Jacobs Geometry
Contents 2 sections

Exam 1 Answers

You: Give me answers to this exam. Let me know if you cannot answer

Leo AI: Here are the answers to the Geometry Test:

Fill in the Blank (Questions 1–25)

  1. Protractor
  2. Angle
  3. Concurrent
  4. Constructions (or Compass-and-straightedge constructions)
  5. Plane
  6. Collinear
  7. Postulate (or Axiom)
  8. Line Segment
  9. Circumference
  10. Perimeter
  11. Biconditional statement
  12. Two (or Exactly two)
  13. Ray
  14. Sum; right; sum; legs (The Pythagorean Theorem: The sum of the squares of the legs of a right triangle is equal to the sum of the squares of the hypotenuse.)
  15. Theorem
  16. Polygon
  17. Consequent
  18. Compass
  19. Circle
  20. Euler diagram (or Venn diagram)
  21. Radius
  22. Proof (or Chain of reasoning/Deductive argument)
  23. 180°
  24. πr2\pi r^2
  25. Antecedent (or Hypothesis)

Question 26: Euler Diagrams

Since I cannot draw directly, here is how you should draw them:

a. If you walk tight ropes, then you are high strung.

  • Draw a large circle labeled “High Strung People” (the consequent).
  • Draw a smaller circle completely inside the large one labeled “Tight Rope Walkers” (the antecedent).
  • Label: The inner circle is a subset of the outer circle.

b. If a triangle has two equal sides, then it is an isosceles triangle.

  • Draw a large circle labeled “Isosceles Triangles.”
  • Draw a smaller circle inside it labeled “Triangles with two equal sides.”
  • (Note: In geometry, these are technically equivalent, so the circles might be equal size, but for conditional logic representation, showing the specific case inside the general category is standard).

Question 27: Egotist Definition

  • Conditional 1: If you are always me-deep in conversation, then you are an egotist.
  • Conditional 2: If you are an egotist, then you are always me-deep in conversation.
  • Relationship: They are converse of each other. (Together they form the biconditional definition).

Question 28: Direct Proof Fill-in

  • If there is a total eclipse of the sun, then the sky becomes dark.
  • If the sky becomes dark, then the crickets think it is night.
  • If the crickets think it is night, then one can count the chirps.
  • Therefore, if there is a total eclipse of the sun, then the temperature can be determined without a thermometer.

Question 29: Logical Conclusion

  • Therefore, Claudia is at least 18 years of age.

Question 30: Area of Shaded Region

  • Given:
    • Inside circle radius (rinr_{in}) = 2 cm.
    • Middle circle diameter (dmidd_{mid}) = 8 cm →\rightarrow radius (rmidr_{mid}) = 4 cm.
    • Outside circle radius (routr_{out}) = 5 cm.
  • Goal: Area of the “middle circle (shaded).”
    • Note: The phrasing “middle circle (shaded)” usually implies the annulus (ring) between the inside and middle circles, OR it implies the shaded region is the ring between the middle and outside. Let’s assume the standard “shaded region” in these concentric problems is the ring between the middle and outer, or the middle minus the inner.
    • Let’s calculate the area of the annulus between the middle circle and the inner circle (if “middle” is the shaded one):
      • Area = π(rmid)2−π(rin)2\pi (r_{mid})^2 - \pi (r_{in})^2
      • Area = π(42)−π(22)=16π−4π=12π\pi (4^2) - \pi (2^2) = 16\pi - 4\pi = 12\pi cm².
    • Alternative Interpretation: If the shaded region is between the middle and outside:
      • Area = π(52)−π(42)=25π−16π=9π\pi (5^2) - \pi (4^2) = 25\pi - 16\pi = 9\pi cm².
    • Most likely: Based on “diameter of the middle circle (shaded),” it likely means the ring between the middle and the outside is NOT shaded, but the middle ring itself is. However, looking at Q31, it asks for “shaded region” in a square/circle. For Q30, if the “middle circle” is the shaded part, it usually means the annulus between radius 2 and 4.
    • Answer: 12π12\pi cm² (Area of circle with r=4 minus area of circle with r=2).

