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14 days ago

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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

?

2 answers:

AnnZ [3.9K]14 days ago

7 0

The area of the washer's top surface, given its outer radius of 10 mm and a central hole diameter of 14 mm, is 160.29 mm².

Further Explanation

Area

Area quantifies the extent of a two-dimensional surface.
Calculating the area depends on the particular shape involved.

For instance:

The area of a rectangle is found by multiplying length by width.
The area of a triangle is computed as one half the base multiplied by the height.
For a circle, area equals π times the radius squared, i.e., πr².
The area of a square is the side length squared, s².

In this problem, the given large washer has an outer radius of 10 mm and an inner radius of 7 mm.

The annulus's area, representing the washer's surface, is calculated by:

Area = πR² − πr²

= π(R² − r²)

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

Using π = 22/7, R = 10 mm, and r = 7 mm, we have:

Area = 22/7 × (10² − 7²)

= 22/7 × (100 − 49) = 22/7 × 51

= 160.29 mm²

Therefore, the washer's top surface area measures 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

babunello [3.6K]14 days ago

4 0

Answer:

The top surface area of the washer equals 160.14 square millimeters.

Step-by-step Explanation:

The washer's top surface forms an annulus, characterized by an outer radius of 10 mm and an inner radius of 7 mm (obtained since the hole's diameter is 14 mm and the radius is half the diameter).

Recall the formula for the area of an annulus:

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

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

Substituting the given values:

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

Thus, the calculation yields:

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

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