Geometry worksheet focusing on calculating areas of parallelograms, trapezoids, rhombuses, and kites using given formulas and dimensions.

Worksheet 11-2: Areas of Parallelograms, Trapezoids, Rhombuses, and Kites

#### Part 1: Find the area of each figure

We will use the following formulas:
1. Parallelogram: \( A = b \cdot h \)
2. Trapezoid: \( A = \frac{1}{2} (b_1 + b_2) \cdot h \)
3. Rhombus/Kite: \( A = \frac{1}{2} d_1 \cdot d_2 \)

---

#### Problem 1:
Figure: Parallelogram with base \( b = 10 \) and height \( h = 4 \).

Solution:
\[
A = b \cdot h = 10 \cdot 4 = 40
\]

Answer: \( \boxed{40} \)

---

#### Problem 2:
Figure: Trapezoid with bases \( b_1 = 7 \), \( b_2 = 11 \), and height \( h = 6 \).

Solution:
\[
A = \frac{1}{2} (b_1 + b_2) \cdot h = \frac{1}{2} (7 + 11) \cdot 6 = \frac{1}{2} \cdot 18 \cdot 6 = 54
\]

Answer: \( \boxed{54} \)

---

#### Problem 3:
Figure: Parallelogram with base \( b = 9 \) and height \( h = 10 \).

Solution:
\[
A = b \cdot h = 9 \cdot 10 = 90
\]

Answer: \( \boxed{90} \)

---

#### Problem 4:
Figure: Trapezoid with bases \( b_1 = 8 \), \( b_2 = 12 \), and height \( h = 3 \).

Solution:
\[
A = \frac{1}{2} (b_1 + b_2) \cdot h = \frac{1}{2} (8 + 12) \cdot 3 = \frac{1}{2} \cdot 20 \cdot 3 = 30
\]

Answer: \( \boxed{30} \)

---

#### Problem 5:
Figure: Trapezoid with bases \( b_1 = 4 \), \( b_2 = 9 \), and height \( h = 9 \).

Solution:
\[
A = \frac{1}{2} (b_1 + b_2) \cdot h = \frac{1}{2} (4 + 9) \cdot 9 = \frac{1}{2} \cdot 13 \cdot 9 = 58.5
\]

Answer: \( \boxed{58.5} \)

---

#### Problem 6:
Figure: Trapezoid with bases \( b_1 = 10 \), \( b_2 = 14 \), and height \( h = 8 \).

Solution:
\[
A = \frac{1}{2} (b_1 + b_2) \cdot h = \frac{1}{2} (10 + 14) \cdot 8 = \frac{1}{2} \cdot 24 \cdot 8 = 96
\]

Answer: \( \boxed{96} \)

---

#### Problem 7:
Figure: Rhombus with diagonals \( d_1 = 10 \) and \( d_2 = 6 \).

Solution:
\[
A = \frac{1}{2} d_1 \cdot d_2 = \frac{1}{2} \cdot 10 \cdot 6 = 30
\]

Answer: \( \boxed{30} \)

---

#### Problem 8:
Figure: Trapezoid with bases \( b_1 = 5 \), \( b_2 = 8 \), and height \( h = 4 \).

Solution:
\[
A = \frac{1}{2} (b_1 + b_2) \cdot h = \frac{1}{2} (5 + 8) \cdot 4 = \frac{1}{2} \cdot 13 \cdot 4 = 26
\]

Answer: \( \boxed{26} \)

---

#### Problem 9:
Figure: Rhombus with diagonals \( d_1 = 12 \) and \( d_2 = 4 \).

Solution:
\[
A = \frac{1}{2} d_1 \cdot d_2 = \frac{1}{2} \cdot 12 \cdot 4 = 24
\]

Answer: \( \boxed{24} \)

---

Part 2: Find the missing part of each figure given the area

#### Problem 10:
Figure: Parallelogram with area \( A = 48 \), base \( b = 12 \), and unknown height \( h \).

Solution:
\[
A = b \cdot h \implies 48 = 12 \cdot h \implies h = \frac{48}{12} = 4
\]

Answer: \( \boxed{4} \)

---

#### Problem 11:
Figure: Trapezoid with area \( A = 42 \), bases \( b_1 = 5 \), \( b_2 = 9 \), and unknown height \( h \).