Question 31: Square and Inscribed Circle

  • Given: Side of square (ss) = 3 cm. The circle is enclosed (inscribed).

a. Area of the square:

  • s2=32=𝟗 𝐜𝐦𝟐s^2 = 3^2 = \mathbf{9 \text{ cm}^2}

b. Radius of the circle:

  • For an inscribed circle, diameter = side of square.
  • d=3d = 3 cm →r=𝟏.𝟓 𝐜𝐦\rightarrow r = \mathbf{1.5 \text{ cm}} (or 32\frac{3}{2} cm)

c. Area of the enclosed circle:

  • πr2=π(1.5)2=𝟐.𝟐𝟓𝝅 𝐜𝐦𝟐\pi r^2 = \pi (1.5)^2 = \mathbf{2.25\pi \text{ cm}^2} (or 9π4\frac{9\pi}{4} cm²)

d. Area of the shaded region (Square - Circle):

  • 9−2.25π cm29 - 2.25\pi \text{ cm}^2

e. Circumference of the circle:

  • 2πr=2π(1.5)=𝟑𝝅 𝐜𝐦2\pi r = 2\pi(1.5) = \mathbf{3\pi \text{ cm}}

f. Perimeter of the square:

  • 4s=4(3)=𝟏𝟐 𝐜𝐦4s = 4(3) = \mathbf{12 \text{ cm}}

Question 32: Triangle Angle Calculation

  • Given: ∠CAB=22∘\angle CAB = 22^\circ, ∠ACB=61∘\angle ACB = 61^\circ. Find ∠ABC\angle ABC.
  • Sum of angles in a triangle = 180∘180^\circ.
  • ∠ABC=180∘−(22∘+61∘)\angle ABC = 180^\circ - (22^\circ + 61^\circ)
  • ∠ABC=180∘−83∘\angle ABC = 180^\circ - 83^\circ
  • Answer: 97∘97^\circ

Question 33: Right Triangle Side

  • Given: Legs are 7 and 25? Or is 25 the hypotenuse?
    • Looking at the layout: “5, 7 units, 25 units”. Wait, the text says “5 7 units 25 units”. Let’s look at the image context. Usually, the longest side is the hypotenuse.
    • If 25 is the hypotenuse and 7 is one leg:
      • a2+72=252a^2 + 7^2 = 25^2
      • a2+49=625a^2 + 49 = 625
      • a2=576a^2 = 576
      • a=576=24a = \sqrt{576} = 24
    • If 5 and 7 are legs? No, 25 is likely the hypotenuse given the numbers.
    • Answer: 24 units

Question 34: Theorem Name

  • Answer: Pythagorean Theorem

Question 35: Indirect Proof (Mrs. Pots)

a. The conditional statement she wanted to prove:

  • If a pan is stainless steel, then it will not rust. (Or: If the pans are stainless steel, they are not supposed to rust).

b. The assumption she made for an indirect proof:

  • She assumed the opposite: The pans ARE stainless steel. (She assumes the claim is true to find a contradiction).

c. The conclusion resulting from the assumption:

  • If they are stainless steel, then they will not rust.

d. The known fact contradictory to part c:

  • The pans are rusting. (This contradicts the conclusion that they won’t rust).

Question 36: Construction (Equilateral Triangle & Bisector)

  • Steps to describe/draw:
    1. Draw segment AB=9AB = 9 cm.
    2. Place compass at AA, open to 9 cm. Draw an arc above the line.
    3. Place compass at BB, open to 9 cm. Draw an arc intersecting the first arc.
    4. Label the intersection point CC.
    5. Draw segments ACAC and BCBC to form the equilateral triangle.
    6. To bisect ABAB: Place compass at AA, draw arcs above and below ABAB (radius > 4.5 cm).
    7. Place compass at BB, same radius, draw arcs intersecting the previous ones.
    8. Draw a line through the two intersection points. This line crosses ABAB at its midpoint, forming the perpendicular bisector.

