Geometry Solver

To solve for the total cost of interior painting, we must calculate the surface area of the walls for each room, subtract the areas of openings (doors and windows), determine the amount of paint required, and add the labor costs.

[Analysis]

The problem requires calculating the wall surface area for various rooms based on a floor plan.

1. Formulas:

   • Wall Area = $\text{Perimeter} \times \text{Height}$.
   • Semicircle Arc Length = $\pi \times r$.
   • Net Wall Area = $\text{Gross Wall Area} - \text{Door Areas} - \text{Window Areas}$.
   • Number of Gallons = $\lceil \frac{\text{Net Area}}{250} \rceil$.
2. Constants:

   • Ceiling Height = $9 \text{ ft}$.
   • Door Width = $2 \text{ ft}$.
   • Window Deduction = $400 \text{ ft}^2$.
   • Labor Cost = $\$5000$.
3. Approach: For each room, we find the perimeter of the existing walls (thick lines), subtract the door widths, multiply by the ceiling height, and then sum the areas.

[Solution]

Step 1: Calculate Wall Area per Room

• Dining & Kitchen (including the semicircle):

  • The room is $12 \text{ ft}$ wide (matching the Garage/Laundry section).
  • Perimeter = Right Wall ($16 \text{ ft}$) + Bottom Wall ($12 \text{ ft}$) + Semicircle Arc ($\pi \times 6 \approx 18.85 \text{ ft}$).
  • Subtract 1 door ($2 \text{ ft}$): $16 + 12 + 18.85 - 2 = 44.85 \text{ ft}$.
  • Area = $44.85 \times 9 = 403.65 \text{ ft}^2$.
• Family Room:

  • Walls include the Top ($15 \text{ ft}$) and Left (against Primary Bed/Hallway, $10 + 6 = 16 \text{ ft}$). Openings to Kitchen and Entry are not painted.
  • Perimeter = $15 + 16 = 31 \text{ ft}$.
  • Area = $31 \times 9 = 279 \text{ ft}^2$.
• Primary Bedroom & Closet:

  • Bedroom Perimeter = $(11 \times 2) + (10 \times 2) = 42 \text{ ft}$. Subtract 3 doors: $42 - 6 = 36 \text{ ft}$.
  • Closet Perimeter = $(5 \times 2) + (7 \times 2) = 24 \text{ ft}$. Subtract 1 door: $24 - 2 = 22 \text{ ft}$.
  • Total Area = $(36 + 22) \times 9 = 522 \text{ ft}^2$.
• Bedroom 1 & Closet:

  • Bedroom Perimeter = $14 + 8 + 12 + 10 = 44 \text{ ft}$. Subtract 2 doors: $44 - 4 = 40 \text{ ft}$.
  • Closet Perimeter = $(2 \times 2) + (6 \times 2) = 16 \text{ ft}$. Subtract 1 door: $16 - 2 = 14 \text{ ft}$.
  • Total Area = $(40 + 14) \times 9 = 486 \text{ ft}^2$.
• Bedroom 2 & Closet:

  • Bedroom Perimeter (with slant) $\approx 32.25 \text{ ft}$. Subtract 2 doors: $32.25 - 4 = 28.25 \text{ ft}$.
  • Closet Perimeter = $(4 \times 2) + (7 \times 2) = 22 \text{ ft}$. Subtract 1 door: $22 - 2 = 20 \text{ ft}$.
  • Total Area = $(28.25 + 20) \times 9 = 434.25 \text{ ft}^2$.
• Hallway:

  • Net Perimeter (after subtracting 4 doors and 2 openings) $\approx 32 \text{ ft}$.
  • Area = $32 \times 9 = 288 \text{ ft}^2$.
• Entry:

  • Net Perimeter (after subtracting front door and 2 openings) $\approx 10 \text{ ft}$.
  • Area = $10 \times 9 = 90 \text{ ft}^2$.
• Primary Bathroom:

  • Perimeter = $(5 \times 2) + (10 \times 2) - 2 = 28 \text{ ft}$.
  • Area = $28 \times 9 = 252 \text{ ft}^2$.
• Bathroom 1:

  • Perimeter = $(7 \times 2) + (5 \times 2) - 2 = 22 \text{ ft}$.
  • Area = $22 \times 9 = 198 \text{ ft}^2$.
• Laundry Room:

  • Perimeter = $(10 \times 2) + (5 \times 2) - 4 = 26 \text{ ft}$.
  • Area = $26 \times 9 = 234 \text{ ft}^2$.
• 3 walls of Garage:

  • Length = $10 + 10 + 10 - 2 = 28 \text{ ft}$.
  • Area = $28 \times 9 = 252 \text{ ft}^2$.

