $3 \text{ cm} \times 4 \text{ cm}$? No, the right side drops.
* Let's split horizontally at height 2.
* Bottom Rectangle: Width 5, Height 2. Area = $5 \times 2 = 10$.
* Top Shape: Sits on top of the bottom rectangle. Width 3. Height $4-2=2$. It's a rectangle $3 \times 2$. Area = 6.
* Is there more? The shape connects (3,4) to (5,2)? If so, there is a triangle on the right.
* Let's look at the diagonal line in the drawing. It goes from the "inner corner" (3,2) to the bottom right (5,0)? Or (3,4) to (5,0)?
* Looking closely at crop 5:
* Left vertical: 4.
* Top horizontal: 3.
* Then a diagonal goes down to the right.
* Bottom horizontal: 5.
* Right vertical: 2.
* This implies the vertices are (0,0), (5,0), (5,2), (3,4), (0,4).
* Let's calculate the area of this polygon.
* Split into a rectangle (0,0) to (3,2)? No.
* Split into:
1. Rectangle on the left: Width 3, Height 4. Area = 12. Vertices (0,0)-(3,0)-(3,4)-(0,4).
2. Trapezoid/Triangle on the right? The remaining part is from x=3 to x=5.
* At x=3, the shape exists from y=0 to y=4? No, the top edge stops at x=3.
* So at x=3, the boundary goes from (3,4) diagonally to... where?
* The rightmost edge is vertical, length 2, at x=5. So it goes from (5,0) to (5,2).
* So the diagonal connects (3,4) to (5,2).
* So the shape is defined by vertices: (0,0), (5,0), (5,2), (3,4), (0,4).
* Let's split this into:
* Rectangle: (0,0) to (3,0) to (3,4) to (0,4). Area = $3 \times 4 = 12$.
* Trapezoid: From x=3 to x=5.
* Left height (at x=3) is 4? No, the rectangle covers up to y=4. The diagonal starts at (3,4).
* Right height (at x=5) is 2.
* So we have a trapezoid with parallel vertical sides? No, the bases are vertical.
* Height of trapezoid (horizontal distance) = $5 - 3 = 2$.
* Parallel Side 1 (at x=3): Length = 4? No, the rectangle already counted the area from y=0 to y=4 at x<3.
* Let's restart the decomposition.
* Draw a vertical line at x=3.
* Left Part: Rectangle $3 \times 4$. Area = 12.
* Right Part: A trapezoid with vertices (3,0), (5,0), (5,2), (3,4).
* Wait, does the shape include the area under the diagonal? Yes.
* So the right part is a trapezoid with parallel vertical sides? No.
* Let's rotate our view. The parallel sides are the vertical lines at x=3 and x=5?
* At x=3, the segment is from y=0 to y=4? No, the diagonal connects (3,4) to (5,2). The bottom is y=0.
* So the right part is a trapezoid with heights 4 and 2, and width 2.
* Area = $\frac{4 + 2}{2} \times 2 = 6$.
* Total Area = $12 (\text{left rect}) + 6 (\text{right trap}) = 18$.
* Let's double check.
* Split into a bottom rectangle and top shapes.
* Bottom Rectangle: $5 \times 2$. Area = 10. (Covers y=0 to y=2 for all x).
* Top Left Rectangle: $3 \times (4-2) = 3 \times 2 = 6$. (Covers x=0 to 3, y=2 to 4).
* Top Right Triangle: Base = $5-3=2$. Height = $4-2=2$? No.
* The point is (3,4) and (5,2).
* Above y=2, we have a triangle with vertices (3,2), (3,4), (5,2).
* Base (vertical) at x=3 is length $4-2=2$. Height (horizontal) is $5-3=2$.
* Area = $\frac{1}{2} \times 2 \times 2 = 2$.
* Total Area = $10 + 6 + 2 = 18$.
* Both methods give 18 sq cm.

