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

2024
Views:

233

Pages:

6
Author:

Beckett Gray

Question 1 To calculate: The area of the shaded region containing one larger circle and three identical smaller circles. Answer Formula used: Area of a circle = 𝜋 𝑟 2 . Area of a circle=πr 2 . Larger circle radius: 𝑅 = 6 R=6. Smaller circle radius: 𝑟 = 2 r=2. Step 1: Calculate the area of the larger circle: 𝐴 large circle = 𝜋 𝑅 2 = 𝜋 ( 6 ) 2 = 36 𝜋 . A large circle =πR 2 =π(6) 2 =36π. Step 2: Calculate the area of one smaller circle: 𝐴 small circle = 𝜋 𝑟 2 = 𝜋 ( 2 ) 2 = 4 𝜋 . A small circle =πr 2 =π(2) 2 =4π. For three smaller circles: 𝐴 three small circles = 3 × 4 𝜋 = 12 𝜋 . A three small circles =3×4π=12π. Step 3: The shaded area is the area of the larger circle minus the area of the three smaller circles: 𝐴 shaded = 𝐴 large circle − 𝐴 three small circles . A shaded =A large circle −A three small circles . 𝐴 shaded = 36 𝜋 − 12 𝜋 = 24 𝜋 . A shaded =36π−12π=24π. Final Answer: 24 𝜋 square units 24πsquare units. Question 2 To draw two congruent circles with radii 6 6, each passing through the center of the other, and find the length of their common chord. Answer Step 1: Concept: The perpendicular from the center of a circle to the chord bisects the chord. Step 2: Triangle 𝐴 𝐷 𝐶 ADC: 𝐴 𝐶 AC is the half-length of the chord. 𝑂 𝐶 = 6 OC=6 (radius), 𝐶 𝐷 = 3 CD=3 (half of the perpendicular distance). Using the Pythagorean theorem: 𝑂 𝐶 2 = 𝐴 𝐶 2 + 𝐶 𝐷 2 . OC 2 =AC 2 +CD 2 . Substitute values: 6 2 = 𝐴 𝐶 2 + 3 2 . 6 2 =AC 2 +3 2 . 36 = 𝐴 𝐶 2 + 9. 36=AC 2 +9. 𝐴 𝐶 2 = 27. AC 2 =27. 𝐴 𝐶 = 27 = 3 3 . AC= 27 =3 3 . Since 𝐴 𝐶 AC is half of 𝐴 𝐵 AB (the full chord): 𝐴 𝐵 = 2 × 𝐴 𝐶 = 2 × 3 3 = 6 3 . AB=2×AC=2×3 3 =6 3 . Final Answer: 6 3 units 6 3 units. Question 3 To find: a) The length of the common external tangent. b) Whether the circles intersect or not. Given: Radii 10 10 and 5 5. Answer Step 1: From the diagram: 𝑂 𝑃 = 13 OP=13 (distance between centers of the circles), Radii 10 10 and 5 5. Using the formula for the length of the common external tangent: 𝐴 𝐵 = ( 𝑂 𝑃 ) 2 − ( 𝑅 1 − 𝑅 2 ) 2 . AB= (OP) 2 −(R 1 −R 2 ) 2 . Substitute values: 𝐴 𝐵 = 1 3 2 − ( 10 − 5 ) 2 . AB= 13 2 −(10−5) 2 . 𝐴 𝐵 = 169 − 25 . AB= 169−25 . 𝐴 𝐵 = 144 = 12. AB= 144 =12. Step 2: Intersection check: If 𝑂 𝑃 < 𝑅 1 + 𝑅 2 OP