Step 2: Total Area and Paint Requirements

• Total Area Needed to be Painted: $403.65 + 279 + 522 + 486 + 434.25 + 288 + 90 + 252 + 198 + 234 + 252 = 3438.90 \text{ ft}^2$.
• Net Area (Subtracting 400 $ft^2$ for windows): $3438.90 - 400 = 3038.90 \text{ ft}^2$.
• Total Gallons of Paint Needed: $\frac{3038.90}{250} = 12.16 \rightarrow 13 \text{ gallons}$.

Step 3: Cost Calculation

• Type of Paint Chosen: Behr Premium Plus Interior Eggshell Enamel (Ultra Pure White).
• Cost per Gallon: $\$32.98$.
• Total Paint Cost: $13 \times 32.98 = \$428.74$.
• Total Cost (Paint + Labor): $428.74 + 5000 = \$5428.74$.

[Answer]

• Total Area Needed to be Painted: $3438.90 \text{ ft}^2$
• Net Area (Minus 400 $ft^2$): $3038.90 \text{ ft}^2$
• Total Gallons of Paint Needed: $13$
• Type of Paint Chosen: Behr Premium Plus Interior Eggshell Enamel
• Cost per Gallon: $\$32.98$
• Cost of Labor: $\$5000$
• Total Cost for Paint: $\$5428.74$

This document provides step-by-step solutions to the geometry problems presented in the image.

[Analysis]

The problems cover several fundamental concepts in geometry:

1. Coordinate Geometry of Circles (Problems 6 & 8): Using the midpoint formula to find the center, the distance formula to find the radius, and the standard form of a circle's equation $(x - h)^2 + (y - k)^2 = r^2$. Transformations involve shifting the center and scaling the radius.
2. Circle Properties (Problem 7): Using the relationship between a chord, the radius, and the distance from the center (forming a right triangle).
3. Sphere Geometry (Problem 9): Applying the Pythagorean theorem to the relationship between the sphere's radius, the distance to an intersecting plane, and the radius of the resulting cross-section circle.
4. Circle Theorems (Problem 10): Utilizing theorems regarding central angles, inscribed angles, tangent-chord angles, and the property that an inscribed angle subtending a diameter is a right angle.

[Solution]

Problem 6

Given: Diameter with endpoints $(-1, 6)$ and $(5, 2)$.

a) Find the Center:
The center $(h, k)$ is the midpoint of the diameter.
$h = \frac{-1 + 5}{2} = \frac{4}{2} = 2$
$k = \frac{6 + 2}{2} = \frac{8}{2} = 4$
Center: $(2, 4)$

b) Find the Radius:
The radius $r$ is the distance from the center $(2, 4)$ to an endpoint, e.g., $(5, 2)$.
$r = \sqrt{(5 - 2)^2 + (2 - 4)^2}$
$r = \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13}$
Radius: $\sqrt{13}$

c) Write the Equation of the Circle:
Using the standard form $(x - h)^2 + (y - k)^2 = r^2$:
$(x - 2)^2 + (y - 4)^2 = (\sqrt{13})^2$
$(x - 2)^2 + (y - 4)^2 = 13$

Problem 7

Given: Radius $r = 30$, Chord $AB = 30\sqrt{3}$.
Find the distance $d$ from the center to the chord.

1. A radius perpendicular to a chord bisects the chord. Half the chord length is $\frac{30\sqrt{3}}{2} = 15\sqrt{3}$.
2. The distance $d$, the half-chord, and the radius form a right triangle.
3. By the Pythagorean theorem:
   $d^2 + (15\sqrt{3})^2 = 30^2$
   $d^2 + 225 \cdot 3 = 900$
   $d^2 + 675 = 900$
   $d^2 = 225$
   $d = \sqrt{225} = 15$
   Distance: $15$

Problem 8

Given: Circle P: $(x - 2)^2 + (y + 3)^2 = 16$.

1. Original Circle P: Center $(2, -3)$, Radius $r_P = \sqrt{16} = 4$.
2. Transformations for Circle Q:

   • Shift center left 7: $h_Q = 2 - 7 = -5$.
   • Shift center up 5: $k_Q = -3 + 5 = 2$.
   • Double the radius: $r_Q = 2 \cdot 4 = 8$.
3. New Equation:
   $(x - (-5))^2 + (y - 2)^2 = 8^2$
   $(x + 5)^2 + (y - 2)^2 = 64$
   Correct Option: B

Problem 9

Given: Plane distance from sphere center $d = 12$ cm. Intersection circle diameter $= 32$ cm.

