6-2 kites and trapezoids answer sheet.pdf - Geometry Worksheet ... - Free Printable

To find the area of each shape, we need to identify the type of quadrilateral and use the correct formula. Here are the common formulas for the shapes shown:

1. Rhombus or Kite: Area = $\frac{1}{2} \times d_1 \times d_2$ (half the product of the diagonals).
2. Trapezoid: Area = $\frac{1}{2} \times (b_1 + b_2) \times h$ (half the sum of the parallel bases times the height).
3. Parallelogram: Area = $b \times h$ (base times height). Note: The height must be perpendicular to the base.

Let's solve them one by one based on the visible numbers in the worksheet.

Problem 1: Rhombus
* Diagonal 1 ($d_1$) = 6 cm
* Diagonal 2 ($d_2$) = 8 cm
* Formula: $A = \frac{1}{2} \times d_1 \times d_2$
* Calculation: $A = \frac{1}{2} \times 6 \times 8 = 3 \times 8 = 24$
* Area = 24 cm²

Problem 2: Kite
* Diagonal 1 ($d_1$) = 5 mm
* Diagonal 2 ($d_2$) = 8 mm
* Formula: $A = \frac{1}{2} \times d_1 \times d_2$
* Calculation: $A = \frac{1}{2} \times 5 \times 8 = 2.5 \times 8 = 20$
* Area = 20 mm²

Problem 3: Trapezoid
* Base 1 ($b_1$) = 7 in
* Base 2 ($b_2$) = 11 in
* Height ($h$) = 4 in
* Formula: $A = \frac{1}{2} \times (b_1 + b_2) \times h$
* Calculation: $A = \frac{1}{2} \times (7 + 11) \times 4 = \frac{1}{2} \times 18 \times 4 = 9 \times 4 = 36$
* Area = 36 in²

Problem 4: Kite
* Diagonal 1 ($d_1$) = 6 ft
* Diagonal 2 ($d_2$) = 10 ft
* Formula: $A = \frac{1}{2} \times d_1 \times d_2$
* Calculation: $A = \frac{1}{2} \times 6 \times 10 = 3 \times 10 = 30$
* Area = 30 ft²

Problem 5: Parallelogram
* Base ($b$) = 9 m
* Height ($h$) = 5 m (The dotted line shows the perpendicular height)
* Formula: $A = b \times h$
* Calculation: $A = 9 \times 5 = 45$
* Area = 45 m²

Problem 6: Trapezoid
* Base 1 ($b_1$) = 8 yd
* Base 2 ($b_2$) = 14 yd
* Height ($h$) = 5 yd
* Formula: $A = \frac{1}{2} \times (b_1 + b_2) \times h$
* Calculation: $A = \frac{1}{2} \times (8 + 14) \times 5 = \frac{1}{2} \times 22 \times 5 = 11 \times 5 = 55$
* Area = 55 yd²

Problem 7: Parallelogram
* Base ($b$) = 12 km
* Height ($h$) = 7 km
* Formula: $A = b \times h$
* Calculation: $A = 12 \times 7 = 84$
* Area = 84 km²

Problem 8: Trapezoid
* Base 1 ($b_1$) = 6 cm
* Base 2 ($b_2$) = 10 cm
* Height ($h$) = 4 cm
* Formula: $A = \frac{1}{2} \times (b_1 + b_2) \times h$
* Calculation: $A = \frac{1}{2} \times (6 + 10) \times 4 = \frac{1}{2} \times 16 \times 4 = 8 \times 4 = 32$
* Area = 32 cm²

Problem 9: Rhombus
* Diagonal 1 ($d_1$) = 9 in
* Diagonal 2 ($d_2$) = 12 in
* Formula: $A = \frac{1}{2} \times d_1 \times d_2$
* Calculation: $A = \frac{1}{2} \times 9 \times 12 = 9 \times 6 = 54$
* Area = 54 in²

Problem 10: Trapezoid
* Base 1 ($b_1$) = 5 ft
* Base 2 ($b_2$) = 9 ft
* Height ($h$) = 6 ft
* Formula: $A = \frac{1}{2} \times (b_1 + b_2) \times h$
* Calculation: $A = \frac{1}{2} \times (5 + 9) \times 6 = \frac{1}{2} \times 14 \times 6 = 7 \times 6 = 42$
* Area = 42 ft²

Problem 11: Composite Shape (Triangle + Rectangle or just a large Triangle?)
Looking closely at problem 11, it appears to be a triangle sitting on top of a rectangle, or perhaps a trapezoid split into parts. However, standard worksheets like this usually present simple polygons. Let's look at the dimensions.
It looks like a triangle with base 14 m and height 6 m, attached to a rectangle? No, looking at the vertices, it is a single polygon. It looks like a trapezoid on its side or a composite shape.
Let's re-examine image crop 4.
Shape 11 is a pentagon-like shape or a composite of a rectangle and a triangle.
Actually, looking at the dotted lines, it seems to be decomposed into a rectangle and a triangle.
Rectangle part: Base = 10 m, Height = 4 m? No, the labels are a bit blurry.
Let's look at the bottom edge: 14 m.
Top edge: 10 m.
Height on the left side: 4 m.
There is a diagonal cut.
Actually, it looks like a Trapezoid with bases 10m and 14m and height 4m?
If it is a trapezoid: $A = \frac{1}{2}(10+14)(4) = 12 \times 4 = 48$.
However, there is a diagonal line drawn inside. And another dimension "6m" pointing to the slanted side or height?
Let's look at Problem 12 first, it might clarify the style.

Problem 12: Kite / Arrowhead
This is a concave kite (dart).
Diagonal 1 (horizontal) = 10 yd + 4 yd = 14 yd? Or is the total length 14?
The segments are labeled 10 yd and 4 yd along the horizontal axis. So total horizontal diagonal $d_1 = 14$ yd.
The vertical diagonal $d_2$ is labeled as 6 yd.
Formula for any quadrilateral with perpendicular diagonals (including kites and darts): $A = \frac{1}{2} \times d_1 \times d_2$.
Calculation: $A = \frac{1}{2} \times 14 \times 6 = 7 \times 6 = 42$.
Area = 42 yd²

Now back to Problem 11.
It looks like a trapezoid where the top base is 10 m, bottom base is 14 m.
The height is indicated by the vertical side on the left which is 4 m? Or is the 4m part of the base?
Let's look at the labels again.
Left vertical side: 4 m.
Bottom horizontal side: 14 m.
Top horizontal side: 10 m.
Right side is slanted.
This is a right trapezoid.
Bases are 10 m and 14 m. Height is 4 m.
Area = $\frac{1}{2} \times (10 + 14) \times 4 = \frac{1}{2} \times 24 \times 4 = 12 \times 4 = 48$.
Wait, there is a label "6 m" near the slanted side. Is that the height? No, the vertical side is clearly marked 4m. The 6m might be the length of the slanted side, which is extra information not needed for area if we have the height.
Or, is the shape composed of a rectangle (10x4) and a triangle?
Rectangle Area = $10 \times 4 = 40$.
Triangle Base = $14 - 10 = 4$. Triangle Height = 4.
Triangle Area = $\frac{1}{2} \times 4 \times 4 = 8$.
Total Area = $40 + 8 = 48$.
This confirms the calculation.
Area = 48 m²

Final Answer:
1. 24 cm²
2. 20 mm²
3. 36 in²
4. 30 ft²
5. 45 m²
6. 55 yd²
7. 84 km²
8. 32 cm²
9. 54 in²
10. 42 ft²
11. 48 m²
12. 42 yd²