Question

this question has two parts. first, answer part a. then, answer part b.
part a
land two farmland lots are created by the boundaries of surrounding roads: adams road and orange road are parallel, and brick road and auten road are parallel.
complete the sentences to prove that ∠abd ≅ ∠cdb.
adams road and orange road are select choice using select choice to by the select choice.
furthermore, brick road and auten road are select choice so by the select choice ∠abd ≅ ∠cbd. because (overline{bd} cong overline{bd}) by
part b
the length of lot a along adams road is 0.8 miles. the length of lot a along brick road is 0.6 miles. assuming that adams road is perpendicular to brick road, approximately how many feet of fencing would be required to fence both lots separately? (assume separate fences for each lot.)
feet

Answer:

Part A Explanation (Geometry - Proving Triangle Congruence or Angle Congruence)

Step 1: Identify Parallel Lines and Transversal

Adams Rd and Orange Rd are parallel (given), and BD is a transversal. Also, Brick Rd and Auten Rd are parallel (given), with BD as a transversal.

Step 2: Use Alternate Interior Angles

For \( \angle ABD \cong \angle CDB \):

• Adams Rd \( \parallel \) Orange Rd (given), BD is transversal. So \( \angle ABD \) and \( \angle CDB \) are alternate interior angles. By the Alternate Interior Angles Theorem, alternate interior angles are congruent when lines are parallel. Thus, \( \angle ABD \cong \angle CDB \) because Adams Rd and Orange Rd are parallel, using the Alternate Interior Angles Theorem.

• Additionally, Brick Rd and Auten Rd are parallel (given), and BD is transversal. For \( \angle ABD \cong \angle CBD \) (wait, maybe a typo, but focusing on \( \angle ABD \cong \angle CDB \)): The key is the parallel lines (Adams || Orange) and transversal BD, so alternate interior angles apply. Also, \( \overline{BD} \cong \overline{BD} \) by the Reflexive Property of Congruence (a segment is congruent to itself).

Part B Explanation (Geometry - Perimeter/Length Calculation)

Step 1: Convert Miles to Feet

• 1 mile = 5280 feet.
• Length of Lot A along Adams Rd: \( 0.8 \) miles \( = 0.8 \times 5280 = 4224 \) feet.
• Length of Lot A along Brick Rd: \( 0.6 \) miles \( = 0.6 \times 5280 = 3168 \) feet.

Step 2: Analyze the Shape (Rectangle or Right Triangle? Wait, Adams Rd is perpendicular to Brick Rd, so Lot A is a rectangle? Wait, Adams Rd ⊥ Brick Rd, so Lot A is a right rectangle (or right triangle? Wait, Olive Rd is a diagonal. Wait, the problem says "separate fences for each lot" – so we need to find the length of Olive Rd (the diagonal) for each lot? Wait, no: "fence both lots separately" – so for Lot A: fence along Adams Rd, Brick Rd, and Olive Rd? Wait, no, the lots are created by Olive Rd. Wait, the problem says: "the length of lot A along Adams Road is 0.8 miles. The length of lot A along Brick Road is 0.6 miles. Assuming that Adams Road is perpendicular to Brick Road, approximately how many feet of fencing would be required to fence both lots separately?"

Wait, Lot A: sides are Adams Rd (0.8 miles), Brick Rd (0.6 miles), and Olive Rd (the diagonal). Similarly, Lot B: since Adams || Orange and Brick || Auten, and Adams ⊥ Brick, so Lot B is also a rectangle? Wait, no, the figure: Lot A is a rectangle with length 0.8 miles (Adams) and width 0.6 miles (Brick), and Olive Rd is the diagonal. Then Lot B: since Adams || Orange and Brick || Auten, and Adams ⊥ Brick, so Lot B is congruent? Wait, no, maybe Lot A and Lot B are right triangles or rectangles. Wait, Adams Rd ⊥ Brick Rd, so Lot A is a right triangle? No, Adams Rd and Brick Rd are perpendicular, so the corner at A is a right angle. So Lot A is a right triangle? Wait, no, Brick Rd and Adams Rd are perpendicular, so the shape is a rectangle, and Olive Rd is the diagonal. So the fencing for Lot A: the three sides? Wait, no, the problem says "fence both lots separately" – so each lot has three sides? Wait, the lots are bounded by Adams Rd, Brick Rd, Olive Rd (for Lot A) and Orange Rd, Auten Rd, Olive Rd (for Lot B)? Wait, no, the figure: Lot A is between Adams Rd, Brick Rd, and Olive Rd. Lot B is between Orange Rd, Auten Rd, and Olive Rd. Since Adams || Orange and Brick || Auten, and Adams ⊥ Brick, so Lot A and Lot B are congruent rectangles? Wait, no, Adams Rd length: 0.8 miles (Lot A along Adams), Brick Rd length: 0.6 miles (Lot A along Brick). So Lot A is a right rectangle with length 0.8 miles, width 0.6 miles, and diagonal Olive Rd.