1. Radius of intersection circle $r_{circle} = \frac{32}{2} = 16$ cm.
2. Let $R$ be the radius of the sphere. Using the Pythagorean theorem:
   $R^2 = d^2 + r_{circle}^2$
   $R^2 = 12^2 + 16^2 = 144 + 256 = 400$
   $R = \sqrt{400} = 20$ cm.
3. Diameter of sphere $= 2R = 2 \cdot 20 = 40$ cm.
   Diameter: $40 \text{ cm}$

Problem 10

Given: $\overleftrightarrow{AB}$ is tangent at $A$; $\overline{AF}$ is diameter; $m\widehat{AG} = 120^\circ$; $m\widehat{CE} = 40^\circ$; $m\widehat{EF} = 25^\circ$.

Step 1: Calculate missing arc measures.

• Since $\overline{AF}$ is a diameter, the semi-circle arcs are $180^\circ$.
• $m\widehat{GF} = 180^\circ - m\widehat{AG} = 180^\circ - 120^\circ = 60^\circ$.
• $m\widehat{AC} = 180^\circ - (m\widehat{CE} + m\widehat{EF}) = 180^\circ - (40^\circ + 25^\circ) = 115^\circ$.

Step 2: Calculate Angle Measures.

1. $m\angle 1$: Central angle subtending $\widehat{GF}$. $m\angle 1 = m\widehat{GF} = 60^\circ$.
2. $m\angle 2$: Central angle subtending $\widehat{AG}$. $m\angle 2 = m\widehat{AG} = 120^\circ$.
3. $m\angle 3$: Inscribed angle subtending $\widehat{FC}$ (where $m\widehat{FC} = 40^\circ + 25^\circ = 65^\circ$). $m\angle 3 = \frac{1}{2}(65^\circ) = 32.5^\circ$.
4. $m\angle 4$: Inscribed angle subtending $\widehat{AG}$. $m\angle 4 = \frac{1}{2}(120^\circ) = 60^\circ$.
5. $m\angle 5$: Inscribed angle subtending $\widehat{AE}$ (where $m\widehat{AE} = 115^\circ + 40^\circ = 155^\circ$). $m\angle 5 = \frac{1}{2}(155^\circ) = 77.5^\circ$.
6. $m\angle 6$: Inscribed angle subtending diameter $\overline{AF}$. $m\angle 6 = \frac{1}{2}(180^\circ) = 90^\circ$.
7. $m\angle 7$: Tangent-chord angle subtending $\widehat{AG}$. $m\angle 7 = \frac{1}{2}(120^\circ) = 60^\circ$.
8. $m\angle 8$: Tangent-chord angle subtending $\widehat{AC}$. $m\angle 8 = \frac{1}{2}(115^\circ) = 57.5^\circ$.

[Answer]

6. a) $(2, 4)$ b) $\sqrt{13}$ c) $(x - 2)^2 + (y - 4)^2 = 13$
7. $15$
8. $B$
9. $40 \text{ cm}$
10.
$m\angle 1 = 60^\circ$
$m\angle 2 = 120^\circ$
$m\angle 3 = 32.5^\circ$
$m\angle 4 = 60^\circ$
$m\angle 5 = 77.5^\circ$
$m\angle 6 = 90^\circ$
$m\angle 7 = 60^\circ$
$m\angle 8 = 57.5^\circ$

Analysis

The provided image contains two geometry problems involving parallelograms. The primary property of a parallelogram required to solve these problems is that opposite sides are equal in length.

• Problem 3: We are given the lengths of opposite sides $AD$ and $BC$. We will set these expressions equal to each other to solve for $x$.
• Problem 4: We are given the lengths of opposite sides $AB$ and $DC$. We will set these values equal to each other to solve for $x$. (Note: The length of side $AD = 24$ is provided but is not necessary for finding $x$).

Solution

Problem 3

Step 1: Identify the property.
In parallelogram $ABCD$, the opposite sides are equal. Therefore, $AD = BC$.

Step 2: Set up the equation.
Substitute the given values from the figure:
$3x - 2 = 13$

Step 3: Solve for $x$.
Add 2 to both sides of the equation:
$3x = 13 + 2$
$3x = 15$

Divide both sides by 3:
$x = \frac{15}{3}$
$x = 5$

Problem 4

Step 1: Identify the property.
In parallelogram $ABCD$, the opposite sides are equal. Therefore, $AB = DC$.

Step 2: Set up the equation.
Substitute the given values from the figure:
$35 = 4x + 3$

Step 3: Solve for $x$.
Subtract 3 from both sides of the equation:
$35 - 3 = 4x$
$32 = 4x$

Divide both sides by 4:
$x = \frac{32}{4}$
$x = 8$

Answer

For Problem 3, $x = 5$.
For Problem 4, $x = 8$.