Question 37: Construction (Obtuse Angle & Bisectors)

  • Steps to describe/draw:
    1. Draw an obtuse angle ∠FAT\angle FAT (greater than 90∘90^\circ).
    2. Bisect ∠FAT\angle FAT:
      • Place compass at AA, draw an arc crossing AFAF and ATAT.
      • From those intersection points, draw two arcs inside the angle that cross each other.
      • Draw ray ASAS from AA through the intersection. ASAS bisects ∠FAT\angle FAT.
    3. Bisect ∠SAT\angle SAT:
      • Place compass at AA, draw an arc crossing ASAS and ATAT.
      • From those new intersection points, draw two arcs inside ∠SAT\angle SAT that cross each other.
      • Draw a ray from AA through this new intersection. This is the bisector of ∠SAT\angle SAT.

You: ok we still need 38 and 39 incorrectly numbered 36)

Leo AI: Here are the answers for the final questions (numbered 38 and 36 in the original document):

Question 38: Simplify the Expressions

a. (4x+2y2)+(3y2−2x)(4x + 2y^2) + (3y^2 - 2x)

  • Combine like terms: 4x−2x+2y2+3y24x - 2x + 2y^2 + 3y^2
  • Answer: 2x+5y22x + 5y^2

b. (4y+2x2)−(3x2−2y)(4y + 2x^2) - (3x^2 - 2y)

  • Distribute the negative sign: 4y+2x2−3x2+2y4y + 2x^2 - 3x^2 + 2y
  • Combine like terms: 2x2−3x2+4y+2y2x^2 - 3x^2 + 4y + 2y
  • Answer: −x2+6y-x^2 + 6y (or 6y−x26y - x^2)

c. (7x2−3y)−(5y−2x2)(7x^2 - 3y) - (5y - 2x^2)

  • Distribute the negative sign: 7x2−3y−5y+2x27x^2 - 3y - 5y + 2x^2
  • Combine like terms: 7x2+2x2−3y−5y7x^2 + 2x^2 - 3y - 5y
  • Answer: 9x2−8y9x^2 - 8y

d. x2+3x2+x2+2xx^2 + 3x^2 + x^2 + 2x

  • Combine like terms: 1x2+3x2+1x2+2x1x^2 + 3x^2 + 1x^2 + 2x
  • Answer: 5x2+2x5x^2 + 2x

Question 36 (Second one, numbered 36 at the bottom): Solve the Equations

a. 3x−10=4(8−x)3x - 10 = 4(8 - x)

  • Distribute: 3x−10=32−4x3x - 10 = 32 - 4x
  • Add 4x4x to both sides: 7x−10=327x - 10 = 32
  • Add 10 to both sides: 7x=427x = 42
  • Divide by 7: x=6x = 6
  • Answer: x=6x = 6

b. 5(x−1)=2(x+3)−325(x - 1) = 2(x + 3) - 32

  • Distribute: 5x−5=2x+6−325x - 5 = 2x + 6 - 32
  • Simplify right side: 5x−5=2x−265x - 5 = 2x - 26
  • Subtract 2x2x from both sides: 3x−5=−263x - 5 = -26
  • Add 5 to both sides: 3x=−213x = -21
  • Divide by 3: x=−7x = -7
  • Answer: x=−7x = -7

c. 5x(x+2)−x(x+6)=2(x2−2)+2x(x+3)5x(x + 2) - x(x + 6) = 2(x^2 - 2) + 2x(x + 3)

  • Expand left side: (5x2+10x)−(x2+6x)=5x2+10x−x2−6x=4x2+4x(5x^2 + 10x) - (x^2 + 6x) = 5x^2 + 10x - x^2 - 6x = 4x^2 + 4x
  • Expand right side: (2x2−4)+(2x2+6x)=4x2+6x−4(2x^2 - 4) + (2x^2 + 6x) = 4x^2 + 6x - 4
  • Set equal: 4x2+4x=4x2+6x−44x^2 + 4x = 4x^2 + 6x - 4
  • Subtract 4x24x^2 from both sides: 4x=6x−44x = 6x - 4
  • Subtract 6x6x from both sides: −2x=−4-2x = -4
  • Divide by -2: x=2x = 2
  • Answer: x=2x = 2