6. Shape 6:
* This is a large triangle with a smaller triangle cut out? Or two triangles?
* It looks like a large right-angled triangle with base 8 and height 6, but the hypotenuse is indented.
* Let's split it into two triangles using the vertical line labeled 4 cm? No, the 4 cm is a vertical segment inside.
* Let's look at the labels.
* Bottom base: 8 cm.
* Left height: 6 cm.
* There is a vertical line segment of length 4 cm dropping from the top vertex? No.
* The shape has vertices: (0,0), (8,0), (something, something), (0,6).
* There is an internal vertex. The lines go from (0,6) to an inner point, and from (8,0) to that inner point.
* The inner point is defined by a horizontal distance and vertical distance?
* Label "4 cm" is on a vertical dashed line? Or solid? It looks like a dimension line for the height of the inner vertex.
* Label "3 cm" is on a horizontal dimension line for the inner vertex.
* So the inner vertex is at (3, 4)? Assuming origin at bottom-left (0,0).
* Wait, the left side is vertical? The diagram shows a vertical line on the left labeled 6 cm. So (0,0) to (0,6).
* The bottom is horizontal labeled 8 cm. So (0,0) to (8,0).
* The shape is bounded by (0,0)-(8,0)-Inner-(0,6)-(0,0).
* Inner vertex position:
* Horizontal distance from left: 3 cm.
* Vertical distance from bottom: 4 cm.
* So Inner Vertex is at (3,4).
* To find the area of this concave quadrilateral, we can subtract the "empty" triangle from the bounding box or split it.
* Method 1: Split into two triangles using a diagonal from (0,0) to (3,4)? No.
* Method 2: Enclose in a rectangle $8 \times 6$? No.
* Method 3: Split the shape into two triangles by drawing a line from (0,0) to the inner vertex (3,4)?
* Triangle 1 (Bottom): Vertices (0,0), (8,0), (3,4).
* Base = 8. Height = 4 (y-coordinate of inner vertex).
* Area = $\frac{1}{2} \times 8 \times 4 = 16$.
* Triangle 2 (Top): Vertices (0,0), (0,6), (3,4).
* Base = 6 (along y-axis). Height = 3 (x-coordinate of inner vertex).
* Area = $\frac{1}{2} \times 6 \times 3 = 9$.
* Total Area = $16 + 9 = 25$ sq cm.

7. Shape 7:
* This is a rectangle with a triangular notch.
* Outer Rectangle: Width = 8 cm, Height = 4 cm. Area = $8 \times 4 = 32$ sq cm.
* Triangular Notch:
* The notch is on the right side.
* The vertical span of the notch is the full height? No, there are segments labeled 1 cm and 1 cm at the top and bottom corners?
* Let's look at the labels.
* Top edge: 8 cm.
* Left edge: 4 cm.
* Right side has a cut-out.
* The vertical positions of the cut-out vertices are given by "1 cm" from top and "1 cm" from bottom?
* The label "1 cm" is next to the small vertical stubs at the top-right and bottom-right.
* So the cut-out starts 1 cm down from the top and ends 1 cm up from the bottom.
* Height of the cut-out triangle base (virtual vertical line) = $4 - 1 - 1 = 2$ cm.
* The depth of the cut-out is labeled "2 cm" horizontally.
* So we are removing a triangle with Base = 2 cm (vertical) and Height = 2 cm (horizontal).
* Area of removed triangle = $\frac{1}{2} \times 2 \times 2 = 2$ sq cm.
* Total Area = Area of Rectangle - Area of Cutout = $32 - 2 = 30$ sq cm.

8. Shape 8:
* This is a semicircle attached to a triangle? Or a sector?
* It looks like a quarter circle and a triangle.
* Let's look at the labels.
* Vertical side: 4 cm.
* Horizontal side: 4 cm.
* There is a curved part. It looks like a quarter circle of radius 4 cm.
* And a triangle?
* The shape is bounded by a vertical line (4cm), a horizontal line (4cm), and a curve?
* No, there is a diagonal line.
* It looks like a square of $4 \times 4$ with a quarter circle drawn inside, and we want the area of the shaded region?
* The shading is the triangle part? Or the curved part?
* Looking at the hatching: The triangle formed by the diagonal and the axes is shaded? No, the region between the chord and the arc?
* Let's look really closely at Crop 8.
* It shows a right angle corner. Vertical leg 4 cm. Horizontal leg 4 cm.
* An arc connects the ends. This forms a quarter circle.
* A straight line (chord) connects the ends.
* The shaded region is the segment of the circle (the area between the chord and the arc).
* Area of Quarter Circle: Radius $r = 4$. Area = $\frac{1}{4} \pi r^2 = \frac{1}{4} \times \pi \times 16 = 4\pi$.
* Area of Triangle (formed by radii and chord): Base = 4, Height = 4. Area = $\frac{1}{2} \times 4 \times 4 = 8$.
* Area of Shaded Segment = Area of Quarter Circle - Area of Triangle.
* Area = $4\pi - 8$.
* Using $\pi \approx 3.14$:
* $4 \times 3.14 = 12.56$.
* $12.56 - 8 = 4.56$ sq cm.
* *Self-Correction*: Sometimes these problems use $\pi = 3.14$ or $\frac{22}{7}$. Let's provide the exact form and the approximate value. Given the other answers are integers, maybe I'm misinterpreting the shape.
* Alternative interpretation: Is it a triangle AND a sector?
* The hatching covers the area bounded by the vertical axis, the horizontal axis, and the ARC? No, the diagonal line is solid. The hatching is between the diagonal and the arc. Yes, it's a circular segment.
* Let's check if the answer is expected to be in terms of Pi. The other answers are integers. This suggests either an approximation or a different shape.
* Could it be a triangle with base 4 and height 4? Area 8. And the curve is just decorative? Unlikely.
* Could it be that the shape is the Triangle ONLY? No, the curve is prominent.
* Could it be the Quarter Circle ONLY? Area $12.56$.
* Let's look at the hatching again. The hatching is clearly in the "bulge" area.
* Let's assume $\pi = 3.14$. Answer: 4.56 sq cm.
* Let's check if there's another interpretation. Maybe the radius is different? No, 4 and 4.
* Maybe the shape is a triangle (area 8) plus something?
* Let's stick with the segment calculation. $4\pi - 8$.

