Exercises 1-12: Use area conjectures to solve for unknown measures in various geometric figures.

To solve the problems, we will use the area formulas for different geometric shapes and apply them step by step. Let's go through each exercise:

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Exercise 1:

Shape: Triangle
Given: Base = 8 cm, Height = 5 cm
Formula: \( A = \frac{1}{2} \times \text{base} \times \text{height} \)
\[ A = \frac{1}{2} \times 8 \times 5 = 20 \, \text{cm}^2 \]

Answer: \( A = 20 \, \text{cm}^2 \)

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Exercise 2:

Shape: Triangle
Given: Base = 11 m, Height = 9 m
Formula: \( A = \frac{1}{2} \times \text{base} \times \text{height} \)
\[ A = \frac{1}{2} \times 11 \times 9 = 49.5 \, \text{m}^2 \]

Answer: \( A = 49.5 \, \text{m}^2 \)

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Exercise 3:

Shape: Rhombus
Given: Diagonals = 15 cm and 20 cm
Formula: \( A = \frac{1}{2} \times d_1 \times d_2 \)
\[ A = \frac{1}{2} \times 15 \times 20 = 150 \, \text{cm}^2 \]

Answer: \( A = 150 \, \text{cm}^2 \)

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Exercise 4:

Shape: Trapezoid
Given: Bases = 6 cm and 14 cm, Height = 6 cm
Formula: \( A = \frac{1}{2} \times (\text{base}_1 + \text{base}_2) \times \text{height} \)
\[ A = \frac{1}{2} \times (6 + 14) \times 6 = \frac{1}{2} \times 20 \times 6 = 60 \, \text{cm}^2 \]

Answer: \( A = 60 \, \text{cm}^2 \)

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Exercise 5:

Shape: Triangle
Given: Area = 39 cm², Base = 13 cm
Formula: \( A = \frac{1}{2} \times \text{base} \times \text{height} \)
\[ 39 = \frac{1}{2} \times 13 \times h \]
\[ 39 = 6.5 \times h \]
\[ h = \frac{39}{6.5} = 6 \, \text{cm} \]

Answer: \( h = 6 \, \text{cm} \)

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Exercise 6:

Shape: Triangle
Given: Area = 31.5 ft², Height = 7 ft
Formula: \( A = \frac{1}{2} \times \text{base} \times \text{height} \)
\[ 31.5 = \frac{1}{2} \times b \times 7 \]
\[ 31.5 = 3.5 \times b \]
\[ b = \frac{31.5}{3.5} = 9 \, \text{ft} \]

Answer: \( b = 9 \, \text{ft} \)

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Exercise 7:

Shape: Trapezoid
Given: Area = 420 ft², Bases = 17 ft and 25 ft
Formula: \( A = \frac{1}{2} \times (\text{base}_1 + \text{base}_2) \times \text{height} \)
\[ 420 = \frac{1}{2} \times (17 + 25) \times LE \]
\[ 420 = \frac{1}{2} \times 42 \times LE \]
\[ 420 = 21 \times LE \]
\[ LE = \frac{420}{21} = 20 \, \text{ft} \]

Answer: \( LE = 20 \, \text{ft} \)

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Exercise 8:

Shape: Trapezoid
Given: Area = 50 cm², Bases = 6 cm and 13 cm
Formula: \( A = \frac{1}{2} \times (\text{base}_1 + \text{base}_2) \times \text{height} \)
\[ 50 = \frac{1}{2} \times (6 + 13) \times h \]
\[ 50 = \frac{1}{2} \times 19 \times h \]
\[ 50 = 9.5 \times h \]
\[ h = \frac{50}{9.5} = \frac{500}{95} = \frac{100}{19} \approx 5.26 \, \text{cm} \]

Answer: \( h = \frac{100}{19} \, \text{cm} \)

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Exercise 9:

Shape: Parallelogram
Given: Area = 180 m², Height = 9 m
Formula: \( A = \text{base} \times \text{height} \)
\[ 180 = b \times 9 \]
\[ b = \frac{180}{9} = 20 \, \text{m} \]

Answer: \( b = 20 \, \text{m} \)

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Exercise 10:

Shape: Triangle
Given: Area = 924 cm², Base = 51 cm
Formula: \( A = \frac{1}{2} \times \text{base} \times \text{height} \)
\[ 924 = \frac{1}{2} \times 51 \times h \]
\[ 924 = 25.5 \times h \]
\[ h = \frac{924}{25.5} = 36 \, \text{cm} \]
Now, find the perimeter \( P \):
The triangle is a right triangle with sides 51 cm, 40 cm, and 24 cm (Pythagorean theorem).
\[ P = 51 + 40 + 24 = 115 \, \text{cm} \]

Answer: \( P = 115 \, \text{cm} \)

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Exercise 11:

Shape: Trapezoid
Given: Area = 204 cm², Perimeter = 62 cm, Bases = 15 cm and 13 cm
Formula: \( A = \frac{1}{2} \times (\text{base}_1 + \text{base}_2) \times \text{height} \)
\[ 204 = \frac{1}{2} \times (15 + 13) \times h \]
\[ 204 = \frac{1}{2} \times 28 \times h \]
\[ 204 = 14 \times h \]
\[ h = \frac{204}{14} = \frac{102}{7} \approx 14.57 \, \text{cm} \]

Answer: \( h = \frac{102}{7} \, \text{cm} \)

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Exercise 12:

Shape: Right Triangle
Given: Hypotenuse = 9 ft, One leg = 6 ft
Formula: Use the Pythagorean theorem to find the other leg \( x \):
\[ 9^2 = 6^2 + x^2 \]
\[ 81 = 36 + x^2 \]
\[ x^2 = 45 \]
\[ x = \sqrt{45} = 3\sqrt{5} \, \text{ft} \]
Now, find \( y \):
The triangle is similar to the larger triangle, so use the proportion:
\[ \frac{x}{6} = \frac{6}{9} \]
\[ \frac{3\sqrt{5}}{6} = \frac{2}{3} \]
\[ y = \frac{6 \times 6}{9} = 4 \, \text{ft} \]

Answer: \( x = 3\sqrt{5} \, \text{ft} \), \( y = 4 \, \text{ft} \)

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Final Answers:

\[
\boxed{
\begin{array}{ll}
1. & 20 \, \text{cm}^2 \\
2. & 49.5 \, \text{m}^2 \\
3. & 150 \, \text{cm}^2 \\
4. & 60 \, \text{cm}^2 \\
5. & 6 \, \text{cm} \\
6. & 9 \, \text{ft} \\
7. & 20 \, \text{ft} \\
8. & \frac{100}{19} \, \text{cm} \\
9. & 20 \, \text{m} \\
10. & 115 \, \text{cm} \\
11. & \frac{102}{7} \, \text{cm} \\
12. & x = 3\sqrt{5} \, \text{ft}, \, y = 4 \, \text{ft}
\end{array}
}
\]