Question

A large washer has an outer radius of 10mm and a hole with a diameter of 14mm. What is the area of the top surface of the washer?

Answer 1

The area of the top surface of a washer with an outer radius of 10 mm and a hole with a diameter of 14 mm is 160.29 mm².  

Further Explanation  

Area  

• Area is a measure of how much space is occupied by a given shape.  
• Area of a substance is determined by the type of shape in question.

For example;  

• Area of a rectangle is given by; Length multiplied by width
• Area of a triangle = 1/2 x base x height  
• Area of a circle = πr² where r is the radius of a circle,  
• Area of a square = S², Where s is the side of the square, etc.  

In this question, we are given a large washer with an outer radius of 10 mm and inner radius of 7 mm.  

Area of annulus region will be given by;  

Area = πR²- πr²

          = π(R²-r²)

Where R is the outer radius while r is the inner radius.

Therefore; Taking π = 22/7, R = 10 mm and r = 7 mm

Area = 22/7( 10²-7²)

        = 22/7 (51)

        = 160.29 mm²

Therefore, the area of the top surface of the large washer is 160.29 mm².

Keywords: Area  

Learn more about;

• Perimeter: brainly.com/question/1322653
• Area : brainly.com/question/1322653

Level: Middle school  

Subject; Mathematics  

Topic: Area and Perimeter

Answer 2

Answer:

The area of the top surface of the washer is:  160.14 square mm.

Step-by-step explanation:

The top of the surface is in the shape of a annulus  with a outer radius of 10 mm and a inner radius of 7 mm ( since the diameter of the hole is: 14 mm and we know that the radius is half of the diameter)

Now, we know that the area of the annulus region is given by:

$Area=\pi (R^2-r^2)$

where R is the outer radius and r is the inner radius.

Here we have:

$R=10\ mm\\\\and\\\\r=7\ mm$

Hence, we have:

$Area\ of\ top\ surface=\pi (10^2-7^2)\\\\i.e.\\\\Area\ of\ top\ surface=\pi (100-49)\\\\i.e.\\\\Area\ of\ top\ surface=\pi\cdot 51\\\\i.e.\\\\Area\ of\ top\ surface=160.14\ mm^2$