9. Shape 9:
* This is a trapezoid? Or a rectangle and triangle?
* Left vertical side: 3 cm.
* Right vertical side: 5 cm.
* Bottom horizontal side: 8 cm.
* Top side is slanted.
* This is a right trapezoid.
* Area = $\frac{a + b}{2} \times h$.
* Parallel sides are the vertical ones? No, the vertical lines are parallel to each other. So they are the bases.
* Base 1 = 3 cm.
* Base 2 = 5 cm.
* Height (distance between parallel sides) = 8 cm.
* Area = $\frac{3 + 5}{2} \times 8 = \frac{8}{2} \times 8 = 4 \times 8 = 32$ sq cm.

10. Shape 10:
* This is an arrow shape.
* It can be split into a rectangle and a triangle.
* Rectangle part:
* The vertical side of the tail is 4 cm.
* The horizontal length of the rectangular part is 6 cm.
* Area = $6 \times 4 = 24$ sq cm.
* Triangle part (the head):
* The base of the triangle is the vertical width of the arrowhead.
* The label "5 cm" indicates the total vertical span of the arrowhead? Or the half-width?
* The dimension line for "5 cm" spans the entire vertical height of the triangular tip.
* So Base = 5 cm.
* The height of the triangle (horizontal length) is labeled "3 cm".
* Area = $\frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 5 \times 3 = 7.5$ sq cm.
* Total Area: $24 + 7.5 = 31.5$ sq cm.

Let's double check Shape 10.
Is the 5 cm the full base? The dimension lines extend from the top tip to the bottom tip of the triangle. Yes.
Is the 4 cm the height of the rectangle? Yes.
Is the 6 cm the width of the rectangle? Yes.
Is the 3 cm the height of the triangle? Yes.
Calculation: $24 + 7.5 = 31.5$.

Let's double check Shape 8.
If the answer key expects an integer, did I miss something?
Maybe the shape is just the triangle? Area 8.
Maybe the shape is the quarter circle? Area ~12.57.
Given the precision of other answers (integers), 4.56 stands out. However, geometry problems involving circles often result in decimals. I will provide the decimal answer rounded to two places.

Let's review Shape 2 again.
Rectangle $4 \times 6 = 24$. Two triangles $2 \times 6 / 2 = 6$ each. Total 36. Correct.

Let's review Shape 5 again.
Area 18. Correct.

Let's review Shape 7 again.
Rect $8 \times 4 = 32$. Cutout triangle base 2, height 2. Area 2. Result 30. Correct.

Let's review Shape 6 again.
Split into two triangles.
Bottom tri: Base 8, Height 4. Area 16.
Top tri: Base 6, Height 3. Area 9.
Total 25. Correct.

Final List:
1. 16
2. 36
3. 32
4. 18
5. 18
6. 25
7. 30
8. 4.56 (or $4\pi - 8$)
9. 32
10. 31.5

For Shape 8, I will write "4.56 sq cm (approx)" to be safe.

──────────────────────────────────────

Final Answer:
1. 16 sq cm
2. 36 sq cm
3. 32 sq cm
4. 18 sq cm
5. 18 sq cm
6. 25 sq cm
7. 30 sq cm
8. 4.56 sq cm (approx)
9. 32 sq cm
10. 31.5 sq cm