Step 3: Calculate the Diagonal (Olive Rd) for One Lot

Using the Pythagorean Theorem: \( c = \sqrt{a^2 + b^2} \), where \( a = 0.8 \) miles, \( b = 0.6 \) miles.

• \( c = \sqrt{0.8^2 + 0.6^2} = \sqrt{0.64 + 0.36} = \sqrt{1} = 1 \) mile. Wait, that's a 3-4-5 triangle scaled: 0.6 (3), 0.8 (4), 1 (5). So the diagonal is 1 mile.

Step 4: Fencing for One Lot

Lot A: sides are 0.8 miles (Adams), 0.6 miles (Brick), and 1 mile (Olive Rd). Wait, but do we fence all three sides? Wait, the problem says "separate fences for each lot" – so for Lot A: fence along Adams Rd, Brick Rd, and Olive Rd. For Lot B: since Adams || Orange and Brick || Auten, and Adams ⊥ Brick, Lot B should have the same dimensions: 0.8 miles (Orange Rd), 0.6 miles (Auten Rd), and 1 mile (Olive Rd). Wait, but maybe the lots are congruent, so each lot has perimeter? Wait, no, the problem says "fence both lots separately" – so total fencing is (fencing for Lot A) + (fencing for Lot B).

• Fencing for Lot A: \( 0.8 + 0.6 + 1 = 2.4 \) miles.
• Fencing for Lot B: \( 0.8 + 0.6 + 1 = 2.4 \) miles. Wait, no, that can't be. Wait, maybe the lots are rectangles, and Olive Rd is the diagonal, so each lot is a right triangle? No, Adams Rd and Brick Rd are perpendicular, so Lot A is a right triangle with legs 0.8 and 0.6 miles, hypotenuse 1 mile. Then fencing for Lot A: 0.8 + 0.6 + 1 = 2.4 miles. Similarly, Lot B: since Adams || Orange and Brick || Auten, and Adams ⊥ Brick, Lot B is also a right triangle with legs 0.8 and 0.6 miles (because the distance between Adams and Orange is the same as Brick and Auten? Wait, no, the length along Adams is 0.8, along Brick is 0.6, so Lot B should have the same legs? Wait, maybe the total fencing is for both lots: (0.8 + 0.6 + 1) + (0.8 + 0.6 + 1) = 4.8 miles. Convert to feet: \( 4.8 \times 5280 = 25344 \) feet. Wait, but let's check:

Wait, the problem says "the length of lot A along Adams Road is 0.8 miles. The length of lot A along Brick Road is 0.6 miles. Assuming that Adams Road is perpendicular to Brick Road, approximately how many feet of fencing would be required to fence both lots separately?"

So Lot A: sides are Adams (0.8), Brick (0.6), and Olive (1). Lot B: since Adams || Orange and Brick || Auten, and Adams ⊥ Brick, Lot B has sides Orange (0.8), Auten (0.6), and Olive (1). So fencing for Lot A: 0.8 + 0.6 + 1 = 2.4 miles. Fencing for Lot B: 0.8 + 0.6 + 1 = 2.4 miles. Total: 4.8 miles. Convert to feet: \( 4.8 \times 5280 = 25344 \) feet.

Part A Answer (Proving \( \angle ABD \cong \angle CDB \))

• Adams Road and Orange Road are parallel, so by the Alternate Interior Angles Theorem, \( \angle ABD \cong \angle CDB \). Also, \( \overline{BD} \cong \overline{BD} \) by the Reflexive Property of Congruence.

Part B Answer (Fencing Length)

First, convert miles to feet:

• \( 0.8 \) miles \( = 0.8 \times 5280 = 4224 \) feet,
• \( 0.6 \) miles \( = 0.6 \times 5280 = 3168 \) feet,
• Diagonal (Olive Rd) for each lot: \( \sqrt{0.8^2 + 0.6^2} = 1 \) mile \( = 5280 \) feet.

Fencing for Lot A: \( 4224 + 3168 + 5280 = 12672 \) feet.
Fencing for Lot B: \( 4224 + 3168 + 5280 = 12672 \) feet.
Total fencing: \( 12672 + 12672 = 25344 \) feet.

Final Answers

Part A (Key Steps)

\( \angle ABD \cong \angle CDB \) because Adams Rd and Orange Rd are parallel, using the Alternate Interior Angles Theorem, and \( \overline{BD} \cong \overline{BD} \) by the Reflexive Property.

Part B

\(\boxed{25344}\) feet (or detailed steps as above).