University:
California State University, Northridge
Course:
MATH 262 | Introduction to Linear Algebra
Academic year:

2024
Views:

96

Pages:

8
Author:

Beckett Gray

Question 1 Coplanar circles that have the same center are called: a) Concentric circles b) Tangent circles c) Congruent circles d) Common circles Answer Step 1: Analyze the type of circle. Step 2: Coplanar means the circles lie on the same plane. Circles that share the same center are called concentric circles. Correct option: a) Concentric circles. Question 2 Find the area of the shaded region. Round to the nearest tenth. (You are looking at a rectangle with 6 circles taken out.) Answer 1 Step 1: Identify the dimensions. Rectangle dimensions: 12 × 8 12×8. Circles tightly fit such that the total horizontal diameter of 3 circles equals 12 12. 3 𝑑 = 12 → 𝑑 = 4 3d=12→d=4 (diameter). Radius 𝑟 = 𝑑 2 = 2 r= 2 d =2. Step 2: Calculate areas. Area of the rectangle: 𝐴 rectangle = 12 × 8 = 96 units 2 . A rectangle =12×8=96units 2 . Area of 6 circles: 𝐴 circle = 𝜋 𝑟 2 = 𝜋 ( 2 ) 2 = 3.14 × 4 = 12.56 units 2 . A circle =πr 2 =π(2) 2 =3.14×4=12.56units 2 . 𝐴 6 circles = 6 × 12.56 = 75.36 units 2 . A 6 circles =6×12.56=75.36units 2 . Shaded region: Subtract circle areas from rectangle area. 𝐴 shaded = 𝐴 rectangle − 𝐴 6 circles = 96 − 75.36 = 20.64 units 2 . A shaded =A rectangle −A 6 circles =96−75.36=20.64units 2 . Final Answer: Rounded to the nearest tenth: 20.6 units². Question 3 Explanation for chord angle theorem: Given two intersecting chords inside a circle, the angle formed is equal to half the sum of the intercepted arcs. Step 1: Theorem used: Angle = Sum of intercepted arcs 2 . Angle= 2 Sum of intercepted arcs . Step 2: Given arcs 4 0 ∘ 40 ∘ and 3 0 ∘ 30 ∘ : Angle = 4 0 ∘ + 3 0 ∘ 2 = 7 0 ∘ 2 = 3 5 ∘ . Angle= 2 40 ∘ +30 ∘ = 2 70 ∘ =35 ∘ . Final Answer: The value of ∠ 1 ∠1 is 3 5 ∘ 35 ∘ . Question 4 Find the area and perimeter of a sector. Solution Given: Radius 𝑟 = 655 m r=655m. Chord 𝐴 𝐵 = 1283.4 m AB=1283.4m. Step 1: Use the cosine rule to find the central angle 𝜃 θ: cos ⁡ ( 𝜃 ) = 𝐴 𝑂 2 + 𝐵 𝑂 2 − 𝐴 𝐵 2 2 × 𝐴 𝑂 × 𝐵 𝑂 . cos(θ)= 2×AO×BO AO 2 +BO 2 −AB 2 . Substitute the values: cos ⁡ ( 𝜃 ) = 65 5 2 + 65 5 2 − 1283. 4 2 2 × 655 × 655 . cos(θ)= 2×655×655 655 2 +655 2 −1283.4 2 . Solving gives 𝜃 = 103.1 3 ∘ θ=103.13 ∘ . Step 2: Calculate the area of the sector: Area of sector = 𝜃 36 0 ∘ × 𝜋 𝑟 2 . Area of sector= 360 ∘ θ ×πr 2 . Substitute: Area = 103.13 360 × 3.14 × ( 655 ) 2 = 760 , 122.50 m 2 . Area= 360 103.13 ×3.14×(655) 2 =760,122.50m 2 . Step 3: Find the perimeter of the sector: Perimeter = 𝐴 𝑂 + 𝐵 𝑂 + Arc length ( 𝐴 𝐵 ) . Perimeter=AO+BO+Arc length(AB). Arc length: Arc length = 𝜃 36 0 ∘ × 2 𝜋 𝑟 . Arc length= 360 ∘ θ ×2πr. Substitute: Arc length = 103.13 360 × 2 × 3.14 × 655 = 2320.98 m . Arc length= 360 103.13 ×2×3.14×655=2320.98m. Total perimeter: Perimeter = 655 + 655 + 2320.98 = 3630.98 m . Perimeter=655+655+2320.98=3630.98m. Final Answer: Area: 760 , 122.50 m 2 760,122.50m 2 . Perimeter: 3630.98 m 3630.98m. Question 5 Solve 𝐶 = 2 𝜋 𝑟 C=2πr for 𝑟 r, and find 𝑟 r if 𝐶 = 15 C=15 inches. Solution (a): Solve 𝐶 = 2 𝜋 𝑟 C=2πr: 𝑟 = 𝐶 2 𝜋 . r= 2π C . Solution (b): Given 𝐶 = 15 C=15: 𝑟 = 15 2 𝜋 = 7.5 𝜋 . r= 2π 15 = π 7.5 . Final Answer: 𝑟 = 7.5 𝜋 r= π 7.5 .