Solution:
\[
A = \frac{1}{2} (b_1 + b_2) \cdot h \implies 42 = \frac{1}{2} (5 + 9) \cdot h \implies 42 = \frac{1}{2} \cdot 14 \cdot h \implies 42 = 7h \implies h = \frac{42}{7} = 6
\]

Answer: \( \boxed{6} \)

---

#### Problem 12:
Figure: Rhombus with area \( A = 48 \), diagonal \( d_1 = 8 \), and unknown diagonal \( d_2 \).

Solution:
\[
A = \frac{1}{2} d_1 \cdot d_2 \implies 48 = \frac{1}{2} \cdot 8 \cdot d_2 \implies 48 = 4d_2 \implies d_2 = \frac{48}{4} = 12
\]

Answer: \( \boxed{12} \)

---

#### Problem 13:
Figure: Trapezoid with area \( A = 64 \), bases \( b_1 = 16 \), \( b_2 = b \), and height \( h = 4 \).

Solution:
\[
A = \frac{1}{2} (b_1 + b_2) \cdot h \implies 64 = \frac{1}{2} (16 + b) \cdot 4 \implies 64 = 2(16 + b) \implies 64 = 32 + 2b \implies 2b = 32 \implies b = 16
\]

Answer: \( \boxed{16} \)

---

#### Problem 14:
Figure: Trapezoid with area \( A = 50 \), bases \( b_1 = 5 \), \( b_2 = 12 \), and unknown height \( h \).

Solution:
\[
A = \frac{1}{2} (b_1 + b_2) \cdot h \implies 50 = \frac{1}{2} (5 + 12) \cdot h \implies 50 = \frac{1}{2} \cdot 17 \cdot h \implies 50 = 8.5h \implies h = \frac{50}{8.5} = \frac{100}{17}
\]

Answer: \( \boxed{\frac{100}{17}} \)

---

#### Problem 15:
Figure: Rhombus with area \( A = 27 \), diagonal \( d_1 = 9 \), and unknown diagonal \( d_2 \).

Solution:
\[
A = \frac{1}{2} d_1 \cdot d_2 \implies 27 = \frac{1}{2} \cdot 9 \cdot d_2 \implies 27 = 4.5d_2 \implies d_2 = \frac{27}{4.5} = 6
\]

Answer: \( \boxed{6} \)

---

Part 3: Find the missing part using the Pythagorean Theorem and find the area

#### Problem 16:
Figure: Parallelogram with base \( b = 8 \), side \( s = 5 \), and height \( h \) to be found.

Solution:
The height \( h \) can be found using the Pythagorean Theorem in the right triangle formed by the height, half the difference of the bases, and the side of the parallelogram.
\[
\text{Half the difference of the bases} = \frac{8 - 3}{2} = 2.5
\]
\[
s^2 = h^2 + 2.5^2 \implies 5^2 = h^2 + 2.5^2 \implies 25 = h^2 + 6.25 \implies h^2 = 18.75 \implies h = \sqrt{18.75} = \frac{5\sqrt{3}}{2}
\]
Now, calculate the area:
\[
A = b \cdot h = 8 \cdot \frac{5\sqrt{3}}{2} = 20\sqrt{3}
\]

Answer: \( \boxed{20\sqrt{3}} \)

---

#### Problem 17:
Figure: Kite with diagonals \( d_1 = 10 \), \( d_2 = x \), and side lengths \( 10 \) and \( 6 \).

Solution:
The diagonals of a kite are perpendicular bisectors of each other. Using the Pythagorean Theorem in one of the right triangles formed:
\[
\left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2 = 10^2 \implies 5^2 + \left(\frac{x}{2}\right)^2 = 10^2 \implies 25 + \left(\frac{x}{2}\right)^2 = 100 \implies \left(\frac{x}{2}\right)^2 = 75 \implies \frac{x}{2} = \sqrt{75} = 5\sqrt{3} \implies x = 10\sqrt{3}
\]
Now, calculate the area:
\[
A = \frac{1}{2} d_1 \cdot d_2 = \frac{1}{2} \cdot 10 \cdot 10\sqrt{3} = 50\sqrt{3}
\]

Answer: \( \boxed{50\sqrt{3}} \)

---

#### Problem 18:
Figure: Trapezoid with bases \( b_1 = 5 \), \( b_2 = 15 \), height \( h \), and slant height \( 13 \).

Solution:
The height \( h \) can be found using the Pythagorean Theorem in the right triangle formed by the height, half the difference of the bases, and the slant height.
\[
\text{Half the difference of the bases} = \frac{15 - 5}{2} = 5
\]
\[
13^2 = h^2 + 5^2 \implies 169 = h^2 + 25 \implies h^2 = 144 \implies h = 12
\]
Now, calculate the area:
\[
A = \frac{1}{2} (b_1 + b_2) \cdot h = \frac{1}{2} (5 + 15) \cdot 12 = \frac{1}{2} \cdot 20 \cdot 12 = 120
\]

Answer: \( \boxed{120} \)

---

Final Answers:

\[
\boxed{40, 54, 90, 30, 58.5, 96, 30, 26, 24, 4, 6, 12, 16, \frac{100}{17}, 6, 20\sqrt{3}, 50\sqrt{3}, 120}
\]