Calculating Rectangle Dimensions

The video tutorial begins with an introduction and acknowledgment of subscribers. It presents a geometry problem involving a rectangle where the length is twice the width, and the area is 288 cm². The tutorial defines variables for length and width, then solves for the width using algebraic substitution and simplification. The width is found to be 12 cm. The video concludes by posing a new question about finding the perimeter and encourages viewers to engage with comments and subscriptions.

What is the main purpose of the video?

• To explain the properties of circles

• To discuss the history of mathematics

• To solve a math problem about a rectangle

• To teach about algebraic expressions

What is the relationship between the length and width of the rectangle?

• The length is three times the width

• The length is half the width

• The length is twice the width

• The length is equal to the width

What is the given area of the rectangle?

• 288 cm²

• 144 cm²

• 576 cm²

• 72 cm²

How is the area of a rectangle calculated?

• Length divided by width

• Length times width

• Length plus width

• Width minus length

What substitution is made to solve for the width?

• Length is replaced with twice the width

• Width is replaced with length

• Width is replaced with twice the length

• Length is replaced with width

What is the first step in solving for the width of the rectangle?

• Divide the area by 2

• Replace length with twice the width

• Add length and width

• Calculate the perimeter

What is the result of dividing the area by 2 in the equation?

• 72

• 144

• 576

• 288

What is the width of the rectangle after solving the equation?

• 10 cm

• 16 cm

• 12 cm

• 14 cm

What mathematical operation is used to find the width from the width squared?

• Addition

• Subtraction

• Square root

• Multiplication

What is the final answer for the width of the rectangle?

• 16 cm

• 14 cm

• 12 cm

• 10 cm