University:
High School
Course:
High School Mathematics
Academic year:

2022
Views:

443

Pages:

3
Author:

Amiah Sosa

Surface Areas & Volumes of Spheres Vocabulary Terms: Great circle: is the cross section of a sphere and a plane that contains the "center" Chord of a sphere: is a segment that starts and ends on the sphere Core Concepts: Surface Area of a Sphere: The surface area S of a sphere where r is the radius of the sphere. S = 4πr² memorize Volume of a Sphere: The volume V of a sphere where r is the radius of the sphere. V = (4/3)πr³ memorize notice the "3" Exercises 1-4: Find the surface area of the solid. TSA = 4πr² = 4π(5)² = 100π cm² TSA = 4πr² = 4π(9)² = 324π in² TSA = 2πr² = 2π(2)² = 8π ft² TSA = 2πr² = 2π(11.2)² = 250.88π m² Exercises 5-8: Find the volume of the sphere. V = (4/3)πr³ = (4/3)π(7.5)³ = 562.5π in³ V = (4/3)πr³ = (4/3)π(13)³ = 2929.33π cm³ V = (4/3)πr³ = (4/3)π(3.5)³ = 57.17π cm³ V = (4/3)πr³ = (4/3)π(10)³ = 166.67π in³ Exercises 9-11: Find the indicated measure. Problem 9: Find the diameter of a sphere with a surface area of 144π square centimeters. TSA = 4πr² 144π = 4πr² √36 = √r² 6 = r d = 2r d = 2(6) d = 12 Problem 10: Find the volume of a sphere with a surface area of 256π square inches. TSA = 4πr² 256π = 4πr² √64 = √r² 8 = r V = (4/3)πr³ V = (4/3)π(8)³ V = 682.67π in³ Problem 11: Find the volume of a sphere with a surface area of 400π square feet. TSA = 4πr² 400π = 4πr² √100 = √r² 10 = r V = (4/3)πr³ V = (4/3)π(10)³ V = 1333